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charles stuart II
17th Century Survival Tips for a Hysterical Age
August 29, 2021 September 5, 2021 / geotrickster	/ 6 Comments
Running themes on this site are historical trickster figures, explorations of books on relevant subject matters in detail, and past parallels to present challenges. Here, I will bring you all three in addition to some original 'artwork' from myself at the end.
Despite being primarily interested in other eras and parts of the world, it should be obvious to regular readers that I have developed a recent fixation on 17th Century Europe and in particular Britain. This is not because it conflates with most of my actual historical interests, as it mostly does not, but because it is the time that is so culturally similar to our own and thus demands closer examination. Those with little to no historical knowledge have a tendency to reach for over-used and often ill-fitting periods, such as the Great Depression and World War II, but the world we live in looks nothing like the Inter-War era in actual substance. This is merely hyperbolic rhetoric from neoliberals who have no comparison point to the fairly regular occurrence of localism re-asserting itself against internationalism.
Today does, however, look a lot like Europe of the Thirty Years War and the Britain of the era of its civil wars. Indeed, the ideologies and struggles of that time plague us still. Our present era, I would argue, is a very Cromwellian one. For those of us who oppose this and find it the potential start of a new dark age, it becomes relevant to familiarize ourselves with how this happened before and how such a time was overcome and displaced. Having already dealt with the lessons that can be learned from the Thirty Years War before, I now wish to move towards the British origin point of so much of present ideological pathologies.
Since the analogy is obviously imperfect, (there is no conventional war yet, for one thing) it should be understood that I am more focusing on the cultural and philosophical life of political society rather than claiming an exact parallel in events. Nevertheless, you may find yourself surprised by the overlaps between then and now. Wokeness, Christian evangelism, universalist liberalism, creationism, and many of the other afflictions of the Anglophone world were born or revived in this time. And now, as the cultural dominance of that world begins to recede in our present era, it comes forth once again with full force and with a cacophonous death rattle…knowing the time to remake the world in its image has come to a close but seeking one last great push.
The Commonwealth and Protectorate's Messianic Endeavor
'If He that strengthens your servants to fight, pleases to give your hearts to set upon these things, in order to His glory, and the glory of your Commonwealth, besides the benefit of England shall feel thereby, you shall shine forth to other nations, who shall emulate the glory of such a pattern, and through the power of God turn into the like.' ~Oliver Cromwell
(an image so good I had to use it twice)
The British Civil Wars began in Scotland and ended in Ireland, though they are often erroneously called 'The English Civil War'. In the end it would be England dictating the peace for the others. The union of the crowns that had begun with James VI of Scotland becoming James I of England upon Elizabeth I's death had finally brought inter-state warfare on the British Isles to a close, but sectarian and domestic political struggles would tear the country apart under his inept successor, Charles I's rule. Parliament would emerge victorious in the resulting civil war, and then eject various other groups from power in Ireland and Scotland. Charles I would be executed, his family driven into exile, and an attempt to set up a republic would ensue. Cromwell himself would end up shutting down parliament and ruling as a dictatorial "Lord Protector" not long after this.
Paul Lay's 'Providence Lost: The Rise and Fall of the English Republic' is an Anglo-centric yet nevertheless engaging read about the state that existed between the fall of Charles I and the restoration of Charles II. He describes a state that began with so much experimental promise but descended into factionalism and moralistic hypochondria. Something akin to if the American Revolution had been immediately co-opted by an alliance of Cotton Mathers and Tipper Gores right after the Treaty of Paris.
Sadly, this could not have been a surprise. Even before the war was over it was soon apparent that, demographically, the Parliamentarians were far more puritan than 'leveller' (the term given to people who wanted a universal male franchise). In a situation that should strike familiarity with anyone who has followed the Syrian Civil War, what looks on the surface like a noble cause can in fact be nothing but a sieve for fanatics and sectarians. People throughout Britain soon learned this as a government that was supposedly committed to freedom of religion began to persecute anyone not clearly of the Puritan ilk, including former allies of theirs like the Quakers.
Messianic regimes, especially new ones, cannot justify themselves without outward expansion. And so, the powerful military edifice built to win the Civil War would be turned onto Spain. In particular, its enormous New World empire. But the invasion of Hispaniola ended disastrously amidst tropical disease and local Spanish soldiers who knew the terrain. As a consolation prize the defeated English swept into barely-defended Jamaica. It would be their only gain from an expedition with dreams of driving the Whore of Babylon out of the New World and introducing a new Protestant reign for Central America.
In a pattern all too familiar to moderns, failure abroad led to a bizarre rise of extremism at home. Cromwell entered a kind of existential crisis. He had not failed in such a way. Surely, it must have been the nation itself that had yet to repent for its wickedness. And so, loyal generals were appointed as satraps throughout the country with explicit instructions to crack down on irreligion, drinking, the arts (especially theater) and even folk festivals. A life simmering within unadorned churches would be the only publicly sanctioned form of culture for the masses. It was this that made the people turn against the government in large numbers. But living in a literal garrison state, there was nothing they could do but grumble. Lay has a particular section that describes the goals of this society which is designed to strike us today:
'The concept of a tirelessly interventionist and inescapable God might be compared to social media, resulting in comparable levels of anxiety and paranoia. Facebook, Twitter, Instagram are realms of round-the-clock surveillance, where one's thoughts and actions, beliefs and appearance are posted and preserved for all to see and subjected to constant comparison and judgement. The shame, vindictiveness, and piety that social media generates would not have felt out of place among seventeenth Century Puritans. But one can opt out of social media, however addictive. There was no such option in the world God had created, nor in the next.'
The various sects denounced each other constantly. To borrow a phrase from Angela Nagle, 'competing over a scarcity of virtue' in order to prove who was the most humble and the most ideological pure. Cromwell himself cared only that people were Protestant, but the forces and style of governance he had unleashed catered to only the worst and most extreme of ideologues. Fortunately, this government did not survive Cromwell. His ineffectual failson and chosen successor fled the country as the disputes between Generals and ex-Parliamentarians threatened to tear the islands apart again. But no one wanted a sequel to full blown civil war. A compromise was reached, if the exiled son of the late king would sign on to acknowledging the existence of Parliament in government, he would be invited back to bring the country together and avoid calamity.
The Restoration Undoes the Era of Hysteria
'The King spent most of his time with confident young men, who abhorred all discourse that was serious, and, in the liberty, they assumed in drollery and raillery, preserved no reverence towards God or man, but laughed at all sober men, and even at religion itself.' ~James Butler, Duke of Ormond.
'He spends all his days
In running in plays
When in his Shop he should be poreng;
And wastes all his Nights
In constant delights
Of Reveling, Drinking, and Whoreing.
~Anon, 'Upon His Majesties' Being Made Free of the Citty.'
One would have expected had he lived the life planned for him King Charles II would have been a diffident, if witty, failson not unlike Richard Cromwell. At least he wasn't that other and most cursed Charles II. The problem with monarchy is the sheer sheltered entitlement it breeds in those growing up expecting to inherit it. Due to the Civil War, however, Charles did not have this luxury. He fled the country, tried to rally support in Scotland only to be held hostage by the fanatic Covenanters and forced by Archibald Campbell, their de facto leader, to sign away much of his powers to them. Then, Cromwell had defeated the Scots (largely due to religious fanatics firing their most experienced commanders for 'drinking and whoring' on the eve of the Battle of Dunbar). Charles had to flee again. Crossing much of the country in disguise as a commoner and having many close calls, the heir to the monarchy slept outside, hid in trees, and developed a knack for socially integrating himself with common people he otherwise would not have. Once he made it out of the country, he would end up living an impoverished yet interesting young adulthood in the Netherlands, France, and Spain. Largely existing as leech on related aristocratic families in those countries.
When he returned to England he did so to a totally changed country. But not more changed than himself. In her book 'A Gambling Man: Charles II's Restoration Game', Jenny Uglow documents how Charles' unconventional and roguish new skill set served him well to meet this particular moment as the restorer of the monarchy in England, Scotland, and Ireland.
Charles II had a victory procession of sorts, but it was not simply pomp. As he disembarked in the country that had once tried to kill him, he was met by crowds of people relieved that there would be no civil war or continuation of Puritan rule. Formerly powerful general submitted to his authority, and he used his political capitol to disband the radical-infested army. This not only removed his most dangerous enemies, but also freed up immense amount of finance for the state. The regicide parliamentarians were hunted down for execution or driven into exile. Archibald Campbell found himself publicly executed in Edinburgh, and Cromwell's body was disinterred for a mock execution and display.
But what followed this score-settling would be even more interesting. Charles was both a monarch who liked to throw massive court parties, indulge in the arts (which he of course re-legalized and came to patron) but also was an accessible 'man of the people.' He was at ease with people of all backgrounds and often struck up conversations with random people he met on the street when walking his dogs (of the breed now named for him). Rumors that he 'rolled from whore to whore' incognito as a commoner abounded and seemed to actually increase many people's affection for him after the dour Cromwellian cultural revolution. When the Great Fire of London threatened to engulf the entire city, he not only led the response in an official capacity, but was smeared in ash and smoke, working the firebreaks and hauling buckets of water with the crowd at the worst of it. I can personally attest from years living in multiple locations in the United Kingdom that there are many pubs named after him to this day. And, of course, there is that Horrible Histories song.
But Charles was not simply a people's partier. He was actually a fairly competent monarch in his own right. His diplomacy showed immense flexibility and his limited naval wars, though often considered indecisive or even losses by conventional historians, did encapsulate his move towards international trade access and naval power. The gains made in this conflict would, in fact, end up with the acquisition of New York and New Jersey, unifying the English colonies in North America into one band of coast-the first springboard for a future great power Britain. These pickups were made possible by the money he made selling indefensible colonial outposts in other places like Tangier and Calais. A keen eye for geography, and the centrality of the offshore stance in Europe but expansion elsewhere, showed the way of the future for an island nation. People back then didn't know it yet, but Charles II's reign would lay the seeds of Britain's future at the expense of its (then) more powerful rivals in France, Spain, and the Netherlands.
Charles would make some major errors too, most importantly designating his thick-headed brother as his official successor knowing it would cause another sectarian crisis. Some people have said this sympathy for a Catholic brother stemmed from Charles' own secret conversion. But Charles, it seems to me, cared little for religion and made this conversion to gain war subsidies from Louis XIV of France. After all, he did get the money and didn't even convert until he was on his death bed. He never ended up converting the court, meaning he got one over on his superpower cousin. But the point here is not to say Charles II was a perfect ruler to whom we should aspire, but rather that he was a cultural force. The right counterbalance at the right time. He singlehandedly ended Britain's first Woke-Evangelical Era not with frothing reactionary policies, but through levity, pragmatism, and disdain for all kinds of cultural extremes. In so doing, a hot mess of a country prone to regular bouts of rebellion and sectarian strife began to transform itself into a future financial and industrial powerhouse.
For a time anyway. All gains are, after all, temporary. Something the Puritan can never understand.
Accepting the Hobbesian Bargain
'The obligations of the subjects to a sovereign is understood to last as long and no longer, than the power lasteth, by which he is able to protect them. For the right men have by nature to protect themselves, when none else can protect them…the end of obedience is protection.' ~Thomas Hobbes
'The losers are the real victors. The victims are the real winners.' This was the sentiment of puritanism, and it is the dominant sentiment in the Anglophone world today. So much of our present-day culture war (which was declared by and waged in service of the right before it became the lefty cause du jour of the present moment) is an all-pervasive and multi-ideological trend. But it is a trend that can be defeated. This can be accomplished by the marriage of two things often not thought of as partners: the marriage of state power and the levity of humor.
Anyone who has ever interacted with ideological cliques such as anarchists knows that it is often the people who fear the state who are the most authoritarian and censorious people imaginable. Radical cliques often degenerate into cults where people psychologically abuse each other for clout and differences between people are not tolerated. Individualism is ineffective for every cause, so when one does not believe in the arbitration of the state, one must create a sect to compensate. The sect, ironically, often tolerates less dissent and divergence than does the state. This is because all they have is ideology, whereas the secular state (whatever form it takes) is a more situational and territorial arbiter. Its concerns (when it is working anyway) is to maintain the peace over its sovereign location and to maximize its autonomy vis-à-vis other states. This is true for all states and state-like entities no matter their internal ideological and traditional structure. Though states that forget this are very likely to degrade the sanity and effectiveness of their governing class and become more like those sectarian cults that spring up in their absence.
Let us return once more to the 17th Century. Thomas Hobbes was an intellectual and instructor who had royalist connections. He missed the civil war due to his job as a private tutor having taken him to France beforehand. When Charles II was in exile in France, Hobbes became his personal instructor. It was at this point that he published his most famous work, Leviathan. Leviathan's blatantly irreligious, pragmatist, and materialist nature would cause scandal in the Stuart court-in-exile…despite the fact that it made an implicit argument for the Stuart style of governance. Fearing retaliation from religious cavaliers, he fled to Cromwell's Protectorate. He reasoned, rightly as it turned out, none of the members of that government had yet read his works. He also made it clear that the necessity of government he wrote about could apply to any form of statecraft. Sovereignty was not held by divine right, but by power over the land and the execution of the prerogatives of the state itself.
When Charles was restored, he invited Hobbes to enter the court. It was there that the already old man, known today as a dour sourpuss due to the nature of his thought, made himself indispensable through his wit, jokes, and ability to disregard superstition and religious dogma (Hobbes himself was almost certainly an atheist in private). This is when people really began reading him.
Hobbes' political thought lacks the subtlety of Han Feizi or even Confucius. In his concept of the mediating sovereign which protects individuals and groups from each other, he is far too supportive of the idea that the subject must support the sovereign no matter what-so long as their security needs are met. He wallows in constant fear of rebellion for obvious reasons given the times he lived in, but the long view of history shows plenty of rebellions that replace an inferior sovereign with a superior one. He does not grapple with the problem, innate to his thinking, of sovereign capriciousness from one head of state to another upon succession and which is particularly common in monarchies.
All of this being said, Hobbes is worth engaging with as his primary observation, that society can only thrive under conditions of sovereignty where a state is the primary mediating influence between actors, is correct for any society larger than that of the tribe. It is also, though this was not Hobbes' intent, a better model of achieving freedom of conscience and securing the ability of divergent people to live with each other than more ideologically motivated models of conversion. Leviathan, it turns out, is a better guardian of private liberty than even the ideal of private liberty itself. Just ask any non-Islamist and non-liberal Syrian today, especially if they come from a minority group.
In his book 'The Two Faces of Liberalism', John Gray examines this lost liberationist aspect of Hobbes. Most useful to us today, he makes a case that the best of liberal values can be saved only by rejecting the worst of them. Specifically, the freedom to live one's life as they please in the cultural and lifestylist sense by sacrificing liberalism's tendency towards universalism and messianic behavior. These two impulses which are endemic in the philosophy are at war with each other, because universalism cannot abide competitors and those who opt out of it, and, on the other side, divergence requires a morally neutral pragmatist state to balance interests without adopting a mission of its own aside from the survival and maintenance of the state itself. This restricts communal projects to the realm of necessary material needs for a community like security and infrastructure.
Whether we like it or not, we live in a very liberal society. And so, to convince those in power to change policies, we must all be liberals to some degree. Gray's reappropriation of Hobbes is a way to do that which makes the messianic culture war obsolete. Indeed, Gray admires Hobbes' Leviathan as a model which could 'Extend the benevolence of indifference' to issues of private lifestylist and social spheres so long as the political order that upholds this indifference is not challenged by the subject. He points out that such arrangements were the norm in the ancient and classical worlds, before messianic religions took hold, and are often still the norm in places like East Asia, were they never came to be powerful at all. They also tend to exist in early modern states like the height of the Ottoman Empire, and, most obviously and perhaps at its greatest extent, in medieval states like that of the Mongols. For a modern example, he cites Singapore as a country that guarantees freedom of religion but bans missionary activity. The liberals did not invent toleration, they merely invented a form which was a successor to the Christian world it was rebelling against. But as such, this toleration inherited many preexisting problems.
Despite liberalism being the pervasive default setting in the Anglosphere, but not in these other examples, Gray wishes to learn from such arrangements as ways to have a collective civically minded state that does not engage in enforcing ideological or cultural uniformity but still maintains a civic unity. I contend that, in order to appeal internally to this Anglosphere's tradition and common historical experience, that the reign of Charles II serves as a potential in-house model for such an arrangement. Not because I am a monarchist (I am definitely not) but because it came from a similar age of hysteria and ended up dissolving many of the problems it inherited. While Charles II is a bit too establishment to fit my mold of previous historical trickster figures, he had a similar personality as those past examples and thus can cross the bridge of communicating these issues between more outsider and insider persuasions. This, naturally, applies to the international system as well as the domestic. Whatever ways we find around our present impasse will differ from the solutions of the past, but we can certainly learn from events that preceded us nevertheless.
Modus Vivendi, as Gray calls his proposal, is not only the acknowledgement that no one way of governance can work for everyone, but that the very idea of political hegemony through one ideology is a potential declaration of war upon much of the domestic population of a state and thus cannot exist in a world where pluralism is the natural state of things. Two Faces of Liberalism is a short book and worth reading in its entirety, so I won't mass quote it here, but there are two passages in particular I wish to conclude with:
'Modus Vivendi expresses the belief that there are many forms of life in which humans can thrive. Among these are some whose worth cannot be compared. Where such ways of life rivals, there is no one of them that is best. People who belong to different ways of life need have no disagreement. They may simply be different. Modus Vivendi is liberal toleration adapted to the historical fact of pluralism.'
'…When liberals set up one regime as a standard of legitimacy for all the rest, pluralists and liberals part company. For pluralists, a liberal regime may sometimes be the best framework for modus vivendi. At other times a non-liberal regime may do as well or better.'
Context reigns. Acknowledging that means there is something we can learn from the experiences of all types of governments. When the chips are down I consider myself more in favor of republics than monarchies, but should I therefore dismiss the experiences of all monarchs or all kingly states? No. Just as it is no great scandal to learn strategy from thinkers of all backgrounds why not also governments? It is this intellectual flexibility that keeps us from falling into the farce of Manichean culture war. That and the right kind of King Charles style levity that acknowledges that while running the state might be serious business, there is no reason it has to be too serious. Platonic absolutes do us no favors here. And those who are interested in working out the practical can do so with anyone else, regardless of that other person's inner life. Speaking from personal experience, I can say that the only times I take part in culture war is defensively or where there is a codified legal imbalance that needs to be adjusted. If people do not seek to inflict their preferences on me then I have no need to do the same to them. But I have never been so insecure as to seek to convert others save on issues of real and pressing policy that affects the entire state. Those are the true structural issues that affect everyone-economic, foreign, and infrastructure policy. And those are the issues that supermajorities of people could, theoretically anyway, come together much more effectively if they were prioritized over the social. Obviously, those content with the status quo therefore have a vested interested in fueling rather than dousing the culture war and its attendant surveillance and cancelling network. But bad policies of the state cannot be challenged by disregarding the state itself, but by coopting or replacing it with another state.
Thoughts to meditate on while regarding the Wrathful Party King Charles Bodhisattva.
And if for some reason you do think it would be nice to have a monarch once again with this personality type, may I recommend the fashionable and party boat owning King of Morocco?


