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Market Dynamics offers a full range of market research services to enable our clients to broaden and retain their business. The company was founded in 2003 by John Gilsenan to provide high-quality market research services to clients in Ireland and the UK.
Services include customer satisfaction, market opportunity surveys, online surveys, strategic plans and white papers. We ensure that the most robust, cost-effective and innovative methodologies are implemented to meet our customers' business requirements and to maximize their return on investment.
Market Dynamics' Director, John Gilsenan, has worked in the international market research industry since 1989. At AGB UK (now part of Taylor Nelson Sofres), the largest market research company in Europe, he focused on industrial market research before joining IDC, the global ICT consultancy, in 1992. At IDC John headed the UK Consulting team before becoming Research Manager of IDC's European Services research group.
In 1998 he moved to Frost & Sullivan where he developed a consulting practice focusing on research into the IT and telecommunications industries. He set up their Internet Research Group in 2000 and went on to become head of ICT Research for the European region. In early 2002, John formed his own research organisation in the UK and a year later returned to Ireland. Market Dynamics was founded in early 2003 to provide market research services and consultancy to clients in Ireland and the UK. He continues to act as a consultant for both Frost & Sullivan and IDC.
John has been quoted in many local and major international publications including the Financial Times, Irish Times, Sunday Tribune, Irish Examiner and Irish Independent.
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heinnews – Taking The Charge Episode 300: Dave and Os say farewell!
Taking The Charge Podcast Episode 300: We say goodbye – episode 300. We look back on 300 episodes but also look ahead, thanks with former long-time co-host Os Davis.


If you have ever wondered how to maneuver through life without all the baggage you carry with you, this is the section for you.
Here we will explore topics like marriage (Spousal Safari), parenting (Parental Pathways), dealing with loss (Restructuring and Restoring After Loss), relationships (Relational Revolution), and so many more.
Each of these topics are full fledged courses in dealing with each of these areas in our lives. Each of these courses consist of lessons with assignments to enhance the learning material and practical ways to apply the lessons in your everyday life.
In these courses you will learn how to use the Personal Tools in a more specific way in the different areas of your life.
Check out this section and find a course that will address an area in your life that you need some input in.
More in this category: « Spousal Safari / Marriage - COMING SOON!


Dr. Phil is coming to Raw. The host of the number one daytime talk show, "Dr. Phil," will guest star on Monday night's hottest show on April 11.
What will happen when one of TV's most recognizable personalities visits the WWE Universe? Find out next Monday night at 8/7C on USA Network.


text:
Harvest Summary of HRW - July 12
Harvest of the 2019 HRW wheat crop continues to move quickly through the southern and central Great Plains as heat and drier weather associated with a high pressure system continues to build and expand. Texas is now 97% complete with harvest, with limited areas in far northwest (higher elevation) parts of the state left to cut. Oklahoma is now 99% complete with areas in the far northwest (higher elevation) and northeast (slowed by rain) parts of the state left to harvest. Kansas is now over 70% harvested with areas in the southern half of the state starting to wind down, while northeast and north central areas of Kansas have moved past the 50% complete mark. Northwest Kansas is significantly behind the rest of the state with still less than 20% of that area cut as of this report. Colorado is now 17% harvested with most cutting being in the southeastern areas of the state, but is moving northward quickly into the east central region. Nebraska has begun harvesting in the southeastern corner of the state at 3% cut. Oregon has begun limited cutting with 5% now harvested.
A total of 180 samples are now in the lab and in various stages of analysis. Those 180 samples represent a geographical area ranging from all of Texas, across all of Oklahoma and into the southern half of Kansas. The test weight average of those samples now averages (1HRW) 60.1 lb/bu (79.1 kg/hl) up from last week's 59.9 lb/bu (78.8 kg/hl), average dockage is up slightly this week at 0.5% from 0.4%, total defects increased slightly (1.5% last week) to 1.6% this week. Average protein remained consistent across the 180 samples this week at 11.3%. A second set of early composited samples made up of sub-samples averaging over 11% protein continue to indicate good protein quality.
180 500 11.9 11.3 0.5 32.0* 388* 1HRW 60.1 79.1 0.2 0.6 0.8 1.6

redpajama_set_name: RedPajamaCommonCrawl


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Drugs & Diseases > Endocrinology
Author: George Ansstas, MD; Chief Editor: George T Griffing, MD more...
Sections Vitamin A Deficiency
Further Outpatient Care
Further Inpatient Care
Inpatient & Outpatient Medications
The word vitamin was originally derived from Funk's term "vital amine." In 1912, he was referring to Christian Eijkman's discovery of an amine extracted from rice polishings that could prevent beriberi. Funk's recognition of the antiberiberi factor as vital for life was indeed accurate. Researchers have since found that vitamins are essential organic compounds that the human body cannot synthesize. Vitamins A, D, K, and E are classified as fat-soluble vitamins, whereas others are classified as water-soluble vitamins. [1, 2]
See 21 Hidden Clues to Diagnosing Nutritional Deficiencies, a Critical Images slideshow, to help identify clues to conditions associated with malnutrition.
Vitamin A was the first fat-soluble vitamin to be discovered. Early observations by ancient Egyptians recognized that night blindness could be treated with consumption of liver. Two independent research teams, Osborne and Mendel at Yale University and McCollum and Davis at the University of Wisconsin, simultaneously discovered vitamin A in 1913. Vitamin A is made up of a family of compounds called the retinoids. The retinoid designation resulted from finding that vitamin A had the biologic activity of retinol, which was originally isolated from the retina.
There are essentially 3 forms of vitamin A: retinols, beta carotenes, and carotenoids. Retinol, also known as preformed vitamin A, is the most active form and is mostly found in animal sources of food. Beta carotene, also known as provitamin A, is the plant source of retinol from which mammals make two-thirds of their vitamin A. Carotenoids, the largest group of the 3, contain multiple conjugated double bonds and exist in a free alcohol or in a fatty acyl-ester form.
In the human body, retinol is the predominant form, and 11-cis -retinol is the active form. Retinol-binding protein (RBP) binds vitamin A and regulates its absorption and metabolism. Vitamin A is essential for vision (especially dark adaptation), immune response, bone growth, reproduction, the maintenance of the surface linings of the eyes, epithelial cell growth and repair, and the epithelial integrity of the respiratory, urinary, and intestinal tracts. Vitamin A is also important for embryonic development and the regulation of adult genes. It functions as an activator of gene expression by retinoid alpha-receptor transcription factor and ligand-dependent transcription factor.
Deficiency of vitamin A is found among malnourished, elderly, and chronically sick populations in the United States, but it is more prevalent in developing countries. Abnormal visual adaptation to darkness, dry skin, dry hair, broken fingernails, and decreased resistance to infections are among the first signs of vitamin A deficiency (VAD). [3]
In a Venezuelan study, Jimenez et al investigated the effect of a single, oral 200,000 IU dose of vitamin A on iron and vitamin A nutritional status, the phagocytic function of neutrophils, and the rate of anemia, in a population of preschool children with a high prevalence of VAD. The study group consisted of 80 children, including 12 controls; no iron supplementation was provided.
In an assessment of the children 30 days after the administration of vitamin A, the authors found that those who had received the supplement exhibited a significant increase in concentrations of hemoglobin (Hb), mean corpuscular Hb, and serum retinol, while the rates of anemia and VAD among the children fell from 17.6% to 13.2% and from 25% to 13.2%, respectively. In addition, the phagocytic capacity of neutrophils increased in the supplement group. The authors concluded that vitamin A supplementation could help to decrease the frequency of VAD and anemia, as well as to increase the immune response, in preschool children. [4]
Once ingested, provitamins A are released from proteins in the stomach. These retinyl esters are then hydrolyzed to retinol in the small intestine, because retinol is more efficiently absorbed. Carotenoids are cleaved in the intestinal mucosa into molecules of retinaldehyde, which is subsequently reduced to retinol and then esterified to retinyl esters. The retinyl esters of retinoid and carotenoid origin are transported via micelles in the lymphatic drainage of the intestine to the blood and then to the liver as components of chylomicrons. In the body, 50-80% of vitamin A is stored in the liver, where it is bound to the cellular RBP. The remaining vitamin A is deposited into adipose tissue, the lungs, and the kidneys as retinyl esters, most commonly as retinyl palmitate.
Vitamin A can be mobilized from the liver to peripheral tissue by a process of deesterification of the retinyl esters. In blood, vitamin A is bound to RBP, which transports it as a complex with transthyretin. The hepatic synthesis of RBP is dependent on the presence of zinc and amino acids to maintain its narrow serum range of 40-50 mcg/dL. Through a receptor-mediated process, the retinol is taken up by the peripheral tissues from the RBP-transthyretin complex.
VAD may be secondary to decreased ingestion, defective absorption and altered metabolism, or increased requirements. An adult liver can store up to a year's reserve of vitamin A, whereas a child's liver may have enough stores to last only several weeks. Serum retinol concentration reflects an individual's vitamin A status. Because serum retinol is homeostatically controlled, its levels do not drop until the body's stores are significantly limited. The serum concentration of retinol is affected by several factors, including RBP synthesis in the liver, infection, nutritional status, and the existing level of other nutrients, such as zinc and iron. [5]
In zinc deficiency, impaired synthesis of proteins occurs with rapid turnover (eg, RBP). In turn, this impairment affects retinol transport by RBP from the liver to the circulation and to other tissues. The mechanism by which iron affects vitamin A metabolism has not been identified, but randomized, double-blind studies have shown that vitamin A supplementation alone is not sufficient to improve VAD in the presence of coexisting iron deficiency.
The bioavailability of the carotenoids varies; this availability depends on absorption and on their yield of retinol. Only 40-60% of ingested beta carotene from plant sources is absorbed by the human body, whereas 80-90% of retinyl esters from animal proteins are absorbed. Carotenoid absorption is affected by dietary factors, including zinc deficiency, abetalipoproteinemia, and protein deficiency.
Because vitamin A is a fat-soluble vitamin, any GI diseases affecting the absorption of fats also affect vitamin A absorption. Patients with cystic fibrosis, sprue, pancreatic insufficiency, inflammatory bowel disorder (IBD), or cholestasis, as well as persons who have undergone small-bowel bypass surgery, are at increased risk for VAD. These patients should be advised to consume vitamin A.
One factor affecting the metabolism of vitamin A is alcoholism. Alcohol dehydrogenase catalyzes the conversion of retinol to retinaldehyde, which is then oxidized to retinoic acid. The affinity of alcohol dehydrogenase to ethanol impedes the conversion of retinol to retinoic acid.
Increased requirements of vitamin A most commonly occur among sick children. The American Academy of Pediatrics has recommended vitamin A supplementation for infants aged 6-24 months who are hospitalized with measles and for all hospitalized children older than 6 months. In the 1960s, the World Health Organization (WHO) undertook the first global survey of VAD with associated xerophthalmia and complicated measles. [6] In 1973, an international vitamin A board was set up to alleviate global malnutrition.
The WHO and the United Nations International Children's Emergency Fund (UNICEF) have issued joint statements recommending that vitamin A be administered to all children, especially those younger than 2 years, who are diagnosed with measles. Coexistent VAD in young children increases the risk of death. A Cochrane Database of Systematic Reviews article concluded that daily treatment with 200,000 IU of vitamin A for at least 2 days reduces mortality rates. [7, 8]
Another Cochrane Review article included 43 randomized trials representing 215,633 children, and provided strong support for the importance of vitamin A supplementation in preventing childhood mortality in children from 6 months to 5 years of age. [9] There is no evidence of vitamin A supplementation related reduction in mortality and morbidity for children one to six months of age. [10]
Pregnant women do not require increased vitamin A supplementation. [11] In fact, the Teratology Society advocates that women be informed of the possible risk of cranial neural crest defects and other malformations resulting from excessive use of vitamin A shortly before or during pregnancy. [12] The recommended daily allowance (RDA) of 800 mcg for all adult females is also appropriate for pregnant women, because their stores of vitamin A meet the fetal accretion rate. The requirements for lactating women have been debated, but the current RDA is 1300 mcg in the first 6 months and 1200 mcg in the second 6 months.
The RDAs of vitamin A for various age groups are as follows:
Infants aged 1 year or younger - 375 mcg
Children aged 1-3 years - 400 mcg
Children aged 7-10 years - 700 mcg
All males older than 10 years - 1000 mcg
All females older than 10 years - 800 mcg
Statistics from the US Centers for Disease Control and Prevention, based on a 1988-1991 survey, showed that age-specific intakes of carotenes were higher among males than females during that period and were higher among adults than children. [13] Significant differences in intake existed among different ethnic groups.
Clinical and subclinical VAD are problems in at least 75 countries. [14] In 1994, the WHO classified countries as having clinical or subclinical, severe, moderate, or mild VAD. Clinical VAD (in which children demonstrate ophthalmic signs and symptoms, including blindness) occurs mainly in countries in Southeast Asia and sub-Saharan Africa. [6] Severe VAD is also found in persons in refugee settlements and in displaced populations. [15, 16, 17]
A study by Stevens et al analyzed trends in the prevalence and mortality rate of vitamin A deficiency in children aged 6-59 months between 1991 and 2013 in low-income and middle-income countries. The study found that between 1991 and 2013, the prevalence of vitamin A deficiency in children in low-income and middle-income decreased from 39% to 29%. Prevalence was highest in sub-Sharan Africa (48%) and south Asia (44%) and 95% of deaths related to vitamin A deficiency occurred in these regions. [18]
Mortality/Morbidity
VAD is uncommon in the general population, but subgroups of patients suffering from fat malabsorption, cholestasis, or IBD or who have undergone small-bowel bypass may have subclinical deficiency with dark-adaptation abnormalities in the range of 60%. Vegans, persons with alcoholism, toddlers and preschool children living below the poverty line, and recent immigrants or refugees from developing countries all have increased risk of VAD secondary to decreased ingestion.
An estimated 250 million children are at risk for vitamin deficiency syndromes. The most widely affected group includes up to 10 million malnourished children, who develop xerophthalmia and have an increased risk of complications and death from measles. Each year, 250,000-500,000 children become blind because of VAD. Improving the vitamin A status of children with deficiencies (aged 6-59 mo) can reduce measles and diarrhea mortality rates by 50% and 33%, respectively, and can decrease risk rates from all causes of mortality by 23%. Routine distribution of vitamin A to children in endemic areas leads to a decrease of childhood mortality of 5-15%. A meta-analysis that included the DEVTA trial and eight other trials resulted in a modest mortality reduction of 11%. [19]
Combs GF Jr. Vitamins. Mahan LK, Escott-Stump S, eds. Krause's Food, Nutrition, and Diet Therapy. 10th ed. Philadelphia, Pa: WB Saunders; 2000.
Pazirandeh S, Burns DL. Overview of fat-soluble vitamins I. UpToDate. 2002.
Russell RM. Vitamin and trace mineral deficiency and excess. Braunwald E, Fauci A, Kasper D, et al, eds. Harrison's Principles of Internal Medicine. 15th ed. New York, NY: McGraw-Hill; 2001. vol 1: 465-6.
Jimenez C, Leets I, Puche R, et al. A single dose of vitamin A improves haemoglobin concentration, retinol status and phagocytic function of neutrophils in preschool children. Br J Nutr. 2009 Dec 15. 1-5. [Medline].
Munoz EC, Rosado JL, Lopez P, et al. Iron and zinc supplementation improves indicators of vitamin A status of Mexican preschoolers. Am J Clin Nutr. 2000 Mar. 71(3):789-94. [Medline]. [Full Text].
Reddy V. History of the International Vitamin A Consultative Group 1975-2000. J Nutr. 2002 Sep. 132(9 Suppl):2852S-6S. [Medline]. [Full Text].
D'Souza RM, D'Souza R. Vitamin A for treating measles in children. Cochrane Database Syst Rev. 2002. (1):CD001479. [Medline].
Gungor S, Olmez A, Pinar Arikan Firat, et al. Serum retinol and beta-carotene levels in subacute sclerosing panencephalitis. J Child Neurol. 2007 Mar. 22(3):341-3. [Medline].
Imdad A, Herzer K, Mayo-Wilson E, Yakoob MY, Bhutta ZA. Vitamin A supplementation for preventing morbidity and mortality in children from 6 months to 5 years of age. Cochrane Database Syst Rev. 2010 Dec 8. 12:CD008524. [Medline].
Imdad A, Ahmed Z, Bhutta ZA. Vitamin A supplementation for the prevention of morbidity and mortality in infants one to six months of age. Cochrane Database Syst Rev. 2016 Sep 28. 9:CD007480. [Medline].
McCauley ME, van den Broek N, Dou L, Othman M. Vitamin A supplementation during pregnancy for maternal and newborn outcomes. Cochrane Database Syst Rev. 2015 Oct 27. CD008666. [Medline].
Rothman KJ, Moore LL, Singer MR, et al. Teratogenicity of high vitamin A intake. N Engl J Med. 1995 Nov 23. 333(21):1369-73. [Medline]. [Full Text].
Alaimo K, McDowell MA, Briefel RR, et al. Dietary intake of vitamins, minerals, and fiber of persons ages 2 months and over in the United States: Third National Health and Nutrition Examination Survey, Phase 1, 1988-91. Adv Data. 1994 Nov 14. 1-28. [Medline]. [Full Text].
Lee V, Ahmed F, Wada S, et al. Extent of vitamin A deficiency among rural pregnant women in Bangladesh. Public Health Nutr. 2008 Jun 12. 1-6. [Medline].
West KP Jr, Mehra S. Vitamin A intake and status in populations facing economic stress. J Nutr. 2010 Jan. 140(1):201S-7S. [Medline].
Semba RD, de Pee S, Sun K, et al. The role of expanded coverage of the national vitamin A program in preventing morbidity and mortality among preschool children in India. J Nutr. 2010 Jan. 140(1):208S-12S. [Medline].
Williams IO, Essien EU, Eka OU. Socioeconomic Factors and Vitamin A Status of Pregnant Women in Calabar Urban, Southeastern Nigeria. Matern Child Health J. 2009 Nov 25. [Medline].
Stevens GA, Bennett JE, Hennocq Q, Lu Y, De-Regil LM, Rogers L, et al. Trends and mortality effects of vitamin A deficiency in children in 138 low-income and middle-income countries between 1991 and 2013: a pooled analysis of population-based surveys. Lancet Glob Health. 2015 Sep. 3 (9):e528-36. [Medline].
Awasthi S, Peto R, Read S, Clark S, Pande V, Bundy D. Vitamin A supplementation every 6 months with retinol in 1 million pre-school children in north India: DEVTA, a cluster-randomised trial. Lancet. 2013 Apr 27. 381(9876):1469-77. [Medline].
Al-Saleh I, El-Doush I, Billedo G. Age and gender-related reference values for serum dl-alpha-tocopherol and all-trans-retinol levels in Saudi population. Int J Vitam Nutr Res. 2007 Sep. 77(5):326-35. [Medline].
Gomes MM, Saunders C, Ramalho A. Placenta: a possible predictor of vitamin A deficiency. Br J Nutr. 2009 Dec 15. 1-5. [Medline].
de Pee S, Dary O. Biochemical indicators of vitamin A deficiency: serum retinol and serum retinol binding protein. J Nutr. 2002 Sep. 132(9 Suppl):2895S-901S. [Medline]. [Full Text].
Weinman AR, Jorge SM, Martins AR, et al. Assessment of vitamin A nutritional status in newborn preterm infants. Nutrition. 2007 Jun. 23(6):454-60. [Medline].
Gorstein JL, Dary O, Pongtorn, et al. Feasibility of using retinol-binding protein from capillary blood specimens to estimate serum retinol concentrations and the prevalence of vitamin A deficiency in low-resource settings. Public Health Nutr. 2008 May. 11(5):513-20. [Medline].
Russell RM. The vitamin A spectrum: from deficiency to toxicity. Am J Clin Nutr. 2000 Apr. 71(4):878-84. [Medline]. [Full Text].
Mayo-Wilson E, Imdad A, Herzer K, Yakoob MY, Bhutta ZA. Vitamin A supplements for preventing mortality, illness, and blindness in children aged under 5: systematic review and meta-analysis. BMJ. 2011 Aug 25. 343:d5094. [Medline]. [Full Text].
Dietary Guidelines for Americans, 2005. Department of Health and Human Services, Department of Agriculture. Jan 2005. Available at http://www.health.gov/dietaryguidelines/dga2005/document/default.htm.
Dietary Supplement Fact Sheet: Vitamin A and Carotenoids. National Institutes of Health, Office of Dietary Supplements. Apr 23 2006. Available at http://ods.od.nih.gov/factsheets/vitamina.asp.
Fletcher RH, Fairfield KM. Vitamin supplementation in disease prevention II. UpToDate. 2002.
Fairfield KM, Fletcher RH. Vitamins for chronic disease prevention in adults: scientific review. JAMA. 2002 Jun 19. 287(23):3116-26. [Medline]. [Full Text].
Willett WC, Hunter DJ. Vitamin A and cancers of the breast, large bowel, and prostate: epidemiologic evidence. Nutr Rev. 1994 Feb. 52(2 Pt 2):S53-9. [Medline].
George Ansstas, MD Assistant Professor of Medicine, Division of Medical Oncology, Washington University School of Medicine
George Ansstas, MD is a member of the following medical societies: American Medical Association
N Gopalswamy, MD Chairman, Professor of Internal Medicine, Department of Gastroenterology, Wright State University, Boonshoft School of Medicine, Veterans Affairs Medical Center
Jigna Thakore, MD Fellow, Department of Gastroenterology, Dayton Veterans Administration Medical Center
Jigna Thakore, MD is a member of the following medical societies: American College of Gastroenterology, American Society for Gastrointestinal Endoscopy
Romesh Khardori, MD, PhD, FACP Professor of Endocrinology, Director of Training Program, Division of Endocrinology, Diabetes and Metabolism, Strelitz Diabetes and Endocrine Disorders Institute, Department of Internal Medicine, Eastern Virginia Medical School
Romesh Khardori, MD, PhD, FACP is a member of the following medical societies: American Association of Clinical Endocrinologists, American College of Physicians, American Diabetes Association, Endocrine Society
George T Griffing, MD Professor Emeritus of Medicine, St Louis University School of Medicine
George T Griffing, MD is a member of the following medical societies: American Association for the Advancement of Science, International Society for Clinical Densitometry, Southern Society for Clinical Investigation, American College of Medical Practice Executives, American Association for Physician Leadership, American College of Physicians, American Diabetes Association, American Federation for Medical Research, American Heart Association, Central Society for Clinical and Translational Research, Endocrine Society
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text:
United States sees unprecedented fall in deaths from cancer
By Sally Robertson, B.Sc.Jan 9 2020
The death rate from cancer in the United States has seen the greatest fall in the year 2016-2017 since rates started to fall in 1992, says the American Cancer Society (ACS).
Image Credit: fusebulb / Shutterstock.com
According to a new ACS report, the overall drop in the death rate from cancer has mainly been driven by accelerating advances in the treatment of lung cancer.
Lead author of the report, Rebecca Siegel, says: "The big news is that despite slowing progress for other leading causes of death, the cancer death rate is not only continuing to drop but had the largest one-year drop ever from 2016 to 2017."
"This is driven by accelerating progress against lung cancer," she added
The decelerated rate of death from lung cancer drove an overall decline in cancer mortality of 2.2% between 2016 and 2017.
There was also a dramatic decline in death from melanoma, the deadliest form of skin cancer, according to the report.
Experts attribute progress to treatment advances
The U.S. has seen the falls in death rates for lung cancer and melanoma since the U.S. Food and Drug Administration approved new therapies for metastatic disease (cancer that has spread to other parts of the body).
In an op-ed piece published in the journal Cancer Letter, Professor of Oncology and Epidemiology at Johns Hopkins University and former chief medical officer at the ACS, Otis Brawley, highlights the advancements that have been made in lung cancer treatment and the impact that this has had on cancer mortality rates.
"As I look through just-published tables of age-adjusted cancer mortality, I recognize an unprecedented development," he writes. "Immunotherapy is showing such a dramatic impact in the treatment of locally advanced and advanced non–small cell lung cancer that this effect elevates the statistics for all lung cancer and ― this I find astonishing ― you can even see its effect in age-adjusted cancer mortality overall."
Chief medical and scientific officer for the ACS, William Cance, also attributed the decline in mortality from lung cancer and melanoma to the progress in treatment made over the last decade:
"They are a profound reminder of how rapidly this area of research is expanding, and now leading to real hope for cancer patients."
To what extent has the death rate slowed?
Between 2008 and 2013, the lung cancer death rate slowed by 3% per year for men and by 2% per year for women. The rates then slowed again between 2013 and 2017. However, at the same time, declines in death rates from breast and colorectal cancers have slowed.
The news this year is mixed. The exciting gains in reducing mortality for melanoma and lung cancer are tempered by slowing progress for colorectal, breast and prostate cancers, which are amenable to early detection."
Rebecca Siegel, American Cancer Society
Lung cancer leading cause of death in U.S.
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Accounting for as much as 27% of all cancer deaths, lung cancer is the leading cause of cancer death in the U.S. and causes more fatalities than breast, prostate, colorectal, and brain cancers combined.
Historically, lung cancer plagued the U.S. for most of the last century. According to the Centers for Disease Control and Prevention (CDC), Americans have been stopping smoking in increasing numbers ever since the Surgeon General Report declared in 1964 that smoking causes lung cancer. However, the death rate from cancer still continued to increase.
Almost three decades after the release of the report, the death rate from cancer finally started to fall in 1992. And compared with 1992, deaths from cancer in 2017 have fallen by 29%, which represents an estimated 2,902,200 fewer cancer deaths, says the ACS.
"This steady progress is largely due to reductions in smoking and subsequent declines in lung cancer mortality, which have accelerated in recent years," writes Siegel and colleagues.
In recent years, many people have taken up vaping, as e-cigarettes have increasingly been marketed as a safer option than conventional cigarettes. A sharp rise in the uptake of vaping and a surge in vaping-related illnesses has sparked concerns amongst public health authorities, and the American Lung Association has warned that "e-cigarettes are not safe."
However, the ACS still advises that vaping is "likely to be significantly less harmful for adults than smoking regular cigarettes."
While the CDC has advised that people who are worried about health risks should avoid vaping, it also advises that anyone who has managed to stop smoking through using e-cigarettes should not go back to traditional cigarettes.
Posted in: Medical Research News | Medical Condition News | Disease/Infection News
Tags: Brain, Cancer, Cancer Treatment, Cell, Colorectal, Epidemiology, Immunotherapy, Lung Cancer, Melanoma, Mortality, Oncology, Prostate, Public Health, Research, Skin, Skin Cancer, Small Cell Lung Cancer, Smoking, Vaping
Sally Robertson
Sally has a Bachelor's Degree in Biomedical Sciences (B.Sc.). She is a specialist in reviewing and summarising the latest findings across all areas of medicine covered in major, high-impact, world-leading international medical journals, international press conferences and bulletins from governmental agencies and regulatory bodies. At News-Medical, Sally generates daily news features, life science articles and interview coverage.
Please use one of the following formats to cite this article in your essay, paper or report:
Robertson, Sally. (2020, January 09). United States sees unprecedented fall in deaths from cancer. News-Medical. Retrieved on January 20, 2020 from https://www.news-medical.net/news/20200109/United-States-sees-unprecedented-fall-in-deaths-from-cancer.aspx.
Robertson, Sally. "United States sees unprecedented fall in deaths from cancer". News-Medical. 20 January 2020. <https://www.news-medical.net/news/20200109/United-States-sees-unprecedented-fall-in-deaths-from-cancer.aspx>.
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Robertson, Sally. 2020. United States sees unprecedented fall in deaths from cancer. News-Medical, viewed 20 January 2020, https://www.news-medical.net/news/20200109/United-States-sees-unprecedented-fall-in-deaths-from-cancer.aspx.
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Frequently Asked Questions about Social Security Disability.
When it comes to disability, looks can be very deceiving. There are many people who look healthy, but can meet Social Security's definition of disabled.
If your disability claim was denied, keep in mind that the Social Security Administration (SSA) denies the majority of claims, particularly at the application and reconsideration levels. It's possible your medical evidence was not presented properly or your application wasn't handled correctly. At Sackett and Associates we can help you appeal your denial.
To qualify for benefits, you must have worked in and paid into Social Security. You must also have a medical condition that meets Social Security's definition of disability. In general, the Social Security Administration (SSA) pays monthly benefits to people who are unable to work for a year or more because of their disability.
Yes. Mental illness is a frequently used basis for getting Social Security Disability benefits.
My doctor says I am disabled so why is Social Security denying my disability claim?
It is not up to your doctor to determine whether or not you are disabled. It is up to Social Security to make their own decision regardless of what your doctor thinks. But if we represent you, we will contact your doctor to get the information necessary to win your case.
Social Security Disability Insurance Benefits (DIB) is for people who suffer with a disability and can no longer work. You may qualify for this program if you have worked a long time and paid sufficient FICA taxes. But, you may still be denied these benefits if SSA determines you are not medically disabled.
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text:
\section{Introduction}

The idea to represent $m$-dimensional vectors by a single string of digits  can be traced back to the work of  A. Vince \cite{Vince93a}  and \cite{Vince93b} who showed  that for any expansive  matrix  $M \in \mathbb{Z}^{m\times m}$   there exists a digit set $\mathcal{D}\subset \mathbb{Z}^m$ such that any integer vector $x \in \mathbb{Z}^m$ can be written in the form   $x= \sum_{k=0}^nM^k {d}_k$, where $d_k\in \mathcal{D}$. In other words, the whole $\mathbb{Z}^m$ is representable in the matrix numeration system $(M,\mathcal{D})$.
On the other hand, if a matrix $M$ has an eigenvalue inside the unit circle, no choice of the digit set $\mathcal{D}\subset \mathbb{Z}^m$ allows to represent all integer vectors as a combination of non-negative powers of $M$ only.

 Many works devoted to positional representations of elements of a commutative finitely generated ring can be interpreted as a special case of the matrix numeration systems.   From this point of view, the  history of the matrix numeration systems had started several decades before the year 1993, the year Vince's results have been published.
 Positional number systems used to represent  the Gaussian integers or,  more general,   elements of a ring of integers in a quadratic number field can be viewed as a matrix number system given by  a  $2\times 2$ matrix.  Number systems of this type   were studied by Penney \cite{Penney}, K\'atai and Szab\'o \cite{KataiSzabo75},  K\'atai and B. Kov\'acs,  \cite{KataiKovacs80}, \cite{KataiKovacs81},  Gilbert \cite{Gilbert81}.
This concept was extended to algebraic fields of higher order by  B. Kov\'acs \cite{KovacsBela81} and   B. Kov\'acs and Peth\H{o} \cite{KovacsPetho83}. A number system with an algebraic base $\beta$ can be interpreted as a matrix number system with the base being the companion matrix of the minimal polynomial of $\beta$. Hence the characteristic polynomial of the matrix is irreducible over $\mathbb{Q}$.
The concept of the so-called canonical number system for irreducible polynomials   was further  generalized to arbitrary  polynomials from  $\mathbb{Z}[x]$
by Peth\H{o} \cite{Petho91}.
The dynamic properties of canonical number systems can be well studied in the formalism of the shift radix systems introduced in \cite{ABBPT05}.
A very general setting  of the canonical number systems was recently  considered in  \cite{EvertseSpol19}.

Our aim is to study the matrix numeration systems. The most important property of a square   matrix  can be deduced from its Jordan  form.
Let us point out that the Jordan form of the companion matrix associated to  a polynomial  has a specific property: only one Jordan block corresponds  to each  eigenvalue. Therefore the study of  matrix numeration systems can display  new phenomena and  is of its own interest.  The  matrix formalism for numeration systems -- under the name {\it numeration systems in lattices}  --  was systematically used  by A. Kov\'acs in  \cite{KovacsAttila03}.  Of course,  A. Kov\'acs, just like Vince, considers integer matrices as they  map  a lattice into itself.
The Vince's results on integer matrices were recently generalized  by J.  Jankauskas and  J. Thuswaldner to matrices   $M \in   \mathbb{Q}^{m\times m}$ with rational entries and without eigenvalues in modulus strictly smaller than 1, see \cite{JT}.

In this  paper,   we generalize the Vince's result in another direction:  our  representation of an integer  vector  can use both positive and negative powers of a matrix $M \in \mathbb{Z}^{m\times m}$ with  $\det M = \Delta \neq 0$. Our aim is to study the set  $$  {\rm Fin}_{\mathcal{D}}(M)= \Bigl\{\sum_{k\in I}M^k {d}_k : I  \text{ finite non-empty subset of } \mathbb{Z} \text{ and } {d}_k \in \mathcal{D} \text{ for each }  k \in I\Bigr\} \subset \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m} \,.
$$
Notice that because the set of positions $I$ is finite, we can treat the position not in $I$ as being occupied by the zero digit. Therefore we always assume that zero is in the digit set $\mathcal D.$
We show that for any non-singular matrix $M \in \mathbb{Z}^{m\times m}$ there exists  a finite set  $\mathcal{D}\subset \mathbb{Z}^m$ such that
$  \mathbb{Z}^{m} \subset   {\rm Fin}_{\mathcal{D}}(M)$.
 For such $\mathcal D$, we also characterise  the matrices $M$  for which  the sets  $ {\rm Fin}_{\mathcal{D}}(M)$   and $ \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m}$  coincide.

\section{The main result}
This section contains the main result and its proof that relies heavily on Proposition~\ref{skoroHotovo}. The proof of this proposition is, however, quite technical, therefore we provide it in its own Section \ref{sectionProposition}.

The method of proving the main theorem is distinct from the one used in \cite{JT}. Their setting allowed the problem to be reduced to a problem in canonical number systems. This approach does not work in our case because we allow eigenvalues that are smaller than one in absolute value.

\begin{thm}\label{main}  Let  $M \in \mathbb{Z}^{m\times m}$  be a non-singular matrix. Then there exists a finite digit set $\mathcal{A}\subset \mathbb Z^m$ such that any vector $z \in \mathbb Z^m$ can be written in the form
$$
z = \sum_{k\in I}  M^kd_k, \ \  \text{ where  }\quad \emptyset\neq I \subset \mathbb Z, \ I \text{\ finite\ \ and }\  d_k \in \mathcal{A} \ \text{for each } k \in I.
$$
\end{thm}

The property $\mathbb Z^m\subseteq \mathrm{Fin}_{\mathcal D}(M)$ easily implies a stronger property. Because $\mathrm{Fin}_{\mathcal D}(M)$ is closed under multiplication by $M^i$ for $i\in\mathbb Z,$ we easily obtain the following corollary.

\begin{coro}
Let $\mathbb Z^m\subseteq \mathrm{Fin}_{\mathcal D}(M)$, then we have $\bigcup_{n\in\mathbb Z}M^n\mathbb Z^m=\mathrm{Fin}_{\mathcal D}(M)$.

In particular, for each $M\in\mathbb Z^{m\times m}$ non-singular, there exists a finite digit set $\mathcal D\subset \mathbb Z^m$, such that $\mathrm{Fin}_{\mathcal D}(M)=\bigcup_{n\in\mathbb Z}M^n\mathbb Z^m$.
\end{coro}

The full proof of Theorem~\ref{main} is rather technical. In order to simplify readability, we put the technical part into Proposition~\ref{skoroHotovo} whose proof will be the content of Section~\ref{sectionProposition}. Nevertheless, we can now present the main idea of our work.

For $\mathcal D\subset\mathbb Z^n$ finite, define $T_d(x) = Mx-d$ and $x_k= T_d(x_{k-1})$ with $x_0 = x.$ We directly obtain
\begin{equation}\label{finite}
    x = M^{-1}d_1 + M^{-2}d_2 + \dots + M^{-k}d_k + M^{-k}x_k.
\end{equation}
Assume that $S\subseteq\mathbb Z^m$ consists of vectors with the property that there exist $d_1,d_2,\dots$ such that $\|x_i\|<C$ eventually (with the constant being universal for the whole $S$). Then the digit set $\mathcal A = \mathcal D+\{d\in\mathbb Z^n : \|d\|<C\}$ suffices for \eqref{finite} to be a finite representation of members of $S$ over $\mathcal A$.
It then suffices to show that for any $x\in\mathbb Z^m$, $M^{-n}x\in S$ for some $n\in\mathbb N$. Indeed, we would have $M^{-n}x = \sum_{i=1}^k M^{-i}d_i,$ i.e. $x = \sum_{i=1}^k M^{n-i}d_i.$

In the following, we will show that an appropriate choice of $S$ is all the elements of $\mathbb Z^m$ whose projection into the expansive eigenspace is bounded by a constant depending on $M$ only. Such a choice ensures that $M^{-n}x\in S$ is achievable for all $x\in\mathbb Z^m$. It then remains to be shown that with a suitable digit set,
a sequence of iterations $T_{d_n}T_{d_{n-1}}\cdots T_{d_1} (x)$  will have a bounded norm eventually for any $x\in S$. The main obstacle are unimodular eigenvalues of $M$, in particular those with different algebraic and geometric multiplicity.

Our  proof uses the real  Jordan form $J$ of the matrix  $M $.
If  $M= P^{-1}JP$,

\begin{equation}\label{prevod}{\text{ the equality  \   $z=  \sum_{k\in I}  M^kd_k $ \ \    gives  \ \     $ Pz =\sum_{k\in I}
J^{k}Pd_k $}}.\end{equation}

Instead of looking for a finite digit set from the lattice $\mathbb Z^m$ which is  suitable for the matrix $M$ and representation of integer vectors,  we look for   a digit set  from the lattice $P\mathbb Z^m$, suitable for $J$ and we represent lattice points of   $P\mathbb Z^m$.  The matrix $P$ transforming $M$ to  its Jordan form  is not determined uniquely. We choose it carefully to approximate in some sense   the lattice $\mathbb Z^m$.    For this purpose,  we introduce the perturbation set
$$\mathcal{E} = \{ \varepsilon  \in \mathbb R^m :  \|\varepsilon \|_\infty <\tfrac13\}.$$
In the article we work with two norms of $\mathbb R^m$:

$\| .\|_\infty$  denotes the  norm   defined by $\|(x_1, x_2,  \ldots, x_m)\|_\infty = \max_{i}|x_i|$ .

 $\| .\|_2$  denotes the Euclidean norm, i.e. $\|(x_1, x_2,  \ldots, x_m)\|_2 = \sqrt{\sum_{j=1}^m|x_j|^2}.$

\begin{defi}\label{close} We say that a lattice $L \subset \mathbb R^m$ is close to the lattice $\mathbb Z^m$, if   \ $\mathbb Z^m \subset  L +\mathcal{E}$. \end{defi}

For every $M\in\mathbb R^{m\times m}$, there exists a non-singular $P\in\mathbb R^{m\times m}$ such that $PMP^{-1}M= J ={\bigoplus_{k}J_k}$ where $J_k$ is a real Jordan block.   Let us recall that  the real Jordan block to $\lambda \in \mathbb C$  is a matrix in the form
$$
\left(\begin{smallmatrix}
R&I&& \\[-7pt]
&R&\ddots& \\\vspace*{-1mm}
&&\ddots&I \\[2mm]
&&&R
\end{smallmatrix}\right)\quad
\qquad \text{with} \
R=\begin{cases}
             \ (\lambda) & \mbox{ if } \lambda\in\mathbb R, \\[2mm]
             \left(\begin{smallmatrix}
               a & b \\
               -b & a
             \end{smallmatrix}\right) &\mbox{ if }  \lambda=a+ib \in\mathbb{C}  \setminus \mathbb{R}
           \end{cases}
$$
Note that $I$ is a unit matrix of order 1 or 2, according to the order of $R$.

The  vector space $\mathbb R^m $ can be decomposed  into  invariant subspaces $\mathbb R^m = V_e\oplus V_u \oplus V_c$ of the matrix  $J$  such that  $J$ restricted to $V_e$ is expansive,
 $J$ restricted to $V_c$ is contractive and $J$ restricted to $V_u$ is orthogonal.  In other words,  all eigenvalues of  $J$ restricted to $V_e$ are in modulus $>1$,  all eigenvalues of  $J$ restricted to $V_u$ are in modulus $=1$,  and  all eigenvalues of  $J$ restricted to $V_c$ are in modulus $<1$.
The subspaces $V_e, V_u$ and $V_c$ may be trivial.   Note that  $\dim V_e  + \dim V_u > 0$ as $M$ is an integer non-singular matrix.
 Any $x \in \mathbb R^m$   can be uniquely written as $x = x_e+x_u+x_c$ with $x_e \in V_e$,  $x_u \in V_u$ and $x_c\in V_c$.   We will denote $x_e =\pi_e(x)$, $x_u =\pi_u(x)$,  and $x_c =\pi_c(x)$,

\bigskip

The proof of the main theorem  will be a consequence of the following proposition.

\begin{prop}\label{skoroHotovo} Let $J\in \mathbb R^{m\times m}$ be a non-singular  matrix in the real Jordan form     and  $\pi_e: \mathbb R^m\to V_e$  be the projection  into the expansive  invariant subspace $V_e$ of  $J$. Let  $L \subset \mathbb R^m$ be a lattice close to  $\mathbb Z^m$. Then   there exists  a finite set   $\mathcal{D} \subset L$  and a constant $C$  such that

\medskip
\noindent  for any
 $x \in \mathbb R^m$  with   $\|\pi_e(x)\|_\infty \leq 1$   and for any sufficiently large $N\in \mathbb N$ we  can write
\begin{equation}\label{Nzmensi}  x = J^{-1}d_1 + J^{-2}d_2  + \cdots + J^{-N}d_N + J^{-N}y,\  \text{ \ for some }      d_1,d_2, \ldots, d_N \in \mathcal{D}\,    \end{equation}
$$ \text{and}    \ \ y \in \mathbb R^m, \ \ \text{with} \ \  \|y\|_\infty \leq C\,.$$
\end{prop}

\begin{proof}[Proof of Theorem \ref{main}]
We  use the statement and notation of Proposition  \ref{skoroHotovo}.     Let  $M =P^{-1}JP$, where $J$ is  the real Jordan canonical form of $M$. The  matrix $P$ is not given uniquely. For example,  any matrix $\alpha P$ with a non-zero $\alpha$ transforms $M$ to its Jordan form, as well. Therefore we can  assume without loss of generality that   the lattice $L=P\mathbb Z^m$ is close to the lattice $\mathbb Z^m$ in the sense of Definition \ref{close}.   As $P^{-1}JP=M$, we get $JP\mathbb Z^m =  PM\mathbb Z^m \subset P\mathbb Z^m$, i.e., $J$ maps the lattice  $L$ into $L$.
For such   $J$ and $L$, we find by  Proposition    \ref{skoroHotovo}, the digit set $\mathcal{D} \subset L$.

   Due to  \eqref{prevod},   we have to show that  there exists a finite  digit set $\mathcal{A}\subset L$  such that
\begin{equation}\label{jinak} \text{ $z \in  L\qquad \Longrightarrow   \quad    \ \
z= \sum_{k\in I}
J^{k}d_k $,  \quad where  $I\subset \mathbb Z$ is finite and  $d_k \in \mathcal{A}$  \ for each $k\in I$.}\end{equation}
 \noindent  Define  $$\mathcal{A} := \mathcal{D} + \mathcal{B}, \quad \text{  where }\
 \mathcal{B} := \{ d \in L :   \|d \|_\infty  \leq C\}.$$

 Obviously,   $\mathcal{B}$ is finite   and  $ 0 \in \mathcal{B}$. Therefore $\mathcal{A}$ is  finite  and  $\mathcal{D}\subset \mathcal{A} \subset L$.    Let us show that our  choice  of
the digit set  $\mathcal{A}$  has the property stated in \eqref{jinak}.

Let    $z \in  L $.     Since $V_e$ is a contractive subspace of the matrix $J^{-1}$, there exists  $j \in \mathbb N$ such that   $\|\pi_e(J^{-j}z)\|_\infty < 1$. Applying  \eqref{Nzmensi} to $x =J^{-j}z$ with  $N\geq j$,    we find $d_1, \ldots, d_n \in \mathcal{D}\subset \mathcal{A}$ and $y$ with  $\|y\|_\infty \leq C$  such that
\begin{equation}\label{cele}J^{-j}z =   J^{-1}d_1 + J^{-2}d_2  + \cdots + J^{-N}d_n + J^{-N}y. \end{equation}

Multiplying \eqref{cele} by $J^N$,  we deduce  $y +d_N = J^{N-j}z - \sum_{k=1}^NJ^{N-k}d_k $.
 As  $J$ maps the lattice  $L$ into $L$,  we have  $y\in L$ and $y \in   \mathcal{B}$.   Obviously,  $y+d_n \in \mathcal{A}$.      Altogether,  $z = \sum_{k=1}^{j-1}J_{-k}d_k + M^{j-N}(d_N+y)$,   and all the coefficients $d_1d_2, \ldots, d_{N-1}$ and $d_{N}+y$ belong to $\mathcal{A}$.

\end{proof}

\section{Proof of Proposition \ref{skoroHotovo}}\label{sectionProposition}
The proof of Proposition \ref{skoroHotovo} will be done in following way. To the matrix $J$ in the real Jordan form, we  find a  finite digit set $\mathcal D\subset L$, such that the iterations of the transformations $Jx-d$ for some $d\in\mathcal D$, have small norm eventually.

First  we find a digit set  in $\mathbb Z^m$ and
then we  replace  each integer digit by a close element   from  the lattice $L$.
When working with  the matrix $J$, the  integer lattice has an important advantage. It can be decomposed  into the direct sum  $\mathbb Z^m = \oplus_{j} \mathbb Z^{m_j}$   where each  lattice   $ \mathbb Z^{m_j}$ is contained in an invariant subspace of a real  Jordan block.  Therefore, one can treat each Jordan block separately.

\subsection{A real Jordan block to an eigenvalue  $\lambda$ on the unit circle.}

Let us assume first that  $\lambda$ is not real. The real Jordan block to such a $\lambda$ has an even size, say $2\ell$,   and the form
\begin{equation}\label{rotace}
J=\left(\begin{smallmatrix}
R&I&& \\[-7pt]
&R&\ddots& \\\vspace*{-1mm}
&&\ddots&I \\[2mm]
&&&R
\end{smallmatrix}\right) \in \mathbb{R}^{2\ell \times 2\ell}\quad
\qquad \text{with } \
R= \left(\begin{smallmatrix}
               \cos \varphi& - \sