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Cersei Lannister es un personaje ficticio de la saga Canción de hielo y fuego de George R. R. Martin y su adaptación televisiva Game of Thrones, donde es interpretada por la actriz bermuda-inglesa Lena Headey. Presentada en 1996 en Juego de tronos, Cersei es miembro de la Casa Lannister, una de las familias más ricas y poderosas del continente de Westeros. Posteriormente aparece en Choque de reyes (1998) y Tormenta de espadas (2000), y se convierte en un personaje de punto de vista prominente a partir de Festín de Cuervos (2005). El personaje aparecerá en el próximo volumen Vientos de invierno.

Cersei es Reina de los Siete Reinos de Westeros por matrimonio con el Rey Robert Baratheon. Su padre, Tywin, arregló el matrimonio después de que fracasara su intento de desposarla con el príncipe Rhaegar Targaryen, a quien idolatraba de niña. Robert tomó el trono con la ayuda de los Lannisters cuando terminó con la dinastía Targaryen. Cersei ha estado involucrada en un romance incestuoso con su hermano gemelo, Jaime, desde la infancia. Los tres hijos de Cersei son de Jaime, lo cual es desconocido para Robert, quien abusa de ella durante su matrimonio. La supuesta ilegitimidad de sus hijos provoca una lucha de poder a raíz de la muerte del rey conocida como la Guerra de los Cinco Reyes. Los principales atributos del personaje de Cersei son su ansia de poder, su punto de vista transgresor, sus intrigas y su amor por sus hijos, a quienes busca proteger.

Cersei es considerado uno de los personajes más complejos de la historia. Headey recibió elogios generalizados de la crítica por su interpretación del personaje en la serie de HBO, Game of Thrones. Fue nominada a cinco premios Emmy a la Mejor Actriz de Reparto en una Serie Dramática y un Globo de Oro a la Mejor Actriz de Reparto por su actuación. Headey y el resto del elenco fueron nominados a siete premios del Sindicato de Actores de la Pantalla por Mejor reparto de televisión en Serie Dramática. En la temporada 5 de la serie, una versión joven del personaje es interpretada por Nell Williams en un flashback.

Concepción y diseño
Cersei es la única hija y la mayor de Tywin Lannister y su esposa, Joanna; su hermano gemelo, Jaime, nació poco después de Cersei. Cersei y Jaime se parecían tanto cuando eran niños que Cersei ocasionalmente vestía la ropa de Jaime y fue confundida con él.

Ya desde su juventud, Cersei admiraba a su padre y sus dotes de mando, deseando probar que era tan buen gobernante como su progenitor, pero cuando consigue el mando no es capaz de manejarlo, ya que no posee la dedicación y prudencia suficientes para la política y evita enfrentarse a situaciones delicadas, prefiriendo rodearse de leales y aduladores. Tampoco es capaz de tolerar las opiniones disidentes y confunde el desacuerdo con rebeldía.

Los gemelos experimentaron sexualmente a una edad temprana, pero fueron descubiertos por un sirviente, quien informó a su madre. Joanna encargó a un guardia que mantuviera a los gemelos separados y mantuvo el asunto en secreto para su padre, amenazando a los gemelos con que le informaría si alguna vez lo repetían.

Poco después, Joanna murió al dar a luz al hermano enano menor de los gemelos, Tyrion. Cersei culpó a Tyrion por la muerte de Joanna y comenzó a abusar de él cuando era un bebé. Martin dijo en Rolling Stone:

Desde que ella era muy joven, Tywin esperaba que su hija se casara con Rhaegar Targaryen, y rechazó la propuesta de la princesa de Dorne de desposar a Oberyn Martell y Elia Martell con Cersei y Jaime, respectivamente. Cersei se enamoró de Rhaegar; sentimientos que ocultó a su hermano como su padre le aconsejó.

A la edad de 11 años, Jaime fue enviado para servir como escudero de Lord Sumner Crakehall. Un año después, Aerys Targaryen rechazó la propuesta de que Cersei se casara con su hijo, humillando a Tywin. Tywin llevó a Cersei a Desembarco del Rey y, en los años siguientes, rechazó todas las ofertas de matrimonio para ella. En 281 d.C., cuando Cersei y Jaime tenían 15 años, Jaime, recién nombrado caballero en el campo de batalla y de camino a Roca Casterly, llegó a Desembarco del Rey para visitar a su hermana, de quien había estado separado durante cuatro años. Ella le informó a Jaime que su padre planeaba casarlo con Lysa Tully y lo persuadió de unirse a la Guardia Real para estar cerca de ella, después de que los gemelos se hubieran acostado juntos por primera vez. Así comenzó una aventura que continuó durante todo el matrimonio de Cersei con Robert Baratheon. Si bien inicialmente estaba feliz de casarse con Robert, él no le devolvió el afecto de Cersei y la engañó repetidamente. Luego reanudó su relación incestuosa con Jaime, lo que resultó en los nacimientos de Joffrey, Myrcella y Tommen.

Posteriormente a su matrimonio, la representada de Cersei como una mujer de carácter narcisista y ambicioso que no se detiene ante nada para cumplir sus objetivos se fortaleció. Ella a admite no amar a ninguna otra persona que no sea ella o sus hijos, pues cree que el amor hace a las personas débiles. Desprecia la sociedad machista que es Poniente pues no le permite alcanzar el poder que por derecho merece, pero también se sirve de sus armas como mujer para manipular y conducir las cosas a su favor. Desprecia al género femenino, al creer a las mujeres como frívolas y débiles pero considerándose ella misma una excepción.

La actriz Lena Headey interpreta al personaje de Cersei en la adaptación televisiva Game of Thrones. Describe así a su personaje: «Está políticamente muy mentalizada, es paranoica y trabaja duramente para conseguir sus objetivos. Ansía el poder, no importa el precio que sea ni a quien afecte. Ama a su hermano, pero siempre y cuando le proporcione apoyo y cariño... yo no llamaría amor a eso».

Como mucho dentro de la narrativa e historia ficticia fantástica, el personaje de Cersei cuenta con paralelismo históricos que datan del medioevo europeo y del renacimiento. La principal comparación histórica y más evidente es hacia la hija natural del Papa Alejandro VI, Lucrezia Borgia, doncella que al igual que Cersei Lannister, se le atribuye una relación íntima y política con su familia inmediata, en especial con su hermano Cezare Borgia.

Historia

Primeros años

Un hecho trascendental se produjo durante su niñez. Cersei acudió en compañía de dos amigas a visitar a una maegi (bruja) que se afirmaba que tenía la capacidad de ver el futuro. Una de ellas huyó antes de llegar, pero Cersei y su otra amiga, Melara Hetherspoon, entraron dentro del tenderete de la bruja. Esta le vaticinó a Cersei su compromiso frustrado con el príncipe Rhaegar Targaryen, que sería reina en un futuro y que tendría tres hijos (aunque no con el propio rey), pero también le dijo que moriría estrangulada por su valonqar (hermano menor). La última en preguntar fue la propia Melara, la amiga de Cersei, que le preguntó si ella se casaría algún día con Jaime, pero la bruja afirmó que no y que ella moriría esa misma noche. Cersei mató a la niña para evitar que hablara de las profecías y por tener ideas por encima de su estatus social. Cersei pasa toda su vida adulta intentando subvertir la profecía y desarrolla una fuerte paranoia, que finalmente hace que la profecía se haga realidad.

La Mano del Rey de Robert, Jon Arryn, descubrió la bastardía de los hijos de Robert y Cersei, pero cayó enfermo antes de poder decírselo al rey. Cersei le ordenó al Gran Maestre Pycelle que dejara morir a Lord Arryn para evitar que divulgara el secreto. Después intentó que Robert nombrara a Jaime como su Mano del Rey, pero él se negó.

Juego de tronos
Al comienzo de la saga, Cersei acompaña al rey Robert rumbo a Invernalia tras la muerte de Jon Arryn, la anterior Mano del Rey. Cersei y Jaime son descubiertos teniendo relaciones sexuales por Bran Stark, el hijo de Lord Eddard Stark, lo que causa que Jaime lo empuje de la torre donde estaban. De regreso a Desembarco del Rey, el príncipe Joffrey tiene un altercado con Arya Stark en el cual termina herido, eso hace que una rencorosa Cersei ordene la muerte de la loba huargo de Sansa Stark, la nueva prometida de Joffrey.

En la capital, cuando su hermano Tyrion es arrestado y Jaime huye tras atacar a Eddard Stark, Robert golpea a Cersei cuando esta le ordena arrestar a Ned y le insulta por su falta de decisión. Ned Stark había estado investigando sobre la muerte de Jon Arryn y termina descubriendo la ilegitimidad de los hijos de Robert y Cersei. Ned confronta a Cersei y le sugiere huir de la ciudad para escapar de la ira de Robert. Cersei respondió arreglando la muerte de Robert y aupando a Joffrey al trono. Se descubre que Cersei ha sobornado a Petyr Baelish y a Janos Slynt, Comandante de los Capas Doradas, los cuales traicionan y arrestan a Ned Stark. Pese a los consejos de su madre, Joffrey decide ejecutar absurdamente a Lord Stark, desencadenando la Guerra de los Cinco Reyes.

Choque de reyes
Cersei gobierna como reina regente en Desembarco del Rey, pero pronto se da cuenta de que su hijo es incontrolable y su desastroso gobierno está sumiendo en el caos a los Siete Reinos. Para resolver la situación, Tywin es nombrado Mano del Rey y Jaime como Lord Comandante de la Guardia Real, Tywin envía entonces a su hijo Tyrion a actuar como Mano del Rey en funciones y meter en vereda a Cersei y al rey-niño. Se producirá entonces entre Cersei y Tyrion una auténtica lucha de poder. En ese tiempo, Cersei comienza a mantener relaciones íntimas con su primo Lancel y con los hermanos Kettleblack, los cuales utiliza de guardaespaldas aunque en realidad estén a sueldo de Lord Petyr Baelish.

Tyrion, para ganarse el favor de la Casa Martell, pacta el matrimonio entre la princesa Myrcella con Trystane Martell, lo que causa la furia de Cersei. La situación entre Tyrion y Cersei se vuelve límite cuando Tyrion pone bajo su custodia al príncipe Tommen, y en represalia, Cersei apresa a Alayaya, una prostituta que creía que era la amante de su hermano, y que utiliza como rehén para asegurarse que su hijo vuelve sano y salvo.

Cuando comienza la Batalla del Aguasnegras, Cersei se refugia en la Fortaleza Roja junto a las demás damas de la corte. Cuando oye que los hombres de Stannis Baratheon están a punto de cruzar las puertas de la capital, ordena que Joffrey sea traído a la Fortaleza Roja, eso hace que los defensores comiencen a desertar a centenares al ver al rey huyendo. La batalla resulta exitosa para el Trono de Hierro cuando Tywin Lannister y sus aliados Tyrell derrotan a Stannis. Como resultado de esta alianza, Joffrey es prometido con Margaery Tyrell.

Tormenta de espadas
Ahora que Tywin comienza a ejercer de Mano del Rey, la capacidad de decisión de Cersei es nula y comienza a preparar la boda entre Joffrey y Margaery Tyrell. Su padre también tiene intención de que Cersei vuelva a casarse pese a la negativa de esta a hacerlo. Tywin sugiere que Cersei case con Oberyn Martell o con un Greyjoy, pero finalmente decide que un matrimonio con Willas Tyrell sería lo mejor pues afianzaría la alianza entre las dos casas. La boda no se llevaría a cabo por la negativa de Lord Mace Tyrell.

Finalmente se produce la boda, pero Joffrey es envenenado durante el banquete nupcial y muere en brazos de Cersei, acusando esta a Tyrion de su muerte y ordenando su arresto. Cersei organiza una farsa de juicio con falsos testigos y acusaciones mientras Tywin, Oberyn Martell y Mace Tyrell actúan como jueces. Viendo que no tiene ninguna posibilidad de salir absuelto, Tyrion escoge un juicio por combate. Ser Gregor Clegane, campeón de Cersei, vence al campeón de Tyrion, el mismo príncipe Oberyn, y Tyrion es declarado culpable, logrando escapar gracias a la ayuda de Jaime. En su huida de Desembarco del Rey, Tyrion asesina a Tywin.

Festín de cuervos
Con la muerte de Tywin, Cersei pasa a ser Señora de Roca Casterly y cabeza de la Casa Lannister (debido a que Jaime es miembro de la Guardia Real y Tyrion es un proscrito) y por fin consigue el poder absoluto que había ansiado desde joven. Cersei modifica el Consejo Privado llenándolo de leales a su causa y después ofrece el puesto de Mano del Rey tanto a Jaime como a su tío Kevan, pero ambos lo rechazan. Cersei nombra entonces a Harys Swyft, a quien considera alguien manipulable, además de un rehén para mantener la buena voluntad de su tío Kevan. Cersei cae en una paranoia total, creyendo que tanto Tyrion como los Tyrell conspiran para hacerle daño a ella o a sus hijos y arrebatarle el poder, incluso ofrece el título de lord a cualquiera que le lleve la cabeza de su hermano, lo que causa una gran matanza de enanos en Desembarco del Rey.

Entre las medidas de Cersei están la reinstauración de la Fe Militante, el antiguo brazo armado de la Fe de los Siete y que fue disuelta por Maegor el Cruel, como manera de solventar las deudas de la Corona con la Fe. Cersei también cancela el pago de la deuda al Banco de Hierro de Braavos lo que hace que este cancele sus préstamos, sumiendo en el caos económico a los Siete Reinos. Con el dinero no pagado, Cersei ordena la construcción de una nueva Armada Real, otorgando el título de Gran Almirante a Aurane Mares, un joven de dudosa lealtad pero que guardaba gran parecido con su amado, el príncipe Rhaegar Targaryen. Mientras todo esto sucede, se descubre que Cersei vive atormentada por una predicción que una maegi le hizo cuando era niña.

Cersei destituye a Harys Swyft como Mano del Rey y nombra a Orton Merryweather, esposo de la que se convierte en su mejor amiga, Taena de Myr. Temerosa del Septón Supremo, Cersei convence a uno de los hermanos Kettleblack de que lo asesine y después que acuda al Muro para eliminar al Lord Comandante Jon Nieve, a cambio le ofrece un título de lord y su cama.

Celosa de la influencia de los Tyrell, Cersei trama toda una conspiración contra Margaery (la reina que ella cree que la destronará), primero tratando de que uno de los Kettleblack la seduzca para así acusarla de adulterio, pero este es arrestado, y bajo tortura, confiesa con quien realmente se ha acostado es con la reina regente, además de eso confiesa que asesinó al anterior Septón Supremo bajo órdenes suyas. Cersei es arrestada por orden del nuevo Gorrión Supremo bajo los cargos de asesinato, adulterio e incesto. Desde prisión, Cersei envía una petición de ayuda a Jaime, que se halla asediando Aguasdulces, pero él ignora su llamada de auxilio.

Danza de dragones
Cersei permanece como prisionera de la Fe de los Siete y decide confesarle al Gorrión Supremo que mantuvo relaciones sexuales con los hermanos Kettleblack y con su primo Lancel, pero no que ordenara el asesinato del rey Robert, ni el del Septón Supremo ni que se acostara con su hermano Jaime, cargos que le valdrían la ejecución. La Fe ordena que Cersei se someta a un paseo de penitencia desde el Gran Septo de Baelor hasta la Fortaleza Roja. Cersei es desnudada y todo el pelo de su cuerpo rapado. Cersei camina soportando los insultos y las miradas lascivas del populacho de Desembarco del Rey y rompe a llorar justo antes de llegar a la Fortaleza Roja.

Cersei cena con su tío Kevan y este nota que su actitud ha cambiado, notándola más sumisa. Ella pide que Taena Merryweather sea llevada a Desembarco del Rey y Kevan le dice que pronto tendrá un juicio por combate, y que Ser Robert Strong, nuevo miembro de la Guardia Real, será su campeón.

Adaptación televisiva

La actriz Lena Headey interpreta a Cersei desde la primera temporada hasta la última temporada. Por otro lado, la actriz Nell Williams la interpretó en un flashback que mostraba un acontecimiento ocurrido en su niñez.

En octubre de 2014, Headey y varios otros miembros clave del reparto, todos contratados por seis temporadas de la serie, renegociaron sus acuerdos para incluir una potencial séptima temporada y aumentos salariales para las temporadas 5, 6 y 7. The Hollywood Reporter calificó los aumentos como "enormes", y señaló que el trato convertiría a los artistas "entre los actores mejor pagados de la televisión por cable". Deadline Hollywood puso el número para la temporada 5 en "cerca de $300,000 por episodio" para cada actor, y The Hollywood Reporter escribió en junio de 2016 que los artistas recibirían "más de $500,000 por episodio" por la temporada 7 y el temporada potencial 8. En 2017, Headey se convirtió en uno de los actores mejor pagados de la televisión y ganará £2 millones por episodio para el programa.

Historia
Gran parte de la historia de Cersei de la primera y segunda temporada de la serie de televisión no ha cambiado con respecto a los libros. Sin embargo, un cambio notable en el programa es que Joffrey autoriza la purga de los bastardos de Robert, en lugar de Cersei.

Temporada 1
Cersei se entera de que su esposo, Robert, corre el peligro de descubrir que los niños que él ve como sus herederos al trono no son suyos. Robert llega a su fin como resultado de un ataque de jabalí en un viaje de caza, antes de que Ned Stark le cuente la verdad sobre sus hijos. Cersei trabaja rápidamente para instalar a su hijo mayor, Joffrey, en el trono, con ella como su principal asesora política y reina regente. Joffrey, sin embargo, muestra rápidamente signos de independencia.

Temporada 2
Tywin decide que Cersei no ejerce suficiente control sobre su hijo y envía a Tyrion como asesor político adicional. Cersei y Tyrion no se llevan bien y constantemente tratan de socavar la autoridad del otro sobre la corona. A partir del final de la temporada 2, Tyrion ha acumulado más influencia sobre el Trono de Hierro, ha enviado a la única hija de Cersei, contra la voluntad de Cersei, de casarse con el Príncipe de Dorne, y está obligando a su primo Lancel, con quien Cersei está teniendo una aventura, para informar sobre ella. La llegada de Tywin con el ejército de Tyrell les quita a ambos hermanos su influencia en el último momento posible.

Temporada 3

Cersei llega a temer que Margaery Tyrell intente usurparla como reina y, sin éxito, intenta alejar a Joffrey de ella. Cuando Tywin se entera del plan de los Tyrell para casar a Sansa Stark con el hermano de Margaery, Loras, interviene haciendo que Tyrion se case con Sansa. Cersei está encantada, pero su alegría se convierte en ira cuando Tywin también le ordena que se case con Loras. Cuando Jaime regresa a Desembarco del Rey, inmediatamente busca a Cersei, quien se sorprende al descubrir que su mano ha sido cortada.

Temporada 4
Cersei hace arrestar a Tyrion después de que Joffrey sea fatalmente envenenado. De luto y vengativa, se da a entender que, culpable o inocente, quiere a Tyrion muerto, mientras que Jaime se niega a creer que Tyrion sea capaz de un regicidio familiar. De hecho, en el juicio de Tyrion, es evidente que Cersei ha manipulado los procedimientos para que los testigos den testimonios incompletos o totalmente falsos para implicar más a Tyrion y Sansa en el asesinato. Cuando Tyrion exige un juicio por combate, Cersei elige a Ser Gregor "La Montaña" Clegane como campeón real, para disminuir aún más las posibilidades de Tyrion, y soborna a Bronn a través del compromiso con una mujer noble, para asegurar su veto como campeón de Tyrion. El deseo de Cersei se hace realidad cuando el campeón de Tyrion, Oberyn Martell, es asesinado por Clegane. Ella todavía se niega a casarse con Loras, amenazando incluso con revelar la verdad de su relación incestuosa con Jaime y sus tres hijos rubios.

Temporada 5
Cersei recibe una amenaza de Dorne, donde Myrcella está comprometida con Trystane Martell, y envía a Jaime a Dorne para recuperarla. Tommen está casado con Margaery, quien intenta manipular a Tommen para que envíe a Cersei a Roca Casterly. Temerosa de perder su poder, Cersei se encuentra con el líder religioso, el Gorrión Supremo, lo nombra Septón Supremo y le da autoridad para restablecer el Militante de la Fe. Para desestabilizar a la Casa Tyrell, Cersei hace arreglos para que el Gorrión Supremo arreste a Loras por su homosexualidad, y que Margaery también sea arrestada por perjurio después de mentir para defender a Loras. Sin embargo, el plan de Cersei fracasa cuando Lancel, ahora miembro de Militante de la fe, confiesa su romance con Cersei, su papel en el asesinato de Robert, y acusa a Cersei de incesto, lo que lleva al Gorrión Supremo a arrestarla. A Cersei finalmente se le permite regresar a la Fortaleza Roja después de confesar haber cometido adulterio con Lancel, pero le cortan el cabello y se ve obligada a caminar desnuda por las calles de Desembarco del Rey como expiación. Cersei regresa, angustiada por el abuso que le han dirigido los habitantes de Desembarco del Rey, pero es consolada por la exitosa reanimación de Ser Gregor Clegane por parte de Qyburn.

Temporada 6

Cersei permanece en la Fortaleza Roja, confinada en sus habitaciones y despojada de su autoridad. Aunque acepta la disculpa de Tommen por su falta de acción en su arresto y camino de expiación, otros miembros de la corte la miran con desprecio. Cersei se entera de Tommen, que el Gorrión Supremo planea que Margaery haga una caminata de expiación antes del juicio de Cersei y conspira con el Pequeño Consejo para que el ejército de Tyrell marche en el Sept de Baelor para asegurar la liberación de Margaery y Loras. Su plan fracasa cuando se revela que Margaery se ha unido a la Fe de los Siete y ha convencido a Tommen de fusionar la Fe y la Corona. A pesar de haber sido despojada de sus aliados, Cersei opta por permanecer en Desembarco del Rey, habiendo nombrado a Ser Gregor como su campeón en su prueba por combate. Sin embargo, Tommen anuncia que el juicio por combate ha sido abolido y reemplazado por el juicio por los Siete. Cersei hace los arreglos para que un alijo de incendios forestales sea detonado debajo del Sept durante el juicio de Loras, que mata al Gorrión Supremo, Margaery, Loras y el resto del Pequeño Consejo. Como resultado de las acciones de Cersei, Tommen se suicida. En el vacío de poder que siguió a la muerte de Tommen, Cersei reclama el Trono de Hierro como la primera reina reinante de los Siete Reinos.

Temporada 7
A pesar de ser nominalmente gobernante de Westeros, el control de Cersei en el trono es tenue, con las Tierras de los ríos en agitación después de la masacre de Arya Stark en la Casa Frey, el Valle y la declaración del Norte del hijo bastardo de Ned Stark, Jon Snow, y Tyrion, Asha Greyjoy, Dorne y The Reach apoyan a Daenerys Targaryen, cuya flota se ha apoderado de Rocadragón. Cersei permanece fría y distante después del suicidio de Tommen, viéndolo como una traición a su familia. Sin hijos a los que proteger, la malicia y la venganza de Cersei han crecido, y busca implacablemente expandir su poder y destruir a sus enemigos. Cersei se acerca al Rey de las Islas del Hierro, Euron Greyjoy, para formar una alianza, pero rechaza la propuesta de matrimonio de Euron porque lo considera indigno de confianza. Más tarde, Euron embosca a la flota de Daenerys y captura a Yara Greyjoy, Ellaria y Tyene Arena, y las presenta como un "regalo" a Cersei. Ella acepta casarse con Euron después de que se gane la guerra; sin embargo, ella continúa su romance con Jaime, ya no intenta ocultar su relación a sus sirvientes. Cersei encarcela a Ellaria y Tyene en las mazmorras, mata a Tyene con el mismo veneno que se utilizó para matar a Myrcella y mantiene a Ellaria con vida para ver morir a su hija.

Tycho Nestoris del Banco de Hierro llega a Desembarco del Rey para solicitar el pago de su deuda. Cersei solicita quince días para cumplir. Luego hace que los ejércitos de Lannister y Tarly ataquen Highgarden, derrotando a los Tyrell. Jaime sin desear matar brutalmente a Olenna Tyrell tras su derrota, le da veneno; esta se regodea con Jaime de que ella fue la responsable del envenenamiento de Joffrey. La riqueza de los Tyrell se transporta de forma segura a Desembarco del Rey, pero la comida que los Lannister roban del Dominio se destruye cuando Daenerys ataca la caravana con su horda y dragón Dothraki, Drogon. Jaime apenas sobrevive a la batalla y anima a Cersei a considerar un alto el fuego.

Tyrion se infiltra en la capital para encontrarse con Jaime, y Jaime le dice a Cersei que Daenerys y Jon quieren reunirse con Cersei para negociar un armisticio mientras la humanidad lucha contra la amenaza de los Caminantes Blancos. Cersei es escéptica pero está de acuerdo y le dice a Jaime que está embarazada. Daenerys y Jon se encuentran con Cersei en Dragonpit en las afueras de Desembarco del Rey y le presentan un espectro capturado de los Caminantes Blancos como evidencia de su amenaza. Aunque aterrorizada, Cersei inicialmente se niega a aceptar un armisticio, ya que Jon se niega a permanecer neutral en la guerra entre los Lannister y los Targaryen. Tyrion habla con Cersei en privado y aparentemente la convence de aliarse con los monarcas rivales. Sin embargo, más tarde le revela a Jaime que tiene la intención de que sus enemigos se aniquilen entre sí y luego ataquen al vencedor, y ha enviado a Euron a Essos para transportar la Compañía Dorada a Westeros. Disgustado por el egoísmo y la miopía de Cersei, Jaime finalmente abandona su lado y va hacia el norte por su cuenta, dejando a Cersei sola.

Temporada 8

A Cersei se le informa que los Caminantes Blancos han traspasado el Muro, para su deleite. Euron Greyjoy regresa a Desembarco del Rey con la Compañía Dorada, y Cersei finalmente tiene relaciones sexuales con él. Cersei le pide a Qyburn que contrate a Bronn para matar a Jaime y Tyrion, usando la ballesta con la que Tyrion había matado a Tywin.

Cersei permite que los civiles busquen refugio en la Fortaleza Roja, aparentemente para protegerlos de una invasión de Desembarco del Rey, pero en realidad planea usarlos como escudos humanos para disuadir un ataque. La Flota de Hierro mata a uno de los dragones de Daenerys, Rhaegal, y se lleva cautiva a su asesora Missandei. Cuando Euron regresa a Desembarco del Rey, Cersei revela que está embarazada y miente que es de él. Daenerys, Tyrion y sus otros asesores viajan a Desembarco del Rey exigiendo la rendición de Cersei, pero Cersei no se inmuta y hace que la Montaña decapite a Missandei frente a Daenerys. Posteriormente, Daenerys lanza un ataque contra la Flota de Hierro, la Fortaleza Roja y las defensas anti-dragón que Cersei ha instalado en los muros de Desembarco del Rey, pero luego continúa quemando el resto de la ciudad. Cersei intenta refugiarse y es encontrada por Jaime, quien se había infiltrado en la Fortaleza Roja por orden de Tyrion de evacuarla. Sin embargo, su paso está bloqueado por escombros. La Fortaleza Roja comienza a colapsar, y Jaime consuela a Cersei en sus momentos finales mientras son aplastados por los escombros que caen. Más tarde, Tyrion encuentra sus cuerpos, cuyo dolor posterior lo lleva a renunciar a su lealtad a Daenerys y persuadir a Jon Snow de que la mate.

Recepción y premios
Cersei ha sido elogiada como uno de los personajes más complejos y polifacéticos en cualquiera de las versiones de la historia. Rolling Stone la clasificó en el puesto número 5 en una lista de los "40 personajes principales de Game of Thrones", describiendo a Cersei como una mujer y matriarca que "lucha por un lugar en la mesa en un mundo hecho para hombres" y haría cualquier cosa por proteger a su descendencia. Como jugadora del juego, Cersei "preferiría morir antes que darte el control". En la lista de Rolling Stone de los "30 mejores villanos de Game of Thrones", Cersei ocupó el primer lugar. Descrita como "el ser humano más peligroso de Westeros" y "uno de los personajes más complejos y fascinantes de la televisión", Cersei es también una "figura extrañamente comprensiva, deformada por haber sido tratada como una yegua de cría cara por hombres poderosos toda su vida y genuina en sus afectos", mientras que su amarga experiencia la convierte en su peor enemiga y en "la villana de su propia historia".

Lena Headey recibió muchos elogios de la crítica por su interpretación de Cersei. James Hibberd de Entertainment Weekly declaró que desde la primera temporada, Headey "entregó perfectamente la mezcla tóxica única de orgullo, venganza, impaciencia, astucia y rencor de Cersei" y Cersei "nunca ha sido nada menos que un personaje completamente realizado". [3] Jeremy Egner del New York Times comentó que Headey "siempre ha sido una de las intérpretes más fuertes del programa". En The New Yorker, Clive James escribió que ella es una "hermosa expresión de terror arbitrario, que combina una gracia bien formada con maldad en la medida justa para asustar a un hombre hasta la muerte mientras lo deja indefenso con el deseo", elogiando la profundidad a la que Heady "transmite la radiante malevolencia de Cersei" a la mente del espectador "que vuelve a despertar una perturbación formativa". Andrew Anthony en The Guardian opinó que la serie era "más a menudo en su mejor momento cuando Headey estaba en la pantalla e irradiaba frío cálculo e intriga despiadada". Describió a Cersei de Headey como "un estudio de la tiranía como un arte seductor. Había algo magnéticamente humano en su inhumanidad. Te atrajeron sus encantos tortuosos incluso cuando retrocediste ante sus actos brutales". George R.R. Martin, autor de las novelas, elogió a Headey en el papel, y señaló que Cersei es "un personaje con muchos colores" y la gama de Headey es Martin consideró a Headey como una "actriz muy comprensiva ... una actriz diferente no podría venderlo", y calificó su interpretación de Cersei como "maravillosa".

Headey recibió un premio EWwy a la Mejor Actriz de Reparto en un Drama en 2012, un Premio Portal a la Mejor Actriz en 2012, y el Premio Women's Image a la Actriz de Serie Dramática en 2014. Recibió cinco nominaciones para el premio Primetime Emmy a la mejor actriz de reparto en una serie dramática en 2014, 2015, 2016, 2018 y 2019. También fue nominada para el Globo de Oro a la Mejor Actriz de Reparto en una Serie, Miniserie o Película para Televisión en 2016, el Critics' Choice Television Award a la Mejor Actriz de Reparto en una Serie Dramática en 2016, Premio Satellite a la Mejor Actriz de Reparto en una Serie, Miniserie o Película para Televisión en 2017, el Premio Golden Nymph a la Mejor Actriz en una Serie Dramática en 2012, y el Premio Saturn a la Mejor Actriz de Televisión en 2012 y 2017 y Mejor actriz de reparto en televisión en 2016.

Apariciones en otros medios
En 2019, Headey y Peter Dinklage aparecieron como Cersei y Tyrion Lannister junto a Elmo en un anuncio de servicio público de Sesame Street sobre la importancia de ser respetuosos entre sí, como parte de la campaña "El respeto nos une".

Referencias

Personajes de Canción de hielo y fuego
Gemelos ficticios
Reinas ficticias
Asesinos en masa ficticios

redpajama_set_name: RedPajamaWikipedia


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text:
\section{Introduction}
In this paper we study the positive definite completion of a \textdef{partial
symmetric Toeplitz matrix, $\mathcal{T} $}.
The main contribution is Theorem \ref{thm:main}, where we present a characterization of
those Toeplitz patterns for which the maximum determinant completion is
Toeplitz, whenever the partial matrix is positive definite completable.
Part of this result answers a conjecture about the existence of a
positive Toeplitz completion with a specific pattern.
A consequence of the main result is an extension to the maximum rank
completion in the positive
semidefinite case, and an application to the \emph{singularity degree}
of a family of \emph{semidefinite programs (SDPs)}.  In the following paragraphs we introduce relevant background information, state the main result, and motivate our pursuit.
\index{$\mathcal{T} $, partial symmetric Toeplitz matrix}

A \textdef{partial matrix} is a matrix in which some of the entries are assigned values while others are unspecified treated as variables.  For instance,
\begin{equation}
\label{eq:M}
\mathcal{M} :=\begin{bmatrix} 6 & 1 & x & 1 & 1 \cr
1 & 6 & 1 & y & 1 \cr
u & 1 & 6 & 1 & z \cr
1 & v & 1 & 6 & 1 \cr
1 & 1 & w & 1 & 6 \end{bmatrix}
\end{equation}
is a real partial matrix, where the unspecified entries are indicated by
letters.  A \textit{completion} of a partial matrix $\mathcal{T} $ is obtained by
assigning values to the unspecified entries.  In other words, a matrix
$T$ (completely specified) is a completion of $\mathcal{T} $ if it coincides with
$\mathcal{T} $ over the specified entries: $T_{ij} = \mathcal{T} _{ij}$, whenever $\mathcal{T} _{ij}$ is
specified.  A \textdef{matrix completion problem} is to determine whether the partial matrix can be completed so as to satisfy a desired
property?  This type of problem has enjoyed considerable attention in
the literature due to applications in numerous areas,
e.g.,~\cite{MR2807419,Rechtparrilofazel}. For example
this is used in sensor network localization
\cite{kriswolk:09,laur:97b}, where the property is that the
completion is a Euclidean
distance matrix with a given embedding dimension.
Related references for matrix completion problems
are e.g.,~\cite{MR1321785,MR2014037,MR2565240,MR2279160,MR1823516}.

The \text{pattern} of a partial matrix is the set of specified entries.
For example, the pattern of $\mathcal{M} $ is all of the elements in diagonals
$-4,-3,-1,0,1,3,4$.  Whether a partial matrix is positive definite completable to some
property may depend on the values assigned to specified entries (the
data) and it may also depend on the pattern of specified entries.  A
question pursued throughout the literature is whether there exist
patterns admitting completions whenever the data satisfy some
assumptions.  Consider, for instance, the property of positive
definiteness.  A necessary condition for a partial matrix to have a
positive definite completion is that all completely specified
principal submatrices are positive definite.  We refer to such partial
matrices as \textdef{partially positive definite}.  Now we ask: what are
the patterns for which a positive definite completion exists whenever a
partial matrix having the pattern is partially positive definite?  In
~\cite{GrJoSaWo:84} the set of such patterns is shown to be fully
characterized by \emph{\textdef{chordal}ity of the graph} of the matrix.

In this work the desired property is \emph{symmetric Toeplitz positive
definite}. In particular, we consider the completion with maximum
determinant over all positive definite completions.  Recall that a real
symmetric $n\times n$ matrix $T$ is Toeplitz if there exist real numbers
$t_0,\dotso,t_{n-1}$ such that $T_{ij} = t_{\lvert i-j\rvert}$ for all
$i,j \in \{1,\dotso,n\}$.  A partial matrix is said to be
\textdef{partially symmetric Toeplitz} if the specified entries are symmetric and consist
of entire diagonals where the data is constant over each diagonal.  The
pattern of such a matrix indicates which diagonals are known and hence
is a subset of $\{0,\dotso,n-1\}$.  Here $0$ refers to the main
diagonal, $1$ refers to the super diagonal and so on.  The subdiagonals
need not be specified in the pattern since they are implied by symmetry.
In fact, since positive definite completions are trivial when the main
diagonal is not specified (and the determinant is unbounded), we assume
throughout that the main diagonal is specified. We therefore only consider
patterns of increasing integers in the set $\{1,\dotso,n-1\}$.  The pattern of $\mathcal{M} $, for instance,
is $\{1,3,4\}$.
\index{$T$, Toeplitz}
\index{Toeplitz, $T$}

For a partial matrix $\mathcal{T} $ with pattern $P$ and $k \in P$, we let $t_k$ denote the value of $\mathcal{T} $ on diagonal $k$ and we refer to $\{t_k : k \in P \cup \{0\} \}$
as the \textit{data} of $\mathcal{T} $.  For $\mathcal{M} $ the data is
$\{t_0,t_1,t_3,t_4\}= \{6,1,1,1\}$.

We say that a partial Toeplitz matrix $\mathcal{T} $ is \textdef{positive
(semi)definite completable} if there exists a positive (semi)definite completion of $\mathcal{T} $.  In this case we denote by
${T^{\star}} $, the unique positive definite completion of $\mathcal{T} $ that maximizes the
determinant over all positive definite completions.  We now state the
main contribution of this paper, a characterization of the Toeplitz patterns
where the maximum determinant completion is itself Toeplitz, whenever the
partial matrix is positive definite completable.
\index{maximum determinant positive definite completion, ${T^{\star}} $}
\index{${T^{\star}} $, maximum determinant positive definite completion}
\begin{thm}
\label{thm:main}
Let $\emptyset \ne \textdef{$P \subseteq \{1,\dotso,n-1\}$}$ denote a
pattern of increasing integers.  The following are equivalent.
\begin{enumerate}
\item
\label{item:parttoep}
Let $\mathcal{T} $ be a partial Toeplitz matrix have pattern P, and let $\mathcal{T} $
be positive definite completable. Then ${T^{\star}} $ is Toeplitz.
\item
\label{item:Prnk}
There exist $r,k \in {\mathbb N} $ such that $P$ has one of the three forms:
\begin{itemize}
\item \textdef{$P_1 := \{k,2k,\dotso,rk\}$},
\item \textdef{$P_2 := \{k,2k,\dotso,(r-2)k,rk\}$, where $n = (r+1)k$},
\item \textdef{$P_3 := \{k,n-k\}$}.
\end{itemize}
\end{enumerate}
\end{thm}
The proof of Theorem \ref{thm:main} is presented in Section~\ref{sec:main}.  Note that for the
partial Toeplitz matrix $\mathcal{M} $ in \eqref{eq:M}, we can
set all the unspecified entries to $1$ and obtain a positive definite
completion. However, the maximum determinant completion is given, to four
decimal accuracy, when $x=z=u=w=0.3113$ and $y=v=0.4247$. But,
this completion is \emph{not}
Toeplitz.  Indeed, the pattern of $\mathcal{M} $ is not among the patterns of Theorem~\ref{thm:main}.

Positive definite Toeplitz matrices play an important role throughout
the mathematical sciences.  Correlation matrices of data arising from
time series,~\cite{MR941464}, and solutions to the trigonometric moment
problem,~\cite{MR1709182}, are two such examples.
Among the early contributions to this area is the following sufficient
condition and characterization, for a special case of pattern $P_1$.
\begin{thm}[\!{\cite{DymGoh:81}}]
\label{thm:banded}
If $\mathcal{T} $ is a partially positive definite Toeplitz matrix with pattern
$P_1$ and $k=1$, then ${T^{\star}} $ exists and is Toeplitz.
\end{thm}
\begin{thm}[\!{\cite[Theorem~1.1]{MR1709182}}]
\label{thm:jonlunnav}
A partially positive definite Toeplitz matrix is positive definite
Toeplitz completable if, and only if, it has a pattern of the form $P_1$.
\end{thm}
In these two results the assumption on the partial matrix is that it is partially positive definite, whereas in Theorem~\ref{thm:main} we make the stronger assumption that a positive definite completion exists.  As a consequence, our characterization includes the patterns $P_2$ and $P_3$.  To the best of our knowledge pattern $P_2$ has not been addressed in the literature.
A special case of pattern $P_3$,
with $k=1$, was considered in~\cite{MR1236734}, where the authors
characterize the data for which the pattern is positive definite completable.
In~\cite{MR2156414} the result is extended to arbitrary $k$ and
sufficient conditions for Toeplitz completions are provided.  Moreover,
the authors conjecture that whenever a partially positive definite
Toeplitz matrix with pattern $P_3$ is positive definite completable then it admits a
Toeplitz completion.  This conjecture is confirmed true in
Theorem~\ref{thm:main} and more specifically in Theorem~\ref{thm:main2}.

Our motivation for the maximum determinant completion comes from
optimization and the implications of the optimality conditions for
completion problems (see Theorem~\ref{thm:maxdet}).  In particular, a
positive definite completion problem may be formulated as an SDP.
The \textit{central path} of standard interior point
methods used to solve SDPs consists
of solutions to the maximum determinant problem. In the recent
work~\cite{SWW:17}, the maximum determinant problem is used to find
feasible points of SDPs when the usual regularity conditions are not
satisfied.  A consequence of Theorem~\ref{thm:main} is that when a
partially Toeplitz matrix having one of the patterns of the theorem
admits a positive semidefinite completion, but not a positive definite
one, then it has a maximum rank positive semidefinite completion that is
Toeplitz. This result, as well as further discussion on the positive semidefinite case, are presented in
Section~\ref{sec:sdp}. The application to finding the \emph{singularity degree}
of a family of  SDPs is presented in Section \ref{sec:sd}.

\section{Proof of Main Result with Consequences}
\label{sec:main}
To simplify the exposition, the proof of Theorem~\ref{thm:main} is
broken up into a series of results.  Throughout this section
we assume that every pattern $P$ is a non-empty subset of
$\{1,\dotso,n-1\}$, consisting of strictly increasing integers, and
$\mathcal{T} $ denotes an $n\times n$ partial symmetric Toeplitz matrix with
pattern (or form) $P$.  We begin by presenting the optimality conditions for the maximum determinant problem.
\index{pattern, $P$}
\index{$P$, pattern}
\begin{thm
\label{thm:maxdet}
Let $\mathcal{T} $ be of the form $P$ and positive definite completable. Then
${T^{\star}} $ exists, is unique, and satisfies $({T^{\star}} )^{-1}_{i,j} = 0$,
whenever $\lvert i - j \rvert \notin P$.
\end{thm}
\begin{proof}
This result is proved for general positive definite completions
in~\cite{GrJoSaWo:84}. See also~\cite{SWW:17}.
\end{proof}
For general positive definite completion problems, this result simply
states that the inverse of the completion of maximum determinant has
zeros in the unspecified (or free) entries.  Since we are interested in
Toeplitz completions, we may say something further using a permutation
under which Toeplitz matrices are invariant.  Let $K$ be the symmetric
$n\times n$ anti-diagonal matrix defined as:
\index{$K$, anti-diagonal permutation}
\index{anti-diagonal permutation, $K$}
\begin{equation}
\label{eq:antidiag}
K_{ij} := \begin{cases}
1 \quad \text{if } i+j = n+1, \\
0 \quad \text{otherwise},
\end{cases}
\end{equation}
i.e.,~$K$ is the permutation matrix that reverses the order of the
sequence $\{1,2,\ldots,n\}$.
\begin{lem}
\label{lem:K}
Let $\mathcal{T} $ be of the form $P$ and positive definite completable.
Let ${T^{\star}} $ be the maximum determinant completion, and let
$K$ be the anti-diagonal permutation matrix in \eqref{eq:antidiag}.
Then the following hold.
\begin{enumerate}
\item
\label{item:KTK}
 ${T^{\star}}  = K{T^{\star}}  K$.
\item
\label{item:PTstar}
If $P$ is of the form $P_2$ with $k=1$, i.e.,~$P = \{1,2,\dotso,n-3,n-1\}$, then ${T^{\star}} $ is Toeplitz.
\end{enumerate}
\end{lem}
\begin{proof}
For Item \ref{item:KTK}, it is a simple exercise to verify that the
permutation reverses the order of the rows and columns and we have
\[
[K{T^{\star}}  K]_{ij} = {T^{\star}} _{n+1-i,n+1-j}, \,
\forall i,j \in \{1,\dotso,n\}.
\]
Moreover,
\[
\lvert n+1-i - (n+1-j)\rvert = \lvert i-j\rvert.
\]
Therefore, it follows that
\[
[K{T^{\star}}  K]_{ij} = {T^{\star}} _{n+1-i,n+1-j} = {T^{\star}} _{ij} = t_{\lvert i-j\rvert},
\,\, \forall \lvert i-j \rvert \in P\cup \{0\}.
\]
Hence $K{T^{\star}}  K$ is a completion of $\mathcal{T} $.  Moreover, $K\cdot K$ is
an automorphism of the cone of positive definite matrices. Hence
$K{T^{\star}}  K$ is a positive definite completion of $\mathcal{T} $, and since $K$ is
a permutation matrix, we conclude that $\det(K{T^{\star}}  K) = \det({T^{\star}} )$.
By Theorem~\ref{thm:maxdet}, ${T^{\star}} $ is the unique maximizer of the
determinant. Therefore ${T^{\star}}  = K{T^{\star}}  K$, as desired.

For Item \ref{item:PTstar}, we let $\mathcal{T} $ be as in the hypothesis and
note that the only unspecified entries are $(1,n-1)$ and $(2,n)$, and
their symmetric counterparts.  Therefore it suffices to show that
${T^{\star}} _{1,n-1} = {T^{\star}} _{2,n}$.  By applying Item \ref{item:KTK} we get
\[
{T^{\star}} _{1,n-1} = [K{T^{\star}}  K]_{1,n-1} = {T^{\star}} _{n+1-1,n+1-(n-1)} = {T^{\star}} _{n,2} = {T^{\star}} _{2,n},
\]
as desired.
\end{proof}

The pattern $\{1,2,\dotso,n-3,n-1\}$ in Lemma \ref{lem:K}, above,
is a special case of pattern $P_2$ with $k=1$.  In fact, we show that a
general pattern $P_2$ may always be reduced to this special case.  A
further observation is that this specific pattern is nearly of the form
$P_1$.  Indeed, if the diagonal $n-2$ were specified, the pattern would be of
the form $P_1$.  In fact, for any pattern of the form $P_2$, if the
diagonal $(r+1)k$ were specified, the pattern would be of the form
$P_1$.  We now state a useful lemma for proving that
Theorem \ref{thm:main}, Item \ref{item:Prnk} implies
Theorem \ref{thm:main}, Item \ref{item:parttoep}, when $P$ is of the
form $P_1$ or $P_2$.

\begin{lem}
\label{lem:block}
Let $\mathcal{S} $ be a partial $n\times n$ positive definite completable symmetric matrix and $Q$ a permutation matrix of order $n$ such that
\[
Q^T\mathcal{S}  Q = \begin{bmatrix}
\mathcal{S} _1 & &&  \\
& \mathcal{S} _2 & &  \\
& & \ddots &  \\
& && \mathcal{S} _{\ell} \\
\end{bmatrix},
\]
for some $\ell \in {\mathbb N} $. Here each block $\mathcal{S} _i$ is a
partial symmetric matrix for
$i \in \{1,\dotso,\ell\}$, and the elements outside of the blocks are
all unspecified.  Then the maximum determinant completion of $\mathcal{S} _i$,
denoted $S^{\star}_i$, exists and is unique.  Moreover, the unique
maximum determinant completion of $\mathcal{S} $ is given by
\[
S^{\star} = Q \begin{bmatrix}
S^{\star}_1 &0 &\cdots&0  \\
0 & S^{\star}_2 &\cdots &0  \\
\vdots &\vdots & \ddots & \vdots \\
0& 0&\cdots& S^{\star}_{\ell} \\
\end{bmatrix}Q^T.
\]
\end{lem}
\begin{proof}
Since $Q^T\cdot Q$ is an automorphism of the positive definite matrices,
with inverse $Q\cdot Q^T$, we have that $Q^T\mathcal{S}  Q$ is positive definite completable and
admits a unique maximum determinant completion, say $\hat{S}$.
Moreover, under the map $Q\cdot Q^T$, every completion of $Q^T \mathcal{S}  Q$
corresponds to a unique completion of $\mathcal{S} $, with the same determinant,
since the determinant is invariant under the transformation $Q\cdot
Q^T$.  Therefore, we have $S^{\star} = Q\hat{S} Q^T$.  Now we show that
$\hat{S}$ has the block diagonal form.  Observe that $\mathcal{S} _i$ is
positive definite completable, take for instance the positive definite submatrices of
$\hat{S}$ corresponding to the blocks $\mathcal{S} _i$.  Thus $S^{\star}_i$ is
well defined, and by the determinant Fischer inequality,
e.g.,~\cite[Theorem 7.8.3]{HoJo:85}, we  have
\[
\hat{S} = \begin{bmatrix}
S^{\star}_1 &0 &\cdots&0  \\
0 & S^{\star}_2 &\cdots &0  \\
\vdots &\vdots & \ddots & \vdots \\
0& 0&\cdots& S^{\star}_{\ell} \\
\end{bmatrix},
\]
as desired.
\end{proof}

In~\cite{MR1709182} it is shown that a partial Toeplitz matrix of the
form $P_1$ with $rk = n-1$ can be permuted into a block diagonal matrix as in
Lemma~\ref{lem:block}.  We use this observation and extend it to
all patterns of the form $P_1$, as well as patterns of the form $P_2$, in the following.

\begin{prop}
\label{prop:P1P2}
Let $\mathcal{T} $ be positive definite completable and of the form $P_1$ or $P_2$. Then ${T^{\star}} $ is Toeplitz.
\end{prop}
\begin{proof}
Let $\mathcal{T} $ be of the form $P_1$ with data $\{t_0,t_k, t_{2k},\dotso,t_{rk}\}$ and let $p \ge r$ be the largest integer so that $pk \le n-1$.  As in~\cite{MR1709182}, there exists a permutation matrix $Q$ of order $n$ such that
\[
Q^T\mathcal{T}  Q = \begin{bmatrix}
\mathcal{T} _0 & &&&&  \\
& \ddots & &&&  \\
& & \mathcal{T} _0 &&&  \\
& && \mathcal{T} _1 &&\\
&&&&\ddots&\\
&&&&& \mathcal{T} _1
\end{bmatrix},
\]
where $\mathcal{T} _0$ is a $(p+1)\times (p+1)$ partial Toeplitz matrix occuring
$n-pk$ times and and $\mathcal{T} _1$ is a $p\times p$ partial Toeplitz.
Moreover, $\mathcal{T} _0$ and $\mathcal{T} _1$ are both partially positive definite.  Let
us first consider the case $p=r$.  Then $\mathcal{T} _0$ and $\mathcal{T} _1$ are actually
fully specified, and the maximum determinant completion of $Q^T \mathcal{T}  Q$,
as in Lemma~\ref{lem:block}, is obtained by fixing the elements outside
of the blocks to $0$.  After permuting back to the original form,
${T^{\star}} $ has zeros in every unspecified entry. Hence it is Toeplitz.
Now suppose $p > r$.  Then $\mathcal{T} _0$ is a partial Toeplitz matrix with
pattern $\{1,2,\dotso,r\}$ and data $\{t_0,t_k,t_{2k},\dotso,t_{rk}\}$
and $\mathcal{T} _1$ is a partial Toeplitz matrix having the same pattern and
data as $\mathcal{T} _0$, but one dimension smaller.  That is, $\mathcal{T} _1$ is a
partial principal submatrix of $\mathcal{T} _0$.  By Theorem~\ref{thm:banded}
both $\mathcal{T} _0$ and $\mathcal{T} _1$ are positive definite completable and their
maximum determinant completions, $\mathcal{T} _0^{\star}$ and $\mathcal{T} _1^{\star}$,
are Toeplitz.  Let $\{a_{(r+1)k}, a_{(r+2)k}, \dotso, a_{pk}\}$ be the
data of $\mathcal{T} _0^{\star}$ corresponding to the unspecified entries and let
$\{b_{(r+1)k}, b_{(r+2)k}, \dotso, b_{(p-1)k}\}$, be the data
corresponding to the unspecified entries of $\mathcal{T} _1$.  By the permanence
principle of \cite{MR892133}, $\mathcal{T} _1$ is a principle submatrix of
$\mathcal{T} _0$ and therefore $b_i = a_i$, for all
$i \in \{(r+1)k,(r+2)k,\dotso,(p-1)k\}$.
By Lemma~\ref{lem:block}, the maximum determinant completion of $Q^T\mathcal{T}  Q$ is obtained by completing $\mathcal{T} _0$ and $\mathcal{T} _1$ to $\mathcal{T} _0^{\star}$ and $\mathcal{T} _1^{\star}$ respectively, and setting the entries outside of the blocks to zero.  After permuting back to the original form we get that ${T^{\star}} $ is Toeplitz with data $a_{(r+1)k}, a_{(r+2)k}, \dotso, a_{pk}$ in the diagonals $(r+1)k, (r+2)k, \dotso,pk$ and zeros in all other unspecified diagonals.

Now suppose that $\mathcal{T} $ is of the form $P_2$.  By applying the same
permutation as above, and by using the fact that $n=(r+1)k$ and each
block $\mathcal{T} _0$ is of size $r+1$, we see that the submatrix consisting only of blocks $\mathcal{T} _0$ is of size
\[
(n-rk)(r+1) = ((r+1)k-rk)(r+1) = k(r+1) = n.
\]
Hence,
\[
Q^T\mathcal{T}  Q = \begin{bmatrix}
\mathcal{T} _0 & &  \\
& \ddots &   \\
& & \mathcal{T} _0   \\
\end{bmatrix},
\]
where $\mathcal{T} _0$ is a partial matrix with pattern $\{1,2,\dotso,r-2,r\}$ and data
\[
\{t_0,t_k,t_{2k},\dotso,t_{(r-2)k},t_{rk}\}.
\]
The unspecified elements of diagonal $(r-1)k$ of $\mathcal{T} $ are contained in
the unspecified elements of diagonal $r-1$ of the partial matrices
$\mathcal{T} _0$.  By Lemma~\ref{lem:K}, the maximum determinant completion of $\mathcal{T} _0$ is Toeplitz with value $t_{(r-1)k}$ in the unspecified diagonal.  As in the above, after completing $Q^T\mathcal{T}  Q$ to its maximum determinant positive definite completion and permuting back to the original form, we obtain the maximum determinant Toeplitz completion of $\mathcal{T} $ with value $t_{(r-1)k}$ in the diagonal $(r-1)k$ and zeros in every other unspecified diagonal, as desired.
\end{proof}

We now turn our attention to patterns of the form $P_3$.  Let $J$ denote
the $n\times n$ lower triangular Jordan block with eigenvalue $0$.  That
is, $J$ has ones on diagonal $-1$ and zeros everywhere else. We also let
$e_1,\dotso,e_n \in {\mathbb R}^n $ denote the columns of the identity matrix,
i.e.,~the canonical unit vectors.  With this notation we have $J = \sum_{j=1}^{n-1} e_{j+1}e_j^T$.  We state several technical results regarding $J$ in the following lemma.
\index{$J$, Jordan block}
\index{Jordan block, $J$}

\begin{lem}
\label{J}
With $J$ defined as above and $k,l \in \{0,1,\dotso,n-1\}$, the following hold.
\begin{enumerate}
\item
\label{item:sumes}
$J^k = \sum_{j=1}^{n-k} e_{j+k}e_j^T$.
\item
\label{item:nnzJ}
$J^k(J^T)^l$ has nonzero elements only in the diagonal $l-k$.
\item
\label{item:JJz}
 If $k < l$, then
\[
J^k(J^T)^l - J^{n-l} (J^T)^{n-k} = 0 \ \iff \ l=n-k.
\]
\end{enumerate}
\end{lem}
\begin{proof}
For Item \ref{item:sumes} the result clearly holds when $k\in \{0,1\}$.
Now observe that for integers of suitable size
$(e_ke_l^T)(e_ie_j^T) \ne 0$ if, and only if, $l=i$ in which case the product is $e_ke_j^T$.  Thus we have
\[
J^2 = \left( \sum_{j=1}^{n-1} e_{j+1}e_j^T\right)\left(\sum_{j=1}^{n-1} e_{j+1}e_j^T\right) = \sum_{j=2}^{n-1} (e_{j+1}e_j^T)(e_je_{j-1}^T) = \sum_{j=1}^{n-2} e_{j+2}e_j^T.
\]
Applying an induction argument yields the desired expression for arbitrary $k$.

For Item \ref{item:nnzJ}, we use the result of item  \ref{item:sumes} to get
\begin{align*}
J^k(J^T)^l &= \left( \sum_{j=1}^{n-k} e_{j+k}e_j^T\right)\left(\sum_{j=1}^{n-l} e_je_{j+l}^T\right), \\
&= \sum_{j=1}^{n-\max \ \{k,l\}} (e_{j+k}e_j^T)(e_je_{j+l}^T), \\
&= \sum_{j=1}^{n-\max \ \{k,l\}} e_{j+k}e_{j+l}^T.
\end{align*}
The nonzero elements of this matrix are contained in the diagonal $j+l - (j+k) = l-k$.

Finally, for Item \ref{item:JJz} we have
\[
J^k(J^T)^l - J^{n-l} (J^T)^{n-k}  = \sum_{j=1}^{n-l} e_{j+k}e_{j+l}^T - \sum_{j=1}^k e_{j+n-l}e_{j+n-k}^T.
\]
This matrix is the zero matrix if, and only if, $l=n-k$.
\end{proof}
We now state a special case of the Schur-Cohn Criterion using the matrix
$J$.  We let ${\mathbb S}^n $ denote the Euclidean space of symmetric matrices,
${\mathbb S}^n_{++} $ the cone of positive definite matrices, and ${\mathbb S}^n_T $ the subset of
symmetric, positive definite, Toeplitz matrices.
\begin{thm}[Schur-Cohn Criterion,~\cite{MR638124}]
\label{thm:schurcohn}
Let $f(z) = a_0 + a_1z + \cdots + a_nz^n$ be a polynomial with real
coefficients $a:=(a_0,\dotso,a_n)$. Let
\[
A(a) := \sum_{j=0}^{n-1} a_jJ^j, \ B(a):= \sum_{j=1}^{n-1} a_j \left( J^T\right)^{n-j}.
\]
Then every root of $f(z)$ satisfies $\lvert z \rvert > 1$ if, and only
if,
\[
{\rm Bez} (a) := A(a)A(a)^T - B(a)^TB(a) \in {\mathbb S}^n_{++} ,
\]
where the matrix ${\rm Bez} (a)$ is the Toeplitz Bezoutian.
Moreover ${\rm Bez} (a)^{-1}$ is Toeplitz.
\end{thm}
The Schur-Cohn criterion is usually stated for the case where the roots
are contained within the interior of the unit disk, but a simple
reversal of the coefficients, as described in Chapter~X of
\cite{MR0225972}, leads to the above statement. For further information
on the \textit{Toeplitz Bezoutian}  ${\rm Bez} (a)$, and for a proof of the fact
that ${\rm Bez} (a)^{-1}$ is
Toeplitz, see  e.g.,~\cite{MR2656823}.

We now present a result on the
maximum determinant completion of partial Toeplitz matrices with pattern $P_3$.
\begin{prop}
\label{prop:P3}
Let $\mathcal{T} $ be positive definite completable and of the form $P_3$. Then ${T^{\star}} $ is Toeplitz.
\end{prop}

\begin{proof}
Let $\mathcal{T} $ be as in the hypothesis with pattern of the form $P_3$ defined the integer $k$.  Furthermore, let ${\mathcal O} \subset {\mathbb R}_{++} \times
{\mathbb R}^2$ consist of all triples $(t_0,t_k,t_{n-k})$ so that the
partial Toeplitz matrix with pattern $P_3$ and data
$\{t_0,t_k,t_{n-k}\}$ is positive definite completable. Then it can be verified that ${\mathcal O}$ is an open convex set, and thus in particular connected. We let ${\mathcal U} \subseteq {\mathcal O}$ consist of those triples $(t_0,t_k,t_{n-k})$ for which the corresponding maximum determinant completion is Toeplitz and we claim that ${\mathcal U} = {\mathcal O}$. Clearly ${\mathcal U} \neq \emptyset$ as $(t_k,0,0) \in {\mathcal U}$ for all $t_k>0$. We show that ${\mathcal U}$ is both open and closed in ${\mathcal O}$, which together with the connectedness of ${\mathcal O}$ yields that ${\mathcal U} = {\mathcal O}$.

First observe that the map $F: {\mathcal O} \to {\mathbb S}^n_{++} $ that takes
$(t_0,t_k,t_{n-k})$ to its corresponding positive definite maximum
determinant completion is continuous; see, for instance,
\cite{MR1824072}. Next, the Toeplitz positive definite matrices,
${\mathbb S}^n_{++}  \cap {\mathbb S}^n_T $, form a closed subset of ${\mathbb S}^n_{++} $ since ${\mathbb S}^n_T $ is closed. Thus ${\mathcal U}
= F^{-1} ({\mathbb S}^n_{++} \cap {\mathbb S}^n_T )$ is closed in $\mathcal{O}$.

To show that ${\mathcal U}$ is also open, we introduce the set,
\[
{\mathcal P} := \{ (p,q,r) \in {\mathbb R}_{++} \times {\mathbb R}^2 : p+qz^k+rz^{n-k} {\rm \ has \ all \ roots \ satisfy \ } |z|>1 \} .
\]
Since the region $\lvert z \rvert > 1$ is an open subset of the complex plane, $\mathcal P$ is an open set.  We consider the map $G: {\mathcal P} \to {\mathbb R} ^3$ defined as
\[
G(p,q,r) = ( [{\rm Bez} (p,q,r)^{-1}]_{11},  [{\rm Bez} (p,q,r)^{-1}]_{k1},  [{\rm Bez} (p,q,r)^{-1}]_{n-k,1} ),
\]
where by abuse of notation ${\rm Bez} (p,q,r)$ is the Toeplitz Bezoutian of Theorem~\ref{thm:schurcohn}:
\begin{align*}
{\rm Bez} (p,q,r) &= (pJ^0+qJ^k+rJ^{n-k})(pJ^0+qJ^k+rJ^{n-k})^T \\
& \qquad - (rJ^k+qJ^{n-k})(rJ^k+qJ^{n-k})^T.
\end{align*}
Then $G$ is continuous and we show that its image is exactly $\mathcal{U}$.  By Theorem~\ref{thm:schurcohn}, for any $(p,q,r) \in \mathcal{P}$ we have
\[
{\rm Bez} (p,q,r) \in {\mathbb S}^n_{++} , \ {\rm Bez} (p,q,r)^{-1} \in {\mathbb S}^n_{++}  \cap {\mathbb S}^n_T .
\]
Thus ${\rm Bez} (p,q,r)^{-1}$ is a completion of the partial matrix having pattern $P_3$ and data $\{[{\rm Bez} (p,q,r)^{-1}]_{11},  [{\rm Bez} (p,q,r)^{-1}]_{k1},  [{\rm Bez} (p,q,r)^{-1}]_{n-k,1}\}$.  It follows that $G(\mathcal{P}) \subseteq \mathcal{O}$.  Moreover, expanding ${\rm Bez} (p,q,r)$ we obtain diagonal terms as well as terms of the form $J^0(J^T)^k$ and $J^0(J^T)^{n-k}$, where the coefficients have been omitted.  By Lemma~\ref{J}, ${\rm Bez} (p,q,r)$ has non-zero values only in entries of the diagonals $0,k,n-k$.  Note that the term $J^k(J^T)^{n-k}$ cancels out in the expansion.  Thus by Theorem~\ref{thm:maxdet}, ${\rm Bez} (p,q,r)^{-1}$ is a maximum determinant completion of the partial matrix with pattern $P_3$ and data $\{[{\rm Bez} (p,q,r)^{-1}]_{11},  [{\rm Bez} (p,q,r)^{-1}]_{k1},  [{\rm Bez} (p,q,r)^{-1}]_{n-k,1}\}$ and $G(\mathcal{P}) \subseteq \mathcal{U}$.  To show equality, let $(t_0,t_k,t_{n-k}) \in \mathcal{U}$ and let $F(t_0,t_k,t_{n-k})$, as above, be the maximum determinant completion of the partial matrix with pattern $P_3$ and data $\{t_0,t_k,t_{n-k}\}$ which is Toeplitz.  Let $f_0$, $f_k$, and $f_{n-k}$ be the $(1,1)$, $(k+1,1)$ and $(n-k+1,1)$ elements of $F(t_0,t_kt_{n-k})^{-1}$ respectively.  Then by the Gohberg-Semencul formula for the inversion of a symmetric Toeplitz matrix (see \cite{MR0353038,MR1038316}) we have
\begin{align*}
F(t_0,t_kt_{n-k})^{-1} &= \frac{1}{f_0}(f_0J^0+f_kJ^k+f_{n-k}J^{n-k})(f_0J^0+f_kJ^k+f_{n-k}J^{n-k})^T \\
& \qquad -  \frac{1}{f_0}(f_{n-k}J^k+f_kJ^{n-k})(f_{n-k}J^k+f_kJ^{n-k})^T, \\
&= {\rm Bez}  \left(\sqrt{f_0}, \frac{f_k}{\sqrt{f_0}}, \frac{f_{n-k}}{\sqrt{f_0}}\right).
\end{align*}
Since $F(t_0,t_kt_{n-k})^{-1} \in {\mathbb S}^n_{++} $, it follows that $\left(\sqrt{f_0}, \frac{f_k}{\sqrt{f_0}}, \frac{f_{n-k}}{\sqrt{f_0}}\right) \in \mathcal{P}$ and
\[
G\left(\sqrt{f_0}, \frac{f_k}{\sqrt{f_0}}, \frac{f_{n-k}}{\sqrt{f_0}}\right) = (t_0,t_k,t_{n-k}).
\]
Therefore $G(\mathcal{P}) = \mathcal{U}$.  Moreover, from the above we have that
\[
G^{-1}(t_0,t_k,t_{n-k}) = \left(\sqrt{f_0}, \frac{f_k}{\sqrt{f_0}}, \frac{f_{n-k}}{\sqrt{f_0}}\right),
\]
with $f_0,f_k$, and $f_{n-k}$ defined above.  Since $G^{-1}$ is continuous, $G^{-1}(\mathcal{U}) = \mathcal{P}$, and $\mathcal{P}$ is an open set, we conclude that $\mathcal{U}$ is open, as desired.
\end{proof}

Now we are ready to prove Theorem~\ref{thm:main}.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
The direction $(\ref{item:Prnk})\!\! \implies\!\! (\ref{item:parttoep})$
follows from Proposition~\ref{prop:P1P2} and Proposition~\ref{prop:P3}.

For the direction $(\ref{item:parttoep})\!\! \implies\!\!  (\ref{item:Prnk})$,
let $\mathcal{T} $ be positive definite completable with pattern $P = \{k_1,\dotso,k_s\}$, $k_0 = 0$, and data $\{t_0,t_1,\dotso,t_s\}$. Assume there exists
data $t_j$ for the diagonal $k_j$, $j \in \{0,\dotso,s\}$, and that ${T^{\star}} $ is Toeplitz.  Then by Theorem~\ref{thm:maxdet}, $({T^{\star}} )^{-1}$ has nonzero entries only in the diagonals $P\cup \{0\}$ (and their symmetric counterparts).  We denote by $a_j$ the value of the first column of $({T^{\star}} )^{-1}$ in the row $k_j+1$ for all $j \in \{0,\dotso,s\}$, and define
\[
A := \sum_{j=0}^s a_j J^{k_j}, \ B:= \sum_{j=1}^s a_j \left( J^T\right)^{n-k_j}.
\]
The Gohberg-Semencul formula gives us that $T^{-1} = \frac{1}{a_0}(AA^T
- B^TB)$.  Substituting in the expressions for $A$ and $B$ and
expanding, we obtain $({T^{\star}} )^{-1}$ is a linear combination of the
following types of terms, along with their symmetric counterparts:
\[
J^{k_j}(J^T)^{k_j}, \quad  J^{k_0}(J^T)^{k_j}, \quad J^{k_j}(J^T)^{k_l} - J^{n-k_j}(J^T)^{n-k_l}, \ j \ne l.
\]
By Lemma~\ref{J}, the first type of term has nonzero entries only on the
main diagonal, and the second type of term has nonzero entries only on
the diagonals belonging to $P$.  The third type of term has nonzero
entries only on the diagonals $\pm \lvert k_j-k_l\rvert$.  As we have
already observed in the proof of Proposition~\ref{prop:P3}, the set of
data for which $\mathcal{T} $ is positive definite completable is an open set.
We may therefore perturb the data of $\mathcal{T} $ so that the entries
$a_0,\dotso,a_j$ of the inverse do not all lie on the same proper linear
manifold. Then terms of the form $J^{k_j}(J^T)^{k_l} -
J^{n-k_j}(J^T)^{n-k_l}$ with  $j \ne l$ do not cancel each other out. We
conclude that, for each pair $j < l$, we have $k_j - k_l \in P$ or $J^{k_j}(J^T)^{k_l} - J^{n-k_j}(J^T)^{n-k_l} = 0$.  By Lemma~\ref{J} the second alternative is equivalent to $l = n-j$.  Using this observation we now proceed to show that $P$ has one of the specified forms.

Let $1\le r \le s$ be the largest integer such that $\{k_1,\dotso,k_r\}$
is of the form $P_1$, i.e.,~$k_2 = 2k_1$, $k_3 = 3k_1$, etc\ldots.  If
$r=s$, then we are done. Therefore we may assume $s \ge r+1$. Now we show
that in fact $s=r+1$. We have that $k_{r+1} - k_1 \in P$ or $k_{r+1} = n-k_1$.
We show that the first case does not hold. Indeed if $k_{r+1} - k_1 \in
P$, then it follows that $k_{r+1} - k_1 \in \{k_1,\dotso,k_r\}$. This
implies that
\[
k_{r+1} \in \{2k_1,\dotso,rk_1, (r+1)k_1\} = \{k_2,\dotso,k_r,(r+1)k_1\}.
\]
Clearly $k_{r+1} \notin \{k_2,\dotso,k_r\}$, and if $k_{r+1} =
(r+1)k_1$, then $r$ is not maximal, a contradiction.  Therefore $k_{r+1}
= n-k_1$.  To show that $s=r+1$, suppose to the contrary that $s \ge
r+2$.  Then $k_{r+2} - k_1 \in P$ or $k_{r+2} = n-k_1$.  The latter does
not hold since then $k_{r+2} = k_{r+1}$.  Thus we have $k_{r+2} - k_1
\in \{k_1,\dotso,k_r, k_{r+1}\}$, which implies that
\[
k_{r+2} \in \{2k_1,\dotso,rk_1,(r+1)k_1,k_{r+1}+k_1\} = \{k_2,\dotso,k_r,k_r+k_1,n\}.
\]
Since $k_{r+2} \notin \{k_2,\dotso,k_r,n\}$, we have $k_{r+2} =
k_r+k_1$. Therefore, since $k_r < k_{r+1}<k_{r+2}$, we have that $0<
k_{r+2} - k_{r+1} <k_1$, and moreover, $k_{r+2} - k_{r+1} \notin P$.  It follows that $k_{r+2} = n-k_{r+1} = k_1$, a contradiction.

We have shown that $P = \{k_1,2k_1,\dotso,rk_1,k_s\}$ with $k_s =
n-k_1$. If $r=1$, then $P$ is of the form $P_3$.  On the other hand if
$r \ge 2$, then we observe that $\{k_s -k_r, \dotso,k_s - k_2\} \subseteq P$, or equivalently,
\[
\{k_s -k_r, \dotso,k_s - k_2\} \subseteq \{k_2,\dotso,k_r\}.
\]
Since the above sets of identical cardinality, distinct increasing elements,
we conclude that $k_s - k_2 = k_r$.  Rearranging, we obtain that $k_s = (r+2)k_1$ and $P$ is of the form $P_2$, as desired.
\end{proof}

\begin{rem}
The results of this section have been stated for the symmetric real case for
simplicity and for application to {\textbf SDP}  in the following Section \ref{sec:sdp}.
With obvious modifications, our results extend to the Hermitian case.
\end{rem}

\section{Semidefinite Toeplitz Completions}
\label{sec:sdp}
In this section we extend the results of Theorem~\ref{thm:main} to
positive semidefinite completions.  In the case where all completions
are singular, the maximum determinant is not useful for identifying a
Toeplitz one, however, a recent result of~\cite{SWW:17} allows us to extend our observations to the semidefinite case.  Given a partial symmetric Toeplitz matrix, $\mathcal{T} $, a positive semidefinite completion of $\mathcal{T} $ may be obtained by solving an SDP feasibility problem.  Indeed, if $\mathcal{T} $ has pattern $P$ and data $\{t_k : k \in P\cup \{0\}\}$, then the positive semidefinite completions of $\mathcal{T} $ are exactly the set
\begin{equation}
\label{eq:feasibleset}
\mathcal{F}  := \{X \in {\mathbb S}^n_+  : \mathcal{A} (X) = b \},
\end{equation}
where $\mathcal{A} $ is a linear map and $b$ a real vector in the image space of $\mathcal{A} $ satisfying
\[
[\mathcal{A} (X)]_{ik} = \langle E_{i,i+k}, X\rangle, \ b_{ik} = t_k, \quad i \in \{1,2,\dotso,n-k\}, \ k \in P\cup \{0\}.
\]
Here $E_{i,j}$ is the symmetric matrix having a one in the entries $(i,j)$ and $(j,i)$ and zeros everywhere else and we use the trace inner product: $\langle X,Y \rangle  = {\rm tr} ( XY )$.  The maximum determinant is used extensively in SDP, for example, the central path of interior point methods is defined by solutions to the maximum determinant problem.  If $\mathcal{F} $ is nonempty but does not contain a positive definite matrix, the maximum determinant may still be applied by perturbing $\mathcal{F} $ so that it does intersect the set of positive definite matrices.  Consider the following parametric optimization problem
\[
X(\alpha) := \arg \max \ \{ \det(X) : X \in \mathcal{F} (\alpha)\},
\]
where $\mathcal{F} (\alpha) := \{X \in {\mathbb S}^n_+  : \mathcal{A} (X) = b + \alpha \mathcal{A} (I)\}$ and $\alpha > 0$.  For each $\alpha > 0$, the solution $X(\alpha)$ is contained in the relative interior of $\mathcal{F} (\alpha)$.  It is somewhat intuitive that if the limit of these solutions is taken as $\alpha$ decreases to $0$, we should obtain an element of the relative interior of $\mathcal{F} (0) = \mathcal{F} $.  Indeed, the following result confirms this intuition.  We denote by $\mathcal{A} ^*$ the adjoint of $\mathcal{A} $.
\begin{thm}
\label{thm:Xalpha}
Let $\mathcal{F}  \ne \emptyset$ and $X(\alpha)$ be as above.  Then there exists $\bar{X}$ in the relative interior of $\mathcal{F} $ such that $\lim_{\alpha \searrow 0}X(\alpha) = \bar{X}$. Moreover, $\bar{Z} := \lim_{\alpha \searrow 0}\alpha (X(\alpha))^{-1}$ exists and satisfies $\bar{X} \bar{Z} = 0$ and $\bar{Z} \in \range(\mathcal{A} ^*)$.
\end{thm}
\begin{proof}
See Section~3 of~\cite{SWW:17}.
\end{proof}
An immediate consequence of this result is the following.
\begin{cor}
\label{cor:sdp}
Let $\mathcal{T} $ be an $n\times n$ partial symmetric Toeplitz matrix of the form $P_1$, $P_2$, or $P_3$.  If $\mathcal{T} $ admits a positive semidefinite completion then it admits a maximum rank completion that is Toeplitz.
\end{cor}
\begin{proof}
Let $\mathcal{T} $ be as in the hypothesis with data $\{t_0,t_1, \dotso, t_s\}$ and let $\mathcal{F} $ be the set of positive semidefinite completions, as above.  If $\mathcal{F}  \cap {\mathbb S}^n_{++}  \ne \emptyset$, then the maximum determinant completion is Toeplitz by Theorem~\ref{thm:main} and is of maximum rank.  Now suppose $\mathcal{F}  \in {\mathbb S}^n_+  \setminus {\mathbb S}^n_{++} $ and observe that for every $\alpha > 0$, $\mathcal{F} (\alpha)$ consists of solutions to to the completion problem having pattern $P_1$, $P_2$, or $P_3$ with data  $\{t_0+\alpha,t_1, \dotso, t_s\}$ and there exists a positive definite completion.  Thus $X(\alpha)$ is Toeplitz for each $\alpha > 0$ and since the Toeplitz matrices are closed, the limit point $\bar{X}$, of Theorem~\ref{thm:Xalpha}, is Toeplitz.  The relative interior of $\mathcal{F} $ corresponds to those matrices having maximum rank over all of $\mathcal{F} $, hence $\bar{X}$ has maximum rank, as desired.
\end{proof}
\begin{rem}
In Theorem~2.2 of \cite{MR1444091} the author gives in the case of two prescribed diagonals (in the strict lower triangular part) necessary and sufficient conditions on the data for the existence of a Toeplitz positive semidefinite completion. In Theorem~10 of \cite{MR2156414} the authors give in the case of pattern $P_3$ necessary and sufficient conditions for the existence of a positive semidefinite completion. If one is able to verify that the conditions are the same, which will require some tenacity, then one would have an alternative proof that for the pattern $P_3$ positive semidefinite completability implies the existence of a Toeplitz positive semidefinite completion. Their results are all stated for the real case, so one advantage of the approach here is that it readily generalizes to the complex Hermitian case.
\end{rem}
While Theorem~\ref{thm:main} characterizes patterns for which the maximum determinant completion is automatically Toeplitz and Corollary~\ref{cor:sdp} addresses the maximum rank completions, one may merely be interested in the existence of a Toeplitz completion when a positive semidefinite one exists. Obviously, the patterns in Theorem \ref{thm:main} fall in this category, but as we see in the following result, there are more.

\begin{thm}
\label{thm:main2}
Define the patterns
\begin{itemize}
\item $P_2' := \{k,2k,\dotso,(r-2)k,rk\}$,
\item $P_3' := \{k,r\}$ where $n \ge k+r$.
\end{itemize}
If $\mathcal{T} $ is an $n\times n$ positive semidefinite completable partial Toeplitz matrix with a pattern in the set $\{P_1,P_2',P_3'\}$, then $\mathcal{T} $ has a Toeplitz positive semidefinite completion.
\end{thm}
\begin{proof}
For pattern $P_1$ this is a consequence of Corollary~\ref{cor:sdp}.
Note that $P_2'$ and $P_3'$ are obtained from $P_2$ and $P_3$,
respectively, by relaxing the restriction on $n$, i.e.,~allowing $n$ to be larger.  Using the results we already have for $P_2$ and $P_3$ we fill in some of the diagonals of $\mathcal{T} $ to obtain a new partial matrix of the form $P_1$.  We show the proof only for patterns of the form $P_2'$ since the same approach may be used for patterns of the form $P_3'$.

Suppose $\mathcal{T} $ has pattern $P_2'$ and consider the partial submatrix containing the first $(r+1)k$ rows and columns.  This partial matrix is Toeplitz, has a positive semidefinite completion, and has pattern $P_2$.  Let $U  := \{1,\dotso,(r+1)k-1\} \setminus P_2$.  The elements of $U$ correspond to the unspecified diagonals of the submatrix.  By Corollary~\ref{cor:sdp}, there exists a Toeplitz completion for this submatrix that assigns the value $a_i$ for every $i \in U$.  Now $U$ is a subset of the unspecified diagonals of $\mathcal{T} $.  We assign the value $a_i$ to the unspecified diagonals of $\mathcal{T} $ for every $i \in U$ thereby obtaining a new partial positive semidefinite Toeplitz matrix, say $\mathcal{T} '$, with pattern $\{1,2,\dotso,(r+1)k\}$.  Since $\mathcal{T} $ and $\mathcal{T} '$ agree on the diagonals of $P_2'$, every completion of $\mathcal{T} '$ is also a completion of $\mathcal{T} $.  The pattern of $\mathcal{T} '$ is of the form $P_1$, hence it admits a positive semidefinite Toeplitz completion, which is also a completion of $\mathcal{T} $, as desired.
\end{proof}

Whether or not Theorem \ref{thm:main2} gives a full characterization of
all patterns for where there is always a Toeplitz completion among all positive semidefinite completions, is an open question.

\section{The Singularity Degree of Some Toeplitz Cycles}
\label{sec:sd}
The \emph{Slater condition} holds for the feasible set of an SDP if it
contains a positive definite matrix.  If the Slater condition does not
hold for an SDP then there is no guarantee of convergence to an optimal
solution using any known algorithm, moreover, it may not be possible to
verify if a given matrix is optimal or not.  One way to regularize an
SDP that does not satisfy the Slater condition is by restricting the
problem to the smallest face of ${\mathbb S}^n_+ $ containing the feasible set.
Since every face of ${\mathbb S}^n_+ $ is a smaller dimensional positive
semidefinite cone, every SDP may be transformed into an equivalent
(possibly smaller dimensional) SDP for which the Slater condition holds.
This transformation is referred to as \emph{facial reduction}, see for
instance \cite{bw1,bw2,DrusWolk:16}.  The challenge, of course, is to
obtain the smallest face.  Most facial reduction algorithms look for
\emph{exposing vectors}, i.e.,~non-zero, positive semidefinite matrices that are orthogonal to the minimal face.  Exposing vectors are guaranteed to exist by the following theorem of the alternative.  Here we let $\mathcal{F} $ be the feasible set of an SDP that is defined by the affine equation $\mathcal{A} (X) = b$, as in \eqref{eq:feasibleset}.
\begin{thm}[{\cite{bw3}}]
\label{thm:fr}
Exactly one of the following holds.
\begin{enumerate}
\item $\mathcal{F}  \cap {\mathbb S}^n_{++}  \ne \emptyset$.
\item  \label{item:expZ}
There exists $Z \in {\mathbb S}^n_+  \cap \range (\mathcal{A} ^*)$ such that $ZX = 0$ for all $X \in \mathcal{F} $.
\end{enumerate}
\end{thm}
This result guarantees the existence of exposing vectors when the Slater
condition does not hold.  By restricting the feasible set of an SDP to
the kernel of an exposing vector, the dimension of the SDP is reduced.
By repeatedly finding exposing vectors and reducing the size of the SDP,
eventually the problem is reduced to the minimal face and the Slater
condition holds.  If the exposing vector obtained at each iteration is
as in Item \ref{item:expZ} of Theorem~\ref{thm:fr} and of maximal rank over all such exposing vectors, then the number of times the original SDP needs to be reduced in order to obtain a regularized SDP is referred to as the \emph{singularity degree}.
This notion and the connection to error bounds for SDP was introduced in
\cite[Sect. 4]{S98lmi}.  For instance, if an SDP satisfies the Slater
condition, then it has singularity degree $0$ and the singularity degree is $1$ if and only if there exists an exposing vector $Z \in {\mathbb S}^n_+  \cap \range(\mathcal{A} ^*)$ such that $\rank(Z) + \rank(X) = n$ for all $X$ in the relative interior of $\mathcal{F} $.

In \cite[Lemma~3.4]{tanigawa} it is shown that for $n\ge 4$, there exists a partial matrix (not Toeplitz) with all entries of the diagonals $0,1,n-1$ specified so that the singularity degree of the corresponding SDP is at least 2.  Here we apply the results of the previous sections to derive the singularity degree (or bounds for it) of a family of symmetric partial Toeplitz matrices with pattern $P = \{1,n-1\}$.  As in much of the matrix completion literature the partial matrix is viewed as arising from a graph and the pattern $P$ corresponds to the graph of a \emph{cycle} with loops.  The following result is useful throughout.
\begin{prop}
\label{Tinverse}
Let $T=(t_{i-j})_{i,j=1}^n$ be a positive definite Toeplitz matrix, and suppose that $(T^{-1})_{k,1}=0$ for all $k\in \{3,\ldots , n-1\}$. Then $T^{-1}$ has the form
\begin{equation}\label{CE} \begin{bmatrix}
a & c & 0& & d \\
c & b & c & \ddots & \\
0& c & b & \ddots &0 \\
& \ddots & \ddots & \ddots & c \\
d & & 0& c & a
\end{bmatrix},\end{equation}
with $b=\frac{1}{a} (a^2+c^2-d^2)$.
\end{prop}

\begin{proof}
Let us denote the first column of $T$ by $\begin{bmatrix} a & c & 0 &
\cdots & 0 & d \end{bmatrix}^T$. By the Gohberg-Semencul
formula we have that
$$ T^{-1} =\frac{1}{a} ( AA^T-B^TB ), $$
where
$$ A=\begin{bmatrix}
a & 0 & 0& & 0 \\
c & a & 0 & \ddots & \\
0& c & a & \ddots &0 \\
& \ddots & \ddots & \ddots & 0 \\
d & & 0& c & a
\end{bmatrix}, B= \begin{bmatrix}
0 & d & 0& & c \\
0 & 0 & d & \ddots & \\
0& 0 & 0 & \ddots &0 \\
& \ddots & \ddots & \ddots & d \\
0& & 0& 0 & 0
\end{bmatrix}.$$
\end{proof}

\begin{ex}
\label{ex:4n}
Let $n=4$ and consider the partial matrix with pattern $P = \{1,3\}$ and data $\{t_0,t_1,t_3\} = \{1+\alpha,\cos(\frac{\theta}{3}), \cos(\theta)\}$ for $\theta \in [0,\pi]$ and $\alpha \ge0$.  Let $\mathcal{F} (\alpha)$ denote the set of positive semidefinite completions for each $\alpha > 0$ as in Section~\ref{sec:sdp}, let $\mathcal{F}  = \mathcal{F} (0)$, and let $\sd(\mathcal{F} )$ denote the singularity degree of any SDP for which $\mathcal{F} $ is the feasible set.  By Corollary~6 of \cite{MR1236734} there exists a positive definite completion whenever $\alpha > 0$ and there exists a positive semidefinite completion (but not a positive definite one) when $\alpha = 0$.  Then by Theorem~\ref{thm:main} the maximum determinant completion is Toeplitz whenever $\alpha  > 0$ and there exists a maximum rank positive semidefinite completion that is Toeplitz when $\alpha = 0$ by Corollary~\ref{cor:sdp}.  Let $X(\alpha)$ denote the maximum determinant positive definite completion when $\alpha > 0$.  Then
\[
X(\alpha) =: \begin{pmatrix} 1 + \alpha &   \cos(\frac{\theta}{3}) & x(\alpha) & \cos(\theta) \cr
\cos(\frac{\theta}{3}) & 1+\alpha  & \cos(\frac{\theta}{3}) & x(\alpha)  \cr
x(\alpha)  &  \cos(\frac{\theta}{3}) & 1+\alpha &  \cos(\frac{\theta}{3}) \cr
\cos(\theta) & x(\alpha)  &  \cos(\frac{\theta}{3}) & 1+\alpha \end{pmatrix}.
\]
Here $x(\alpha)$ denotes the value of the unspecified entry.  Using the symbolic package in MATLAB, we obtain
\[
x(\alpha) = \frac 12 \left( \sqrt{ \alpha(\alpha+2) + (4\cos^2(\frac{\theta}{3}) - 1)^2 } - (1+\alpha) \right).
\]
Taking the limit as $\alpha$ decreases to $0$, we get
\[
\bar{X} := \lim_{\alpha \searrow 0} X(\alpha) = \begin{pmatrix} 1 &   \cos(\frac{\theta}{3}) & \cos(\frac{2\theta}{3}) & \cos(\theta) \cr
\cos(\frac{\theta}{3}) & 1  & \cos(\frac{\theta}{3}) & \cos(\frac{2\theta}{3})  \cr
\cos(\frac{2\theta}{3})  &  \cos(\frac{\theta}{3}) & 1 &  \cos(\frac{\theta}{3}) \cr
\cos(\theta) & \cos(\frac{2\theta}{3})  &  \cos(\frac{\theta}{3}) & 1 \end{pmatrix}.
\]
This matrix has maximum rank over all positive semidefinite completions when $\alpha = 0$ due to Corollary~\ref{cor:sdp}.  Specifically, $\bar{X}$ has rank 2 whenever $\theta \in (0,\pi]$ and rank 1 when $\theta = 0$.
To derive the singularity degree of $\mathcal{F} $ we need to find the maximal rank of an exposing vector having the properties of Theorem~\ref{thm:fr}.  To this end let $Z(\alpha) := \alpha X(\alpha)^{-1}$ and let $\bar{Z} = \lim_{\alpha \searrow 0} Z(\alpha)$.  By Theorem~\ref{thm:Xalpha}, $\bar{Z}$ exists and is an exposing vector for $\mathcal{F} $ (as long as it is not the zero matrix) as in Theorem~\ref{thm:fr}.  By Proposition~\ref{Tinverse} we have
\begin{equation}
\label{eq:Zalpha}
Z(\alpha) =: \begin{pmatrix} a(\alpha) &   c(\alpha) & 0 & d(\alpha) \cr
c(\alpha) & b(\alpha)  & c(\alpha) & 0 \cr
0 &  c(\alpha) & b(\alpha) &  c(\alpha) \cr
d(\alpha) & 0 &  c(\alpha) & a(\alpha) \end{pmatrix},
\end{equation}
where $b(\alpha) = \frac{1}{a(\alpha)}(a(\alpha)^2 + c(\alpha)^2 - d(\alpha)^2)$.  Let $a,b,c,$ and $d$ be the limit points of $a(\alpha),b(\alpha),c(\alpha),$ and $d(\alpha)$ respectively, as $\alpha$ decreases to $0$.  Then
\[
\bar{Z} = \begin{pmatrix} a &   c & 0 & d \cr
c & b  & c & 0 \cr
0 &  c & b &  c \cr
d & 0 &  c & a \end{pmatrix}.
\]
We observe that if $b \ne 0$, then $\rank(\bar{Z}) \ge 2$ and if $b = 0$ then $\rank(\bar{Z}) \le 1$.  The first observation is trivial, while for the second observation, suppose $\bar{Z} \ne 0$ from which we get that $a  > 0$.  Then since $\bar{Z}$ is positive semidefinite, we have $c = 0$ and from the equation $b = \frac{1}{a}(a^2 + c^2 - d^2)$ we get
\[
0 = \frac{1}{a}(a^2 -d^2),
\]
which implies that $a = d$ and $\rank(\bar{Z})= 1$.  Now since $X(\alpha)Z(\alpha) = \alpha I$ we have
\[
X(\alpha) \begin{bmatrix}
c(\alpha) \\
b(\alpha) \\
c(\alpha) \\
0
\end{bmatrix} = \alpha \begin{bmatrix}
0 \\
\alpha \\
0 \\
0
\end{bmatrix}.
\]
Solving for $b(\alpha)$ we obtain the expression
\[
b(\alpha) = \begin{cases}
\frac{\alpha(\cos(\frac{\theta}{3}) + \cos(\theta))}{(1+\alpha)(\cos(\frac{\theta}{3}) + \cos(\theta))-2x(\alpha)\cos(\frac{\theta}{3})}, \quad &\theta \in [0,\pi], \ \theta \ne \frac{3\pi}{4}, \\
\frac{\alpha(1+\alpha+x(\alpha))}{(1+\alpha)(1+\alpha +x(\alpha)) - 2\cos^2(\theta)}, &\theta = \frac{3\pi}{4}.
\end{cases}
\]
Evaluating the limits we get that $b = 0$ if $\theta = \pi$ and $b$ is non-zero for all other values of $\theta$ in $[0,\pi]$.  It follows that $\rank(\bar{Z}) \ge 2$ when $\theta \in (0,\pi)$ and since $\rank(\bar{X}) =2$ for these values of $\theta$, we conclude that $\sd(\mathcal{F} ) = 1$ when $\theta \in (0,\pi)$.  When $\theta = 0$ it can be derived that $a = b= \frac 34$ and $c = d = -\frac 38$.  Then $\bar{Z}$ is a rank 3 matrix and $\sd(\mathcal{F} ) = 1$.  For the case $\theta = \pi$ we have that $\rank(\bar{Z}) \le 1$ and now we show that every exposing vector for $\mathcal{F} $ that lies in ${\mathbb S}^n_+  \cap \range( \mathcal{A} ^*)$ has rank at most 1.  Indeed, for $\theta = \pi$ we have
\[
\bar{X} = \begin{pmatrix} 1 &   \frac12 & - \frac12 & -1 \cr  \frac12 & 1  & \frac12 & - \frac12 \cr
- \frac12 &  \frac12 & 1 &  \frac12 \cr -1 & - \frac12 &  \frac12 & 1 \end{pmatrix} .
\]
Now a basis for the kernel of $\bar{X}$ is formed by the vectors
\[
v := \begin{bmatrix}
1 \\ 0 \\ 0 \\ -1
\end{bmatrix}, \ u := \begin{bmatrix}
1 \\ -1 \\ 1 \\0
\end{bmatrix}.
\]
Observe that $\range(\mathcal{A} ^*)$ consists of all the matrices with entries $(1,3)$ and $(2,4)$ identically $0$.  Now if $Z$ is any exposing vector for $\mathcal{F} $, we have $\bar{X}Z = 0$ and hence $Z = \lambda (vv^T) + \mu (uu^T)$ for some $\lambda, \mu \in {\mathbb R} $.  But since $uu^T \notin \range(\mathcal{A} ^*)$, it follows that $\rank(Z) \le 1$ and $\sd(\mathcal{F} ) \ge 2$.  We conclude this example by summarizing our observations:
\[
\sd(\mathcal{F} ) = \begin{cases}
1, \quad & \theta \in [0,\pi),\\
\ge 2, &\theta =\pi.
\end{cases}
\]
\end{ex}

Some of the observations of this example extend to general $n \ge 4$.  First we show that the partial matrix admits a unique positive semidefinite completion, which is Toeplitz.

\begin{prop}
\label{prop:cosn}
Consider the partial symmetric $n\times n$ Toeplitz matrix with pattern $P=\{ 1, n-1\}$ and data
\[
\{ t_0, t_1 , t_{n-1} \} = \{ 1, \cos \theta , \cos((n-1)\theta) \} ,
\]
where $\theta \le \frac{\pi}{n-1}$.
Then the unique positive semidefinite completion is
$$ \left( \cos \left((i-j)\theta \right) \right)_{i,j=0}^{n-1} = B^T B  , $$ where
$$ B= \begin{pmatrix} 1 & \cos \theta & \cos (2\theta )& \cdots & \cos ((n-1)\theta) \cr 0
& \sin\theta& \sin (2\theta ) & \cdots & \sin ((n-1)\theta) \end{pmatrix}. $$
\end{prop}

\begin{proof} Let us denote the first column of a positive semidefinite completion by $\begin{pmatrix} \cos \theta_0 & \cos \theta_1 &  \cos \theta_2 & \cdots &  \cos \theta_{n-1} \end{pmatrix}^T$, where $\theta_0 = 0 , \theta_1 =\theta$, $\theta_{n-1} = (n-1)\theta$ and $\theta_2, \ldots , \theta_{n-2} \in [0,\pi ]$. If we look at the principal submatrix in rows and columns 1, $n-1$ and $n$, we get the positive semidefinite submatrix
$$ \begin{pmatrix} 1 & \cos (\theta_{n-2} ) & \cos((n-1)\theta ) \cr \cos( \theta_{n-2} ) & 1 & \cos (\theta ) \cr
 \cos((n-1)\theta )  & \cos(\theta) & 1 \end{pmatrix} . $$ \cite[Proposition 2]{MR1236734} yields that $ (n-1)\theta \le \theta_{n-2} +\theta$. Thus
 \begin{equation}\label{co} \theta_{n-2} \ge (n-2)\theta . \end{equation}
Next, consider the $(n-1)\times (n-1)$ upper left corner with data
\[
\{ t_0, t_1 , t_{n-2} \} =\{ 1, \cos \theta , \cos \theta_{n-2} \}.
\]
By \cite[Corollary 2]{MR1236734} we have that
\begin{equation}\label{cor2} 2 \max \{ \theta_{n-2} , \theta \} \le  (n-2) \theta + \theta_{n-2}. \end{equation}
This implies that \begin{equation}\label{cor2a}\theta_{n-2} \le (n-2)\theta. \end{equation} Combining this with
\eqref{co} we have $\theta_{n-2} = (n-2)\theta$. If instead we looked at the principal submatrix in rows and columns 1,2, and $m$ and combine it with the $(n-1)\times (n-1)$ lower right corner, we obtain that also in the $(n,2)$th position we necessarily have $\cos ((n-2)\theta )$. Thus along the $(n-2)$th diagonal the value is $\cos ((n-2)\theta )$.

One can repeat this argument for smaller matrices (or invoke induction) and obtain that in the $k$th diagonal necessarily all entries equal $\cos \theta_k=\cos (k\theta )$, $k=2,\ldots , n-2$.
\end{proof}

Now we show that in case $\theta = \frac{\pi}{n-1}$, we have $\rank(\bar{X}) + \rank(\bar{Z}) \le 3 < n$ for all $n \ge 4$.

\begin{ex}
\label{ex:generaln}
Let $n\ge 4$ and consider the $n\times n$ symmetric partial Toeplitz matrix with pattern $P=\{ 1, n-1\}$ and data $\{ t_0, t_1 , t_{n-1} \} = \{ 1+\alpha , \cos (\frac{\pi}{n-1}) , -1 \}$ where $\alpha > 0$.  As in Example~\ref{ex:4n}, we let $\mathcal{F} $ denote the set of positive semidefinite completions when $\alpha = 0$ and we let $X(\alpha)$ denote the maximum determinant completion when $\alpha > 0$.  By Proposition~\ref{prop:cosn}, $\mathcal{F} $ is the rank $2$ matrix
\[
\bar{X} := \left( \cos \left(\frac{(i-j)\pi}{n-1}\right) \right)_{i,j=0}^{n-1},
\]
and by Theorem~\ref{thm:Xalpha}, $\bar{X} = \lim_{\alpha \searrow 0} X(\alpha)$.  If $Z(\alpha) = \alpha X(\alpha)^{-1}$ and $\bar{Z}$ is the limit of $Z(\alpha)$ as $\alpha$ decreases to $0$, we show that $\rank(\bar{Z}) \le 1$.  Let $a,b,c,$ and $d$ be the limit points of $a(\alpha),b(\alpha),c(\alpha),$ and $d(\alpha)$ respectively.  By Proposition~\ref{Tinverse}, $Z(\alpha)$ is as in \eqref{eq:Zalpha}.  We claim that if $a =0$ then $\bar{Z} = 0$.  Indeed, by the fact that $\bar{Z}$ is positive semidefinite we have $c=d=0$.  Moreover,
\[
0 = {\rm tr} (\bar{X}\bar{Z} ) = (n-2)b,
\]
which implies that $b = 0$ and consequently $\bar{Z} = 0$.  Thus we may assume $a>0$ and the equation
\[
b = \frac 1a (a^2 + c^2 - d^2),
\]
holds.  From $ \bar{X} \bar{Z} = 0$ and the above equation, we obtain
$$ 2\cos \left(\frac{\pi}{n-1}\right)c + b = 0 ,\ a + \cos \left(\frac{\pi}{n-1}\right)c -d = 0.$$
This gives $c=-b/(2\cos(\frac{\pi}{n-1})), \ d=a -\frac{b}{2}$, and thus
$$ b=a + \frac{1}{a} \frac{b^2}{4\cos^2(\frac{\pi}{n-1})} -\frac{1}{a} \left(a-\frac{b}{2}\right)^2.$$
After rearranging, we obtain
$$ \frac{b^2}{a} \left(\frac14 - \frac{1}{4\cos^2(\frac{\pi}{n-1})} \right)=0 . $$
Consequently $b=0$, and $\rank( \bar{Z}) \le 1$ follows.

Numerical experiments suggest that $\bar{Z}$ is the rank 1 matrix with $(\bar{Z})_{11}=(\bar{Z})_{nn}=(\bar{Z})_{1n}=(\bar{Z})_{n1}= \frac{n-1}{4}$ and all other entries equal to 0.

\end{ex}

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redpajama_set_name: RedPajamaArXiv


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The other is wealth creation. When I started, I did not possess almost any starting capital – so I decided to focus on the latter for the most part.
At the time, my thinking was that even if I can have a starting capital of say 10 000 EUR for capital multiplication approach, and can invest into all sorts of assets to get a pretty good return on investment of 7% yearly, and if I add 10 000 EUR to my investment portfolio every year, then with compound interest and 30 years timeline – I would get about one million EUR in my bank account and a decent yearly income to boot.
Which is great, but I would probably have a lower quality of life while saving all that money, but a much higher one later down the road. Since I prefer to live in the NOW, I decided that this was just not my path to wealth. Besides, one million might sound like a lot, but to me – it is in the same category as 10 000 EUR, so I would instead aim for 100 000 000 right away.
As a result, I want to focus on wealth creation – that is to work on things that have a higher risk, lower cost of entry and a potential "black swan size" reward. That is why I have been starting companies, and getting equity in other companies and not investing/saving as much. That is also why I have been focusing on generating more income from more sources in addition to salary: Consulting, Mentoring, Rental, etc.
In a sense, I am betting that one of my businesses or equity holdings will bring more than one million within 30 years and that I will get to use the money that I make now to have a relatively higher quality of life, while at the same time growing my monthly income.
That said, I still invest into capital multiplication nowadays, as there have been some good results from my "wealth creation" approach, and I can put some small amounts of money away to invest on a monthly basis. For instance, from any card transaction that I do, I move a fraction of the value into my savings account and then invest that monthly.
1. Go where nobody goes, and do the opposite of what everyone is doing.
Market crashing – buy. Market growing – sell. That might fall into the "basics" category for you, but still – most of us fail to do it, and I have failed to do that many times as well. Still, even when it comes to startups – I like the ones nobody else notices at first. The ones that do not go to startup events for example.
2. Learn the basic concepts of investment.
For that, I would recommend to read about Warren Buffet & Fooled By Randomness.
3. Generate passive income sources.
In addition to my primary income source, I try to find ways to have alternative income sources, and hopefully some of them in the passive category. Even if it is not a lot of money, to begin with – even a 5 EUR monthly passive income is impressive. Especially when compounded. There are a lot of sources of passive income out there. Most of them require your time in the beginning.
4. Invest with your time.
Again, as I chose to focus on wealth creation mostly – I invest with my time, my knowledge, and my network to get both equity and financial returns. Starting a business would also go into this category. When that works out, you can start investing your capital, which I now also do.
5. Find your path to wealth and money.
Everyone is different, and I think it is essential to figure out who your "real you" is and then base your decisions on that. If you save money, would you feel like you are losing out on life or will you be okay with it? If you invest into startups, would you think its too much of a risk? Etc.


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La giada giadeite è un minerale il cui nome deriva dallo spagnolo "pedra de ijada", ossia pietra del fianco, dato il suo presunto potere benefico sui lombi e sui reni.

La giadeite (che, assieme alla nefrite, viene comunemente chiamata giada) fa parte dei pirosseni alcalini assieme all'egirina, con cui forma soluzioni solide. È un inosilicato importante in ambito metamorfico perché, se reagisce con acqua, dà il glaucofane, che assieme al granato forma le eclogiti. Un'altra reazione che coinvolge la giadeite, sempre in ambito metamorfico, è quella con il quarzo a dare il feldspato plagioclasio albite.

Il colore della giadeite presenta toni bianchi e verdi del tutto particolari, uniformi o a macchie, con lucentezza grassa e aspetto traslucido che la rendono particolarmente apprezzata dai popoli orientali. Si rinviene soprattutto nelle aree alluvionali della Birmania.

Abito cristallino

Origine e giacitura
Zona pian del Re Monviso valle Po CN

Forma in cui si presenta in natura

Altri progetti

Collegamenti esterni

Inosilicati