text:
package com.mikephil.charting.utils;

/**
 * Point encapsulating two double values.
 *
 * @author Philipp Jahoda
 */
public class PointD {

    public double x;
    public double y;

    public PointD(double x, double y) {
        this.x = x;
        this.y = y;
    }

    /**
     * returns a string representation of the object
     */
    public String toString() {
        return "PointD, x: " + x + ", y: " + y;
    }
}

redpajama_set_name: RedPajamaGithub


text:
DataSnapshots persistence
=================
Enables data snapshots persistence

F.Cerbelle

##Requirements

##Instructions


Every year the president of the U.S. gives a status update to Congress and the American people with a State of the Union address—but now we shift the focus to the humanities nationwide and in particular, writing at Washington College.
WC's literary tradition is evident across the board, from a curriculum that stresses writing with required writing intensive courses and first-year courses like English 101 and GRW to the esteemed Sophie Kerr prize. Programs for writers abound with offerings including the Poetry Club, "The Collegian," the Kiplin Hall trip, "The Pegasus," internships and more.
With only so many hours in a day, students make decisions, leading some student-run organizations to suffer more so than others. The Writers' Union, for example, is on hiatus according to Dr. James Hall, associate professor of English. He said that they plan to recharge next semester.
Humanities as a whole—including disciplines like philosophy, English, and history—have seen a nationwide decline thanks to the great STEM (science, technology, engineering, and math) push. In 2014, the humanities hit a record low in America, making up only 6.1 percent of all bachelor's degrees according to Humanities Indicators.
"There's a lot of pressure to figure out what you're going to do after college. The STEM sciences look more like sure bets," Dr. Moncrief said.
English has not seen a significant decrease at WC in fact the creative writing major remains the College's most popular one by nearly double the runner-up. In his 40 years at WC, Dr. Richard De Prospo, professor of English and American studies, has only seen a surge in quality among student writers. "The quality of student writing has improved, but the problem may be there's a war against the humanities now," he said.
There's a stereotype that English majors will not see the immediate economic success of their math and science peers. Although there is a gap, according to "Inside Higher Ed," it is not as significant as one may think. Those with a bachelor's degree in the humanities see about a 5.4 percent rate of unemployment following graduation, compared to 4.6 percent for other degrees. This figure drops to 3.1 percent, lower than the 3.4 percent for other degrees, with advanced degrees. Salaries for humanities graduates are about $7,000 lower than those with another degree but are still significantly higher than the median salary levels for all Americans.
In the words of President Barack Obama and many presidents before him, "The state of our Union is strong." Many agree the same could be said about WC's status as a writer's college, but the state of humanities nationwide is largely uncertain.


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Murali Sreeshankar (born 27 March 1999) is an Indian athlete who competes in the long jump event. He holds the national record of 8.36 metres set  in 2022.

Career
In March 2018, Sreeshankar cleared a 7.99 m jump at the Federation Cup in Patiala. He was named in the Indian contingent for the 2018 Commonwealth Games but had to pull out 10 days before the April event after being diagnosed with appendicitis. Following an emergency appendix surgery, he was placed on a liquid diet, which resulted in significant weight loss and an inability to walk properly. Two months later, he participated in the 2018 Asian Junior Athletics Championships in Gifu despite having "less strength, speed and focus" from before the illness and won bronze with a jump of 7.47 metres. At the 2018 Asian Games in Jakarta, he "struggled with run-up issues" and finished sixth in the final with 7.95 metres.

In September 2018, Sreeshankar broke the national record at the National Open Athletics Championships in Bhubaneswar where he achieved a jump of 8.20 metres. It was also the world leading jump of the season among under-20 athletes and made him the first Indian athlete to qualify for the 2019 World Athletics Championships scheduled to be held in September–October in Doha. At the World Championships, Sreeshankar failed to qualify for the finals, achieving a best leap of 7.62m, with the qualification mark set at 8.15.

Sreeshankar qualified for the 2020 Summer Olympics by recording a jump of 8.26m, a new national record, at the Federation Cup in Patiala in March 2021. At the Olympics, he registered a jump of 7.69m in the qualifying round and failed to enter the final.

At the 2022 World Athletics Championships, he qualified for the final round and finished seventh with a jump of 7.96 m. At the 2022 Commonwealth Games, he won the silver medal with a jump of 8.08 metres and became the first male long jumper from India to win a silver medal at the Games.

Personal life
Sreeshankar is coached by his father S. Murali who is a former triple jump athlete and silver medalist at the South Asian Games. Sreeshankar would accompany his father at practices as a four-year-old, when his father noticed his potential as a sprinter. Sreeshankar became a state-level under-10 champion in 50 metres and 100 metres, but switched from sprinting to long jump at the age of 13. His mother K. S. Bijimol had won silver medal in 800 metres at the 1992 Asian Junior Athletics Championships. His sister Sreeparvathy is a heptathlete. As of August 2019, Sreeshankar is pursuing BSc in mathematics at the Government Victoria College, Palakkad.

Sreeshankar is part of the Target Olympic Podium (TOP) scheme started by the Ministry of Youth Affairs and Sports. He is sponsored by JSW Sports and supported by JSW's Inspire Institute of Sport in Bellary district, Karnataka.

References

External links

1999 births
Living people
Indian male long jumpers
Athletes from Palakkad
Athletes (track and field) at the 2018 Asian Games
Asian Games competitors for India
Athletes (track and field) at the 2020 Summer Olympics
Olympic athletes of India
Athletes (track and field) at the 2022 Commonwealth Games
Commonwealth Games silver medallists for India
Commonwealth Games medallists in athletics
21st-century Indian people
Medallists at the 2022 Commonwealth Games

redpajama_set_name: RedPajamaWikipedia


Linnea Ericsson Carey with Caddie R. Photo (c) Jenny Abrahamsson.
In September, Caddie R (Come Back II x Lucky Light) was sold from Denmark's Linnea Ericsson-Carey as well as her husband Shane Carey to Belgium. Euro Horse-rider Jos Verlooy took over the reins on the 11-year-old stallion, and after some recent successful weeks showing Caddie R in the States the horse has been sold on.
It will be Jennifer Gates who takes over the ride on Caddie R, that has jumped up to five-star level.


Tune in on Thursday, October 30th as we dedicate 24 hours of programming to legendary trumpeter Clifford "Brownie" Brown in celebration of the 84th anniversary of his birth. In the course of only four years, from 1952 to 1956, Brown led and contributed to some of the greatest ensembles and recordings in jazz history, leaving his imprint on the genres of bebop and hard bop, as well as the art of jazz trumpet.
Born in 1930 in Delaware, Brown drew much of his influence from the innovative Fats Navarro, who mentored Brown in his early years in Philadelphia. Much like Navarro, Brown became quickly known for his signature warm, round tone, his tasteful improvisation, and his distinct bebop sensibility. By the early 1950s, Brown was performing with the likes of Tadd Dameron, Lionel Hampton, and Art Blakey, among others. Soon, Brown had formed a quintet with drummer Max Roach, a band now renowned as one of, if not the premiere hard bop bands of the time. Tragically, in 1956, Brown died at the young age of 25 in a car accident with pianist Richie Powell and Powell's wife Nancy. Join us at WKCR this Thursday as we honor the short-lived, but incredibly significant musical career of Clifford Brown with 24 hours of uninterrupted bop!


Use your Teleportr address while shopping online.
Tracked and insured, of course.
… and we spring into action!
The Teleportr app automatically informs our delivery team, without you lifting a finger.
until your parcel is in your hands – for free.
Apps that use your location drain your battery? That doesn't have to be!
Do we know where you are the whole time? No, we don't (and don't want to)!


The heads light on your vehicle are essential to safety when driving. Other than blown bulbs they have very few issues. However one thing that does happen over time is corrosion due to UV. What happens is the UV breaks down the outer layer of the plastic lens causing the head light lens to turn dull and cloudy. This decreases the light intensity projected onto the road by the head lights, also it ages the appearance of the vehicle.
To rectify the damage due to the UV Pickards uses a three step process to restore the lens to like new condition.
Step 1: Sand the outer layer of the lens, gradually increasing the grade of the paper used, making sure the lens is kept wet.
Step 2: Cut and polish the lens using two grades of polish.
Step 3: Adding a head light lens protestant to decrease the effect of UV on the plastic lens.
I find that after headlight restoration the vehicle looks significantly better, giving the vehicle a much cleaner look and alot of customer satisfaction.
So when your car is next booked in at Pickards Automotive ask us about having your head lights restored.


text:
A winged helmet is a helmet decorated with wings, usually on both sides. Ancient depictions of the god Hermes, Mercury and of Roma depict them wearing winged helmets, and in the 19th century the winged helmet became widely used to depict the Celts. It was also used in romantic illustrations of legendary Norse gods and heroes. The motif, along with the horned helmet, became a clichéd signifier of the Northern warrior.

Historical evidence

There is some limited evidence of such decorative motifs being used on actual helmets in the ancient world, but these may have functioned as ceremonial rather than functional objects. Attic helmets decorated with wings of sheet bronze were worn by the Samnites and other Italic peoples before their conquest by Rome. A number of such helmets have been excavated and can be seen in various museums.

Helmets decorated with animal motifs, no doubt including wings, were described by Diodorus Siculus as being worn by Celts:

On their heads they wear bronze helmets which possess large projecting figures lending the appearance of enormous stature to the wearer. In some cases, horns form one part with the helmet, while in other cases it is relief figures of the foreparts of birds or quadrupeds.

An actual example of this type of Celtic helmet was discovered in Romania, dating to the third century BC: it has a high-mounted crest formed as a bird, possibly an eagle or a raven, with large wings spread out to either side. The crest was cunningly made, the wings articulated at the body so that they would have flapped up and down as the wearer moved. Today this kind of helmet is commonly believed to have been worn by the Celts. The Celts, however, mostly wore plainer helmets of conical or sub-conical shape, such as the 'Montefortino' type later adopted by the Romans.

The ancient depictions of Mercury with a winged helmet are taken to symbolize speed. In modern comic book mythology, this has evolved into the wings present in the helmet or head-portion of the costume of various versions of The Flash, Captain America, and Thor. Another famous comic character who always wears a winged helmet is Asterix, after the inspiration by various modern paintings and sculptures of the Gallic chieftain Vercingetorix.

Vendel Period helmets from Sweden often feature bird-motifs. On these helmets, however, there is a stylized noseguard or more often a winged man whose wings form "eye-brows" around the eye holes of the helmets.

See also
Horned helmet
Horned God
Petasos
Golden hat
Pointy hat

References

Bibliography
Connolly, P. (1981) Greece and Rome at War. Macdonald Phoebus, London.

External links

Helmets
Flight folklore


text:
New rights for unwed couples who separate
Former cohabitees may be able to claim a share of their former partner's home or pension in future. Lisa Bachelor reports
Sun 5 Aug 2007 04.18 EDT First published on Sun 5 Aug 2007 04.18 EDT
Cohabiting couples who split up will get new rights to claim money from former partners if proposals by the Law Commission are adopted.
The commission, an independent body set up by Parliament to recommend legal reforms, reached the end of a two-year consultation last week and recommended that unmarried couples who have broken up after living together be able to claim similar financial remedies as married couples would in a divorce.
One partner should be able to claim a lump-sum payment from the other, the right to live in the family home and a share of their other half's pension. The commission suggests this should apply unless both partners opt out.
There are estimated to be more than 2 million cohabiting couples in England and Wales, and that is set to double by 2050. Over a million children are being raised by unmarried couples.
'At the moment cohabiting couples have rights under a hotchpotch of laws which in some cases are quite archaic,' says Suzanne Kingston, head of family at law firm Dawsons. 'These new proposals offer much broader provision in terms of the potential awards a court will be able to consider.'
Alex Carruthers, founding partner of law firm Hughes Fowler Carruthers, agrees: 'If there aren't any children [in a cohabiting relationship], any claims which parties do have will normally be in respect of their home. These claims are notoriously difficult to evaluate and the whole area of law is vague.'
However, the proposals stop short of giving cohabiting couples the same financial rights as married couples. On divorce a number of factors, including the income of each spouse, their financial needs, their age and the contributions both parties have made to the marriage (or are likely to make) are taken into account in assessing a settlement. These factors have been expanded by the outcome of high-profile cases that have resulted in a 50/50 entitlement on divorce.
Unlike divorce, cohabitees would not be expected to make maintenance payments to a former partner and there would be no principle that the parties should share their assets equally.
Cohabitees filing for a settlement from their former partner would, instead, have to prove that they would suffer a continuing economic disadvantage on separation. The Law Commission believes that by making adequate provision for the adult parties, the scheme would give more leeway to the court to provide better protection for their children.
The commission has recommended that the law apply to those who have children together or have been cohabiting for at least two years, unless they opt out. Suzanne Todd, a partner at law firm Withers, thinks this could be problematic: 'This could still result in injustice for people whose relationships break down short of the two-year mark, particularly in the current climate where house prices can increase significantly in less than two years.'
Could it affect you?
John and Lucy, in their twenties, have been living together for two years in rented accommodation. They have both worked full-time and kept their finances separate. Would either be entitled to money from the other?
No, because: on separation neither is left with a retained benefit or a continuing economic disadvantage. Neither has retained any benefit in the property, as it is rented, and the relationship has been economically neutral for each in terms of their earnings and earning capacity.
Alice and Greg, in their thirties, had been living together for three years in Greg's house when they started a family. They agreed that Alice should not return to work once their child, Tom, was born. They are now separating. Under the current law it is unlikely that Alice could get a share in the family home, and Greg has no obligation to provide for her. The couple share the obligation of meeting the basic cost of caring for Tom. Would Alice be better off under the new scheme?
Yes, because: Greg would be required to share Alice's economic disadvantage caused by her contributions to the relationship, while his existing obligation to pay child support would be unaffected.
Source: The Law Commission

redpajama_set_name: RedPajamaCommonCrawl


A site collection is a grouping of sites that includes a top-level site and all subsites. Subsites are created from the top-level site through the Site Actions menu and are generally related to the top-level site, as well as to each other. Subsites can inherit their security settings and navigational scheme from their parent, or they can define their own.
Under the SharePoint Site Management section, select Create Site Collection.
Choose the Web application in which the site collection will be created from the drop-down list as shown in Figure 4-1.
Specify the Title and Description for the site (see Figure 4-2).
Specify the URL path to the new site.
Choose the template to use for site creation. The template list consists of all installed site templates and site definitions, including those installed with SharePoint Services and any custom templates that have been installed.
(Optional) Specify the Secondary Site Collection Administrator.
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Click OK to create the site.
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On the New SharePoint Site page, specify a Title and (optionally) a Description for the new site.
Specify the URL for the new site. The first part of the URL has been set by the administrator.
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The Set Up Groups for this Site page appears, and you can set up the groups that will have access to the site. You can either add users to existing groups or add new ones.
Often, high-profile sites, such as CEO or CIO sites, benefit from the disaster-recovery advantages of being in a standalone content database. If you need a site collection to be in its own content database, you can create a new site collection from the command line using Stsadm.exe.
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Open SharePoint Central Administration and go to the Application Management tab.
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Quota templates are created at the farm level; therefore any new templates are available across the entire farm. However, because they are applied at the Web application level, all site collections in a Web application share the same quota template, unless it is over-ridden for an individual site. This multitiered approach allows for a great amount of flexibility in the management of quota templates.
Open SharePoint Central Administration and click the Application Management tab.
In the SharePoint Site Management section, choose Quota Templates (see Figure 4-3).
If there are existing templates, you are given the option to edit one of these templates. Use caution when doing so because the changes will affect all sites already using this template. You should usually select the option to create a new template.
If there are existing templates, you can use one as a starting point for your new template. If you choose an existing template in the drop-down list, the storage limit and warning values are filled in according to the chosen template. If there is an existing template that is similar to the one you want to create, choose it; otherwise choose to start from a new blank template.
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In the SharePoint Site Management section, choose Quota Templates.
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Not Locked is the default state, in which content can be added, edited, and deleted by anyone with the proper permissions. The Adding Content Prevented lock allows Web Parts and content to be modified or deleted, but nothing can be added. A Read-Only lock prevents any modifications to sites whatsoever, whereas a No Access lock prevents access to the site collection altogether.
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Follow these same steps any time that you need to change the lock status of a site collection; for example, when clearing a lock set by an application or by exceeding a quota.
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It is possible, and in most cases recommended, to specify both primary and secondary site collection administrators so that if the primary administrator is unavailable for any reason, the secondary administrator will still receive administrative alerts. One change from the 2003 version is that IT administrators, such as local computer and domain administrators, no longer automatically have site collection administrator privileges, unless explicitly given those privileges in SharePoint Services.
Users with Full Control or Hierarchy Manager permissions on a particular site can create subsites from that site. Subsites and workspaces can either inherit permissions from the parent or be given a completely independent set of permissions. Workspace is the term that is applied to a site created for the purpose of collaborating on a particular task, such as a document or a meeting. For all practical purposes, workspaces are simply sites.
From the Home page of the parent site, click the Site Actions drop-down menu located in the upper right section of the page.
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In the section labeled Navigation, specify whether to include links to the new site on the Quick Launch and the Top Link Bar of the parent site, as well as whether to display the Top Link Bar from the parent.
Finally, select the Basic Meeting Workspace template from the Meetings tab, and click Create.
The new subsite is shown in Figure 4-5.


text:
\section*{Abstract}
Consider the $n$th degree polynomial equation,
$X^n+A_{n-1}X^{n-1}+\cdots+A_1X+A_0=0$ over the ring of
$2\times2$ complex matrices.
If this equation has more than $\binom{2n}{2}$ solutions, then it
has infinitely many solutions. We show here that for any
$n,m \in\mathbb{N}$ such that $m\leq\binom{2n}{2}$, there
exists an $n$th degree polynomial equation with exactly $m$
solutions.
\section{Introduction}
Let $R$ be a ring and let
\begin{equation*}
f(X)=X^n+A_{n-1}X^{n-1}+\dots+A_1X+A_0
\end{equation*}
be a polynomial in the indeterminate $X$ with coefficients in $R$
(with powers of $X$ on the right side of the coefficients).
It is well known that if $R$ is a field, the number of solutions in
$R$ of $f(X)=0$ is $\leq n$. If $R$ is a division ring, then (\cite{BW})
the number of solutions in $R$ of $f(X)=0$ is either $\leq n$ or
infinite. In case $R$ is the ring of $k\times k$ matrices over $\mathbb{C}$,
a number of authors (e.g., \cite{FS}, \cite{G}, \cite{LR}, \cite{GLR})
have studied solutions of $f(X)=0$. In particular, Fuchs
and Schwartz have shown that in the generic case, $f(X)=0$ has
$\binom{kn}{k}$ diagonalizable solutions.

In this paper, we consider the case where $R$ is the ring of $2\times 2$
matrices over the complex numbers, $M_2(\mathbb{C})$. It is true
(Proposition \ref{inf}) that if the equation $f(X)=0$ has more than
$\binom{2n}{2}$ solutions then it has infinitely many. The main result
of this paper is that every number of solutions $\leq \binom{2n}{2}$ can
arise. That is,
\begin{thm}\label{main}
Given $m,n\in\mathbb{N}$, $m\leq \binom{2n}{2}$, there exists an $n$th degree equation
over $2\times2$ complex matrices that has exactly $m$ solutions.
\end{thm}

The paper is organized as follows.
Section \ref{solving} recalls (from \cite{FS})
a general method of solving such equations and proves that
if there are more than $\binom{2n}{2}$ solutions, there are
infinitely many.
Section \ref{proof} provides a general
algorithm for finding an $n$th degree equation with exactly
$m$ solutions provided that $m\leq\binom{2n}{2}$, except when
$m$ is 4 or 16. Section \ref{cases} deals with these special cases.
The proof of theorem \ref{main} is in
Section \ref{thm}.

\section{Solving Matrix Polynomial Equations}\label{solving}
Let $X$ satisfy
\begin{equation}\label{general}
X^n+A_{n-1}X^{n-1}+\dots+A_1X+A_0=0
\end{equation}
where $X,A_{n-1},\dots,A_0\in M_2(\mathbb{C})$.
Since $X$ is a matrix over the complex
numbers, $X$ must have an eigenvalue, $\lambda$, and a
corresponding eigenvector, $v$. Then we have that
\begin{align*}
0&=(X^n+A_{n-1}X^{n-1}+\dots+A_1X+A_0)v\\
&=X^nv+A_{n-1}X^{n-1}v+\dots+A_1Xv+A_0v\\
&=\lambda^n v+A_{n-1}\lambda^{n-1}v+\dots+A_1\lambda v+A_0 v\\
&=(\lambda^nI+\lambda^{n-1}A_{n-1}+\dots+\lambda A_1+A_0)v
\end{align*}
meaning that the matrix
$M(\lambda)=\lambda^nI+\lambda^{n-1}A_{n-1}+\dots+\lambda A_1+A_0$ must have a
non-trivial nullspace.
We shall call $M(t)$ the {\em corresponding polynomial matrix}.
In order for $M(\lambda)$ to have a non-trivial nullspace, we need
\begin{equation}
\det(M(\lambda))=0.
\end{equation}
Thus we need to solve the equation $\det(M(t))=0$ to get the possible
eigenvalues. Since $\det(M(t))$ is a $2n$th degree polynomial, it
has $2n$ roots, counting multiplicities.
We call these roots, $\lambda_1,\dots,\lambda_{2n}$, the {\em critical values}
of the equation since they are the only possibilities for eigenvalues
of the solutions.

If we assume that $\lambda$ is an eigenvalue of $X$, then it must have
an eigenvector, $v$. This vector can be any non-zero
vector in the nullspace
of $M(\lambda)$. We call the nullspace of $M(\lambda)$ the
{\em $\lambda$-critical space} and any nonzero
vector in it a {\em $\lambda$-critical vector}.

Using this, we can now prove proposition \ref{inf}.

\begin{prop}\label{inf}
If (\ref{general}) has finitely many solutions, then there are at most
$\binom{2n}{2}$ of them.
\end{prop}
\begin{proof}
The idea is that
if there are $p$ distinct critical values then there are at most
$\binom{p}{2}$ diagonalizable solutions of (\ref{general});
for each repeated root of $M(t)$
there is at most one non-diagonalizable solution;
and
non-repeated roots do not have associated non-diagonalizable solutions.
Then the number of non-diagonalizable solutions is at most $2n-p$, so the total
number of solutions is at most $\binom{p}{2}+2n-p$. Since $p\leq 2n$, we have
\begin{align*}
  \binom{p}{2} + 2n - p = 2n +\frac{p(p-3)}{2}\leq
  2n + \frac{2n(2n-3)}{2} = \binom{2n}{2}.
\end{align*}

We will now show that there are at most $\binom{p}{2}$
diagonalizable solutions.
A diagonalizable solution is formed by taking
two distinct critical values and their corresponding critical vectors,
and using them to define a matrix. If there are $p$
distinct critical values, then there are at most $\binom{p}{2}$
diagonalizable solutions.

We discount the case in which a critical value, $\lambda$, has a critical
space with dimension greater than one. This is because in this case we can
create infinitely many solutions by taking a critical value different from
$\lambda$ and its critical space together with $\lambda$ and any of the
one-dimensional subspaces of the $\lambda$-critical space. (If there is no
critical value distinct from $\lambda$ then there is only one-diagonalizable
solution anyway.)

We will now show that if a critical value, $\lambda$, is not repeated,
then there is no non-diagonalizable solution with $\lambda$ as its eigenvalue.
This is Theorem VIII.4 in \cite{G}. If $X$ is a non-diagonalizable solution to (\ref{general}) with eigenvalue
$\lambda$, then the characteristic polynomial of $X$ has $\lambda$
as a double root. However, $\det(M(t))$ is a multiple of the characteristic
polynomial of $X$. Explicitly:
\begin{eqnarray*}
M(t)&=&\sum_{i=1}^{n}
       \left(\sum_{j=0}^{i-1} A_{n-i+j+1}X^j\right)t^{n-i}\cdot(tI-X)\\
\det(M(t))&=&\det\left(\sum_{i=1}^{n}
             \left(\sum_{j=0}^{i-1} A_{n-i+j+1}X^j\right)t^{n-i}\right)
           \cdot\det(tI-X)
\end{eqnarray*}
where $A_n=I$.
Thus if $X$ is non-diagonalizable, then its eigenvalue must be a repeated
critical value of $M(t)$.

We now show that each repeated critical value has at most 1 non-diagonalizable
solution associated with it.

 Let $f(X) = X^n + a_{n-1}X^{n-1} + ... + A_0.$  Assume $Y$ and $Z$ are distinct nilpotent solutions of (\ref{general}). Then, since
$Y$ and $Z$ are two by two matrices, $Y^2 = Z^2 = 0$,
and so $0 = f(Y) = A_1Y + A_0 = f(Z) = A_1Z + A_0.$ We will show that (\ref{general})
has infinitely many solutions.  If $A_1 = 0$ we have $0 = A_0$.  Then any solution
of $X^2 = 0$ is a solution of (\ref{general}) and so there are infinitely many solutions.  Now
assume $A_1 \ne 0$.  We have $A_1Y + A_0 =  A_1Z + A_0$.  Multiplying on the right by $Z$ gives $A_1YZ + A_0Z = A_1Z^2 + A_0Z$ so that $A_1YZ = 0$.  Similarly, $A_1ZY = 0$
and so $A_1(YZ - ZY) = 0$.  Now $tr(YZ - ZY) = 0 $ and since $A_1 \ne 0$ we have $det(YZ - ZY) = 0$.  Thus $YZ-ZY$ is nilpotent and so  $0 = (YZ - ZY)^2 = YZYZ + ZYZY.$  But $(Y+Z)^4 = YZYZ + ZYZY$ so $Y+Z$ is nilpotent and hence $0 = (Y+Z)^2 = YZ + ZY.$  Then $\mu Y + (1- \mu)Z$ is a nilpotent solution of (\ref{general}) for any $\mu \in \mathbb{C}.$

Now if $Y$ is a non-diagonalizable $2$ by $2$ matrix, it has a single eigenvalue $\lambda$.  If $Y,Z$ are two distinct non-diagonalizable solutions of (1) with eigenvalue $\lambda$, then $Y - \lambda I, Z - \lambda I$
are distinct nilpotent solutions of $(X + \lambda I)^2 + A_1(X + \lambda I) + A_0 = 0$ so this equation and hence equation (\ref{general}) have infinitely many solutions.

Thus, if (1) has finitely many solutions, the number of non-diagonalizable solutions of (1) is bounded by the number of repeated roots, say $k$, of $M(t) = 0$.  Since the number of distinct roots is $\le 2n-k$, the number of diagonalizable solutions is $\le {{2n-k} \choose 2}$ and the total number of solutions is $\le {{2n-k} \choose 2} + k \le {{2n} \choose 2}.$

\end{proof}

\section{Finding Equations with Exactly $m$ Solutions}\label{proof}
In this section, we will show how to find equations
with exactly $m$ solutions, all of which are
diagonalizable.

The crux of being able to find our equation with
exactly $m$ solutions is choosing
the critical vectors for the equation to have. We want to
choose the vectors in such a way that exactly $m$ pairs of
these vectors are linearly independent. However, it is not
always possible to find a set of $2n$ vectors with this
property.

The first subsection shows a way to create an equation
that has only $p$ distinct critical values ($p\leq2n$)
but does not have any
two-dimensional critical space or non-diagonalizable
solutions. The second subsection shows how to choose the
configuration of the $p$ critical vectors such that
exactly $m$ pairs are linearly
independent. The third subsection shows how to choose the
critical values and vectors such that a polynomial with
the desired properties exists.
\subsection{Cutting down the number of critical values}
Since we may not always want $2n$ distinct critical values,
this lemma shows a way to make an equation with only $p$
distinct critical values by making one of the critical
values have multiplicity $\bar{p}=2n-p+1$ while keeping
the critical space at dimension 1 and not creating any
non-diagonalizable solutions. How to choose $p$ and $\bar{p}$
is discussed in lemma \ref{vectors}.

Let $A_i = \left[\begin{array}{rr}
a_{11}^{(i)} & a_{12}^{(i)} \\
a_{21}^{(i)} & a_{22}^{(i)}\end{array}\right]$.

\begin{lemma}\label{values}
Let $n, \bar{p}\in\mathbb{N}$, $2n\geq \bar{p}$.
Assume that either
\begin{enumerate}
\item $\bar{p}\leq n$, and
$a_{11}^{(i)}=a_{21}^{(i)}=0$ for $0\leq i\leq \bar{p}-1$\nonumber
\item $\bar{p}>n$, and
$a_{11}^{(i)}=a_{21}^{(i)}=0$ for $0\leq i\leq n-1$ and
$a_{22}^{(i)}=0$ for $0\leq i\leq \bar{p}-n-1$
\end{enumerate}
Then 0 is a critical value of equation (\ref{general})
with multiplicity $\geq\bar{p}$ and
$\left[\begin{array}{r}1\\0\end{array}\right]$
is a 0-critical vector of (\ref{general}).

Furthermore, if condition 1 holds and
the multiplicity of 0 is exactly $\bar{p}$, or if condition 2 holds and
$a_{12}^{(0)} \neq 0$ then
the 0-critical space of (\ref{general}) is exactly
$\textrm{span}\left\{\left[\begin{array}{r}1\\0\end{array}\right]\right\}$,
and there are no nilpotent solutions.
\end{lemma}
\begin{proof}\qquad
\paragraph{Case 1:} $\bar{p}\leq n$

The polynomial matrix, $M(t)$, is
\begin{displaymath}
\left[\begin{array}{rr}
t^n + a_{11}^{(n-1)}t^{n-1} + \dots + a_{11}^{(\bar{p})}t^{\bar{p}} &
      a_{12}^{(n-1)}t^{n-1} + \dots + a_{12}^{(0)} \\
      a_{21}^{(n-1)}t^{n-1} + \dots + a_{21}^{(\bar{p})}t^{\bar{p}} &
t^n + a_{22}^{(n-1)}t^{n-1} + \dots + a_{22}^{(0)}
\end{array}\right].
\end{displaymath}
The determinant of this is
\begin{displaymath}
t^{\bar{p}}\cdot det
\left[\begin{array}{rr}
t^{n-\bar{p}} + a_{11}^{(n-1)}t^{n-\bar{p}-1} + \dots + a_{11}^{(\bar{p})} &
           a_{12}^{(n-1)}t^{n-1}     + \dots + a_{12}^{(0)} \\
           a_{21}^{(n-1)}t^{n-\bar{p}-1} + \dots + a_{21}^{(\bar{p})} &
t^{n}    + a_{22}^{(n-1)}t^{n-1} + \dots + a_{22}^{(0)}
\end{array}\right].
\end{displaymath}
Thus $0$ is an critical value with multiplicity at least $\bar{p}$.

\paragraph{Case 2:} $\bar{p}>n$

The polynomial matrix, $M(t)$, is
\begin{displaymath}
\left[\begin{array}{rr}
t^n &       a_{12}^{(n-1)}t^{n-1} + \dots + a_{12}^{(0)} \\
0   & t^n + a_{22}^{(n-1)}t^{n-1} + \dots + a_{22}^{(\bar{p}-n)}t^{\bar{p}-n}
\end{array}\right].
\end{displaymath}
As this matrix is upper triangular, its determinant is
$t^{2n}+a_{22}^{(n-1)}t^{2n-1}+\ldots + a_{22}^{(\bar{p}-n)}t^{\bar{p}}$.
Thus $0$ is an critical value with multiplicity at least $\bar{p}$.

Note that in the first case, the multiplicity of 0 is exactly $\bar{p}$
if and only if
$\det\left(\left[\begin{array}{rr}
a_{11}^{(\bar{p})} & a_{12}^{(0)} \\
a_{21}^{(\bar{p})} & a_{22}^{(0)}
\end{array}\right]\right)\neq0$. In particular, this means that
$\left[\begin{array}{r}a_{12}^{(0)}\\a_{22}^{(0)}\end{array}\right]\neq0$.
In the second case, we required that $a_{12}^{(0)}\neq 0$, so it is also
true here that
$\left[\begin{array}{r}a_{12}^{(0)}\\a_{22}^{(0)}\end{array}\right]\neq0$

To find the 0-critical space, consider
$M(0)=\left[\begin{array}{rr} 0&a_{12}^{(0)}\\0&a_{22}^{(0)}\end{array}\right]$.
We can see that $v=\left[\begin{array}{r}1\\0\end{array}\right]
\in\textrm{null}(M(0))$, so $\left[\begin{array}{r}1\\0\end{array}\right]$
is a 0-critical vector.
If the multiplicity of 0 is exactly $\bar{p}$
then $a_{12}^{(0)}$ and $a_{22}^{(0)}$ are not both $0$ so
the 0-critical space is exactly
$\textrm{span}\{v\}$.

To show that when the multiplicity of 0 is exactly $\bar{p}$ there are no
nilpotent solutions,
assume $X$ is a nilpotent solution to (\ref{general}),
$X=\left[\begin{array}{rr}x_{11} & x_{12}
\\ x_{21} & x_{22}\end{array}\right]$.
Since $X$ has eigenvalue $0$ with eigenvector
$v=\left[\begin{array}{r}1\\0\end{array}\right]$,
$X$ must have the form
$\left[\begin{array}{rr}0 & x_{12} \\ 0 & x_{22}\end{array}\right]$.
Since $X$ is non-diagonalizable, there exists a vector
$w= \left[\begin{array}{r}w_1\\w_2\end{array}\right]$
such that
$Xw=\left[\begin{array}{r}1\\0\end{array}\right]$.
Since $X^2=0$, we get
\begin{eqnarray*}
0&=&X^nw+A_{n-1}X^{n-1}w+\dots+A_1Xw+A_0w\\
 &=&A_1Xw+A_0w\\
 &=& \left[\begin{array} {rr} 0 & a_{12}^{(1)} \\
0 & a_{22}^{(1)} \end{array}\right]Xw
+
\left[\begin{array} {rr} 0 & a_{12}^{(0)} \\
0 & a_{22}^{(0)} \end{array}\right]
w\\
&=&\left[\begin{array} {rr} 0 & a_{12}^{(1)} \\
0 & a_{22}^{(1)} \end{array}\right]v
+
\left[\begin{array} {rr} 0 & a_{12}^{(0)} \\
0 & a_{22}^{(0)} \end{array}\right]
w\\
&=&\left[\begin{array}{r}0\\0\end{array}\right]
+
\left[\begin{array} {r} a_{12}^{(0)}w_2 \\ a_{22}^{(0)}w_2 \end{array}\right].
\end{eqnarray*}
Since $a_{12}^{(0)}$ and $a_{22}^{(0)}$ are not both $0$, $w_2=0$,
so $w=\left[\begin{array}{r}w_1\\0\end{array}\right]\in\textrm{span}\{v\}$.
Thus
$\left[\begin{array}{r}1\\0\end{array}\right]
=Xw=\left[\begin{array}{r}0\\0\end{array}\right]$, a contradiction.
\end{proof}

\subsection{Cutting down the number of linearly independent pairs of critical vectors}
This lemma shows how many of the
critical values should be distinct, and how
to construct a configuration of vectors
such that exactly $m$ pairs are linearly independent.

Recall that the critical value $0$ plays a distinguished role as
it may have multiplicity $>1$.

\begin{lemma}\label{vectors}
For any $m\in\mathbb{N}$, $m\neq 4,16$, let $p$ be the integer such that
\begin{equation}\label{p}
{p-1 \choose 2}<m\leq{p \choose 2}.
\end{equation}
There exists an equivalence relation $\sim$ on $\mathbb{N}_p$, where
$\mathbb{N}_p=\{0,1,2,\dots,p-1\}$,
such that
\begin{enumerate}
\item the set
$\{(x,y)\in\mathbb{N}_p\times\mathbb{N}_p | x<y\text{, and } x\not \sim y\}$
has exactly $m$ elements
\item $0$ is in an equivalence class by itself.
\item No equivalence class has more than $\lceil\frac{p}{2}\rceil$ elements
\end{enumerate}
\end{lemma}
\begin{proof}
We will define $\sim$ to satisfy 1 and 2 in each of several cases.
We will defer the verification that $\sim$ also satisfies 3 until
the end of the proof.

When we define $\sim$, we want the number of pairs
that are not equivalent to be $m$,
so we want the number of pairs that are equivalent to be
 ${p \choose 2}-m$.
Let $a$ and $b$ be the integers such
that ${p \choose 2}-m=3a+b$ where $0\leq b<2$.

Now, by (\ref{p}),
\begin{eqnarray}
{p-1 \choose 2} &<& m\nonumber \\
1-p={p-1 \choose 2}-{p \choose 2} &<& m-{p \choose 2}\nonumber\\
3a+b+1={p \choose 2}-m+1 &<& p\label{pineq}
\end{eqnarray}
so $3a+b+1\in\mathbb{N}_p$.

\paragraph{Case 1:} $b=0$\\
Let $\sim$ be the equivalence relation corresponding
to the following partition of $\mathbb{N}_p$
{\setlength\arraycolsep{0pt}
\begin{eqnarray*}
\{&\{&0\},\\
  &\{&1,2,3\},\{4,5,6\},\dots,\{3a-2, 3a-1, 3a\},\\
  &\{&3a+1\},\{3a+2\},\dots,\{p-1\}\}.
\end{eqnarray*}}
Note that since $3a+b+1\in\mathbb{N}_p$, this is a partition of $\mathbb{N}_p$.
Since each set of three creates 3 equivalent pairs, and there are $a$
groups of three, there are $3a={p \choose 2}-m$ pairs that are
equivalent and $m$ pairs are not.

\paragraph{Case 2:} $b=1$\\
Let $\sim$ be the equivalence relation corresponding
to the following partition of $\mathbb{N}_p${\setlength\arraycolsep{0pt}
\begin{eqnarray*}
\{&\{&0\},\\
  &\{&1,2,3\},\{4,5,6\},\dots,\{3a-2, 3a-1, 3a\},\\
  &\{&3a+1,3a+2\}\\
  &\{&3a+3\},\{3a+4\},\dots,\{p-1\}\}.
\end{eqnarray*}}
Note that since $3a+b+1\in\mathbb{N}_p$, this is a partition of $\mathbb{N}_p$.
Each set of three creates 3 equivalent pairs, and the set of two
creates 1 equivalent pair. There are $a$
sets of three, and 1 set of two so there are $3a+1={p \choose 2}-m$
pairs that are equivalent and $m$ pairs are not.

\paragraph{Case 3a:} $b=2$ and $a \geq 2$\\
Let $\sim$ be the equivalence relation corresponding
to the following partition of $\mathbb{N}_p$
{\setlength\arraycolsep{0pt}
\begin{eqnarray*}
\{&\{&0\},\\
  &\{&1,2,3\},\{4,5,6\},\dots,\{3a-8, 3a-7, 3a-6\},\\
  &\{&3a-5, 3a-4, 3a-3, 3a-2\},\\
  &\{&3a-1,3a\},\{3a+1,3a+2\},\\
  &\{&3a+3\},\{3a+4\},\dots,\{p-1\}\}.
\end{eqnarray*}}

Each set of three creates 3 equivalent pairs,
each set of two creates 1 equivalent pair,
and the set of four creates 6 equivalent pairs.
Thus there are $3(a-2)+6+2=3a+2={p \choose 2}-m$ pairs that are equivalent
and $m$ pairs that are not.

\paragraph{Case 3b:} $b=2$ and $p>3a+4$
Let $\sim$ be the equivalence relation corresponding
to the following partition of $\mathbb{N}_p$
{\setlength\arraycolsep{0pt}
\begin{eqnarray*}
\{&\{&0\},\\
  &\{&1,2,3\},\{4,5,6\},\dots,\{3a-2, 3a-1,3a\},\\
  &\{&3a+1,3a+2\},\{3a+3,3a+4\},\\
  &\{&3a+5\},\{3a+6\},\dots,\{p-1\}\}.
\end{eqnarray*}}

Each set of three creates 3 equivalent pairs,
each set of two creates 1 equivalent pair,
Thus there are $3a+2={p \choose 2}-m$ pairs that are equivalent and $m$ pairs
that are not.

Note that the two parts of the $b=2$ case are not
mutually exclusive, and they do
not cover all the $b=2$ cases. The cases that are not covered are the ones
in which $b=2$, $a<2$, and $p\leq 3a+4$.

If $b=2$, $a=0$ and $p\leq 3a+4$,then $p\leq 4$. But by (\ref{pineq}),
$3=3a+b+1<p$ so $p=4$ and hence $m=\binom{4}{2}-2=4$. Thus $m=4$ is
one of our special cases.

If $b=2$, $a=1$, and $p\leq 3a+4$, then $p\leq 7$. But by (\ref{pineq}),
$6=3a+b+1<p$, so $p=7$ and hence $m=\binom{7}{2}-5=16$
So $m=16$ is our other special case.

Our construction of the equivalence relations satisfied the first
two conditions. We have yet to prove that no equivalence class
has more than $\lceil\frac{p}{2}\rceil$ elements.

First of all, notice that none of our constructed equivalence classes
have more than $4$ elements, so if $p\geq 7$, then we are done.

The only case in which we have more than $3$ elements in one equivalence
class is when $b=2$ and $a\geq2$. But then, by (\ref{pineq}),
\begin{equation*}
9\leq 3a+b+1 < p.
\end{equation*}
Thus case 3a does not occur when $p<7$, so when $p\leq 9$,
 we will not have any equivalence classes with more than
3 elements. Thus if $p\geq 5$, we are done.

If $p\leq4$, then by (\ref{pineq}), $3a+b+1<p \leq 4$ so $3a+b\leq2$
and $a=0$. However, if $a=0$ then none of the other cases require
an equivalence class of three elements. Thus if $p\leq4$, all equivalence
classes have at most two elements, so for $p\geq3$, we are done.

If $p\leq2$, then $m=\binom{p}{2}$, so $a=b=0$. This falls into case
1 with none of the elements being equivalent to anything other than
itself.

Now we have satisfied all three conditions, so except
for our special cases ($m=16$ and $m=4$), we can create the
required equivalence relation.
\end{proof}

\subsection{Choosing the numbers}
The previous lemma ensured that no equivalence class has more
than $\lceil\frac{p}{2}\rceil$ elements. Since $n,p\in\mathbb{N}$ and
$p \leq 2n$, this means that no equivalence class has more
than $n$ elements.

\begin{lemma}\label{entries}
Given
\begin{enumerate}
\item $n\in\mathbb{N}$
\item $p\in\mathbb{N}$, $2 \leq p \leq 2n$
\item an equivalence relation, $\sim$ on $\mathbb{N}_p$,
such that 0
is only equivalent to itself, and no equivalence class has more
than $n$ elements
\end{enumerate}
it is possible to choose 2-dimensional vectors
$v_0, v_1, \dots, v_{p-1}$ and scalars
$\lambda_0, \lambda_1, \dots, \lambda_{p-1}$ such that
\begin{enumerate}
\item $v_i$ and $v_j$ are linearly dependent if and only if $i\sim j$
\item there exists an $n$th degree
equation over 2 by 2 matrices that has critical values
$\lambda_0, \lambda_1, \dots, \lambda_{p-1}$,
and the critical space corresponding
to $\lambda_i$ is $span(v_i)$.
\item the equation has no non-diagonalizable solutions.
\end{enumerate}
\end{lemma}

\begin{proof}\quad
\paragraph{Choosing the critical vectors}
Let
$v_0 = \left[\begin{array}{r} 1\\0 \end{array}\right]$
and let all the others be of the form
$v_i = \left[\begin{array}{r} 1\\y_i \end{array}\right]$
where $y_i\neq0$ and $v_i=v_j\Leftrightarrow i\sim j$.

\paragraph{Choosing the critical values}
Let $\lambda_0=0$ and
let $\lambda_i$ be an $n$th root of $y_i$.
Furthermore, take the $\lambda_i$ to be distinct.
This is possible because there are at most $n$ numbers in the same
equivalence class, so no more than $n$ of the $y_i$ will be the same.

\paragraph{Tying it together and proving that it works}
Let $\bar{p}=2n-p+1$. If $\bar{p}\leq n$, we use case 1 of
lemma \ref{values} and our equation will have a corresponding
polynomial matrix of the form
\begin{equation*}
\left[\begin{array}{rr}
t^n + a_{11}^{(n-1)}t^{n-1} + \dots + a_{11}^{(\bar{p})}t^{\bar{p}} &
      a_{12}^{(n-1)}t^{n-1} + \dots + a_{12}^{(0)} \\
      a_{21}^{(n-1)}t^{n-1} + \dots + a_{21}^{(\bar{p})}t^{\bar{p}} &
t^n + a_{22}^{(n-1)}t^{n-1} + \dots + a_{22}^{(0)}
\end{array}\right].
\end{equation*}
We need to choose the $a_{ij}^{(k)}$
so that when we plug in $\lambda_i$ for $t$, we
get a matrix with a null space spanned by $v_i$.
First let us focus on finding the $a_{1j}^{(k)}$, i.e., on finding the
first row of each $A_k$. (Finding the $a_{2j}^{(k)}$ is similar.)
Since the first component of
$M(\lambda_i)v_i$ must be 0, we obtain the equation
\begin{equation*}
a_{11}^{(\bar{p})}\lambda_i^{\bar{p}} + \dots +
a_{11}^{(n-1)}\lambda_i^{n-1} +
a_{12}^{(0)}y_i +\dots +
a_{12}^{(n-1)}\lambda_i^{n-1}y_i=-\lambda_i^n.
\end{equation*}
The corresponding system of $p-1$ linear equations (in the $a_{ij}^{(k)}$)
may be written in the following matrix form

\begin{displaymath}
\left[\begin{array}{cccccccc}
\lambda_1^{\bar{p}} & \lambda_1^{\bar{p}+1} & \dots
       & \lambda_1^{n-1}
& y_1 & \lambda_1 y_1 & \dots
       & \lambda_1^{n-1}y_1 \\
\lambda_2^{\bar{p}} & \lambda_2^{\bar{p}+1} & \dots
       & \lambda_2^{n-1}
& y_2 & \lambda_2 y_2 & \dots
       & \lambda_2^{n-1}y_2 \\
&&&&\vdots&&&\\
\lambda_{p-1}^{\bar{p}} & \lambda_{p-1}^{\bar{p}+1} & \dots
       & \lambda_{p-1}^{n-1}
& y_{p-1} & \lambda_{p-1} y_{p-1} & \dots
       & \lambda_{p-1}^{n-1}y_{p-1} \\
\end{array}\right]
\left[\begin{array}{r}
a_{11}^{(\bar{p})} \\ a_{11}^{(\bar{p}+1)} \\ \vdots \\a_{11}^{(n-1)} \\
a_{12}^{(0)} \\ a_{12}^{(1)} \\ \vdots \\a_{12}^{(n-1)}
\end{array}\right]
=
\left[\begin{array}{r}
-\lambda_1^n \\ -\lambda_2^n \\ \vdots \\ -\lambda_{p-1}^n
\end{array}\right].
\end{displaymath}

(We do not need to include the $0$th equation
because lemma \ref{values} ensures that $\lambda_0$ and $v_0$ will
be a critical value/critical vector pair.)
Because we chose each $\lambda_i$ to be an $n$th root of $y_i$,
we have that $y_i=\lambda_i^n$. Thus this equation can be written

\begin{equation*}
\left[\begin{array}{rrrr}
\lambda_1^{\bar{p}} & \lambda_1^{\bar{p}+1} & \dots & \lambda_1^{2n-1}  \\
\lambda_2^{\bar{p}} & \lambda_2^{\bar{p}+1} & \dots & \lambda_2^{2n-1}  \\
&\vdots&&\\
\lambda_{p-1}^{\bar{p}}  & \lambda_{p-1}^{\bar{p}+1}
& \dots & \lambda_{p-1}^{2n-1}
\end{array}\right]
\left[\begin{array}{r}
a_{11}^{(\bar{p})} \\ a_{11}^{(\bar{p}+1)} \\ \vdots \\a_{11}^{(n-1)} \\
a_{12}^{(0)} \\ a_{12}^{(1)} \\ \vdots \\a_{12}^{(n-1)}
\end{array}\right]
=
\left[\begin{array}{r}
-\lambda_0^n \\ -\lambda_1^n \\ -\lambda_2^n \\ \vdots \\ -\lambda_{p-1}^n
\end{array}\right].
\end{equation*}
Note that the coefficient matrix is square ($2n-\bar{p}=p-1$).
The determinant of this matrix is
\begin{eqnarray*}
&&\det\left(\left[\begin{array}{rrrr}
\lambda_1^{\bar{p}} & \lambda_1^{\bar{p}+1} & \dots & \lambda_1^{2n-1}  \\
\lambda_2^{\bar{p}} & \lambda_2^{\bar{p}+1} & \dots & \lambda_2^{2n-1}  \\
&\vdots&&\\
\lambda_{p-1}^{\bar{p}}  & \lambda_{p-1}^{\bar{p}+1}
& \dots & \lambda_{p-1}^{2n-1}
\end{array}\right]\right)\\
&=&
\prod_{i=1}^{p-1} \lambda_i^{\bar{p}}
\cdot
\det\left(\left[\begin{array}{rrrrr}
1 & \lambda_1 & \lambda_1^{2} & \dots & \lambda_1^{p-2}  \\
1 & \lambda_2 & \lambda_2^{2} & \dots & \lambda_2^{p-2}  \\
&\vdots&&\\
1 & \lambda_{p-1} & \lambda_{p-1}^{2}
& \dots & \lambda_{p-1}^{p-2}
\end{array}\right]\right)\\
&=&
\prod_{i=1}^{p-1} \lambda_i^{\bar{p}}
\cdot
\prod_{i=1}^{p-1} \prod_{j=i+1}^{p-1} (\lambda_j-\lambda_i)
\end{eqnarray*}
(with the last equality following from
the formula for the determinant of a
Vandermonde matrix). Since none of
$\lambda_1,\dots,\lambda_{p-1}$ are zero, and all of our $\lambda_i$ are
distinct, the determinant is non-zero
so the system of equations is consistent.

If $\bar{p}>n$, then take $a_{11}^{(i)}=a_{21}^{(i)}=0$
for $0\leq i\leq n-1$,
$a_{22}^{(i)}=0$ for $0\leq i\leq \bar{p}-n-1$,
as specified by lemma \ref{values}.
Also take $a_{12}^{(i)}=0$ for $p-1\leq i\leq n-1$.
This does not conflict with the restriction that $a_{12}^{(0)}\neq 0$
because we required $2\leq p$.
The matrix polynomial looks like
\begin{equation*}
\left[\begin{array}{rr}
t^n &       a_{12}^{(p-2)}t^{p-2} + \dots + a_{12}^{(0)} \\
0   & t^n + a_{22}^{(n-1)}t^{n-1} + \dots + a_{22}^{(\bar{p}-n)}t^{\bar{p}-n}
\end{array}\right].
\end{equation*}
To find the coefficients of the first row, we set up the equations
in the matrix form
\begin{equation*}
\left[\begin{array}{rrrrr}
y_1 & y_1 \lambda_1 & y_1\lambda_1^2 & \dots & \lambda_1^{p-2}y_1\\
y_2 & y_2 \lambda_2 & y_2\lambda_2^2 & \dots & \lambda_2^{p-2}y_2\\
&&\vdots&&\\
y_{p-1} & y_{p-1} \lambda_{p-1}
& y_{p-1}\lambda_{p-1}^2 & \dots & \lambda_{p-1}^{p-2}y_{p-1}
\end{array}\right]
\left[\begin{array}{r}
a_{12}^{(0)} \\ a_{12}^{(1)} \\ \vdots \\ a_{12}^{(p-2)}
\end{array}\right]
=
\left[\begin{array}{r}
-\lambda_1^n \\ -\lambda_2^n \\ \vdots \\ -\lambda_{p-1}^n
\end{array}\right].
\end{equation*}

Since $y_i = \lambda_i^n$ for $1\leq i\leq p-1$, we see that the first
column of the coefficient matrix is the negative of the vector on the
right-hand side of the equation. Thus
\begin{equation*}
  \left[\begin{array}{c}
      a_{12}^{(0)} \\a_{12}^{(1)}\\ \vdots \\a_{12}^{(p-2)}
  \end{array}\right]
  =
  \left[\begin{array}{c}
      -1\\0\\\vdots\\0
  \end{array}\right]
\end{equation*}
is a solution to the system of equations. Note that this shows
 $a_{12}^{(0)}\neq 0$ so the hypothesis of Lemma \ref{values} is satisfied.

To find the coefficients of the second row, we use the matrix equation
\begin{equation*}
\left[\begin{array}{rrr}
\lambda_1^{\bar{p}-n}y_1 & \dots & \lambda_1^{n-1}y_1  \\
\lambda_2^{\bar{p}-n}y_2 & \dots & \lambda_2^{n-1}y_2  \\
&\vdots&\\
\lambda_{p-1}^{\bar{p}-n}y_{p-1} & \dots & \lambda_{p-1}^{n-1} y_{p-1}
\end{array}\right]
\left[\begin{array}{r}
a_{22}^{(\bar{p}-n)} \\ \vdots \\a_{22}^{(n-1)}
\end{array}\right]
=
\left[\begin{array}{r}
-\lambda_1^ny_1 \\ -\lambda_2^ny_2 \\ \vdots \\ -\lambda_{p-1}^n y_{p-1}
\end{array}\right]
\end{equation*}

and the determinant of this matrix is

\begin{eqnarray*}
&&\det\left(\left[\begin{array}{rrr}
\lambda_1^{\bar{p}-n}y_1 & \dots & \lambda_1^{n-1}y_1  \\
\lambda_2^{\bar{p}-n}y_2 & \dots & \lambda_2^{n-1}y_2  \\
&\vdots&\\
\lambda_{p-1}^{\bar{p}-n} y_{p-1}& \dots & \lambda_{p-1}^{n-1} y_{p-1}
\end{array}\right]\right)\\
&=&\det\left(\left[\begin{array}{rrrr}
\lambda_1^{\bar{p}} & \lambda_1^{\bar{p}+1} & \dots & \lambda_1^{2n-1}  \\
\lambda_2^{\bar{p}} & \lambda_2^{\bar{p}+1} & \dots & \lambda_2^{2n-1}  \\
&\vdots&&\\
\lambda_{p-1}^{\bar{p}}  & \lambda_{p-1}^{\bar{p}+1}
& \dots & \lambda_{p-1}^{2n-1}
\end{array}\right]\right)\\
&=&
\prod_{i=1}^{p-1} \lambda_i^{\bar{p}}
\cdot
\det\left(\left[\begin{array}{rrrrr}
1 & \lambda_1 & \lambda_1^{2} & \dots & \lambda_1^{p-2}  \\
1 & \lambda_2 & \lambda_2^{2} & \dots & \lambda_2^{p-2}  \\
&\vdots&&\\
1 & \lambda_{p-1} & \lambda_{p-1}^{2}
& \dots & \lambda_{p-1}^{p-2}
\end{array}\right]\right)\\
&=&
\prod_{i=1}^{p-1} \lambda_i^{\bar{p}}
\cdot
\prod_{i=1}^{p-1} \prod_{j=i+1}^{p-1} (\lambda_i-\lambda_j).
\end{eqnarray*}

Since $\lambda_1,\dots\lambda_{p-1}$ are all distinct and
non-zero, the determinant is non-zero so this system is solvable.

Because we chose $p-1$ non-zero distinct critical values for the
equation, 0 cannot have a multiplicity of greater
than $2n-(p-1)=\bar{p}$. Lemma \ref{values} assures us that 0
will have multiplicity at least $\bar{p}$, so 0 must have
multiplicity exactly $\bar{p}$. Then we can use the second
part of lemma \ref{values}. Thus the 0-critical space is
1-dimensional, and (\ref{general}) has no nilpotent solutions.
\end{proof}
\section{Special Cases}\label{cases}
For \textbf{$m=4$}, we must have $4=m\leq\binom{2n}{2}$,
so $n\geq 2$. Then we may use the equation whose corresponding
polynomial matrix is
\begin{equation*}
\left[\begin{array}{rr}
(t-1)(t+1)^{n-1} & 0\\
0 & (t-2)(t+2)^{n-1}
\end{array}\right].
\end{equation*}
To show that this has no non-diagonalizable solution with
an eigenvalue $-1$ in general, use a change
of variables $u=t+1$ so we are looking for a non-diagonalizable
solution with eigenvalue 0.
The polynomial matrix becomes
\begin{equation*}
\left[\begin{array}{rr}
(u-2)u^{n-1} & 0\\
0 & (u-3)(u+1)^{n-1}
\end{array}\right]
\end{equation*}
the 0-critical space of this equation is
$\textrm{span}\left\{\left[\begin{array}{r}1\\0\end{array}\right]\right\}$
so $X$ must have the form
$\left[\begin{array}{rr}0 & x_{12} \\ 0 & x_{22}\end{array}\right]$.
Since $X$ is nilpotent, $x_{22}=0$.
Then we have that
\begin{eqnarray*}
0&=&A_1 X +A_0\\
&=& \left[\begin{array}{rr}a_{11}^{(1)} & 0 \\ 0 & a_{22}^{(1)}\end{array}\right]
 \left[\begin{array}{rr}0 & x_{12} \\ 0 & 0\end{array}\right]+
 \left[\begin{array}{rr}0 & 0 \\ 0 & -3\end{array}\right]\\
&=&\left[\begin{array}{rr}0 & a_{11}^{(1)} x_{12} \\ 0 & -3\end{array}\right].
\end{eqnarray*}
This is a contradiction, thus there are no non-diagonalizable solutions.
The argument for why there is no non-diagonalizable solutions
with $-2$ as an eigenvalue is similar.

For \textbf{$m=16$}, we must have $16=m\leq\binom{2n}{2}$, so $n\geq4$.
Then we may use the equation whose corresponding
matrix polynomial is
\begin{equation*}
\left[\begin{array}{rr}
(t-3)(t+3)(t-1)(t+1)^{n-3} & 0\\
0 & (t-4)(t+4)(t-2)(t+2)^{n-3}
\end{array}\right].
\end{equation*}
It is straightforward why there are exactly 16 diagonalizable
solutions,
and the reason why there are no non-diagonalizable solutions
is the same as for when $m=4$.
\section{Conclusion}\label{thm}
This section proves the main result of this paper, theorem \ref{main}.
We assume $m,n\in\mathbb{N}$, $m\leq \binom{2n}{2}$,
and show that there exists an $n$th degree equation
over $2\times2$ complex matrices that has exactly $m$ solutions.

If $m \neq 4, 16$, use lemma (\ref{vectors}) and $m$ to
define the number of distinct
critical values that we want our equation to have, $p$, and the
equivalence relation, $\sim$, that we want to determine which
critical spaces will be distinct.

All the conditions of lemma (\ref{entries}) are satisfied
by the conclusions of lemma (\ref{vectors}) except $p\leq 2n$.
This condition is satisfied because
$\binom{p-1}{2} < m \leq \binom{2n}{2}$ so
$p-1 < 2n$ so we get $p\leq 2n$ as desired.
We can then use lemma (\ref{entries}) to produce our equation.

Since lemma (\ref{entries}) promises that we will not have any
non-diagonalizable solutions and that all our critical spaces
will be one-dimensional, all our solutions will have two
different eigenvalues. By our choice of $\sim$, we have exactly
$m$ ways to choose two critical value/critical vector pairs such
that the critical vectors are linearly independent. Thus we have exactly
$m$ solutions.
The special cases of $m=4,16$ have been dealt with in section
\ref{cases} so the proof is complete.

redpajama_set_name: RedPajamaArXiv


text:
Q: How to run more than one docker-compose simultaneously with docker toolbox I have installed docker toolbox on windows 10 (https://docs.docker.com/toolbox/toolbox_install_windows/)
Then I have created a directory for my project, with docker-compose.yml file (https://docs.docker.com/compose/)
I run MINGW64 then cd to the directory and hit docker-compose up to start the images for the project. Note: MINGW64 terminal is now showing the logs and I can't hit other commands.
Now, say I have to create another project: I will create another directory with another docker-compose.yml file.
How to run both the projects simultaneously? In this situation, I can't write on MINGW64 terminal to run the second project while the first project is running.
Hwo to run two projects (with two different docker-compose.yml files) simultaneously, with docker toolbox?
Thank you

A: If by simultaneously, you mean from a single command, you could go for something like that :
$ docker-compose -f path/to_your/first_compose.yml up -d && docker-compose up -d

Provided that you have your second compose file in current folder.
If you mean from a single docker-compose call, I don't think there's a way to do it.
The fact that you didn't get interaction on the CLI back is due to the way you launched the your compose. By providing the '-d' flag, you detach the containers' output from current shell. Example with a gitlab setup with 3 containers:
$ docker-compose -f dev/docker/docker-gitlab/docker-compose.yml up -d
Creating docker-gitlab_postgresql_1 ... done
Creating docker-gitlab_redis_1      ... done
Creating docker-gitlab_gitlab_1     ... done
$ do_something # you get back interactive shell

You can still access the output of the containers with docker-compose if required, the same as with docker, with the log command.

redpajama_set_name: RedPajamaStackExchange


text:
Northern Ky. teacher wins $11 million judgment against gossip Web site | Lexington Herald Leader
Northern Ky. teacher wins $11 million judgment against gossip Web site
By Brett Barrouquere - Associated Press
LOUISVILLE — A gossip Web site has been hit with an $11 million judgment for libel and slander after posting false accusations that a Northern Kentucky teacher who works on the side as a Cincinnati Bengals cheerleader was exposed to two venereal diseases.
The judgment against Dirty World Entertainment Recordings, which runs the site Thedirty.com, came Thursday after the site declined to answer a lawsuit brought by Sarah Jones, a high school teacher whose picture was posted on the site along with the accusation.
U.S. District Judge William O. Bertelsman added an annual interest rate of 0.25 percent to the $1 million in compensatory damages and $10 million in punitive damages. The operator of the Arizona-based Web site, Hooman Karamian, who uses the online name "Nik Ritchie," did not respond to the lawsuit.
Jones' attorney, Eric Deters of Independence, said the lawsuit could have been resolved if the posting had simply been removed.
Another Democrat, a Kentucky farmer, will face McGrath in Senate primary
By Daniel Desrochers
Michael Broihier announced his U.S. Senate campaign in hopes to beat former pilot Amy McGrath and U.S. Senate Majority Leader Mitch McConnell
For 25 years, artists kept Lexington hopping with a unique downtown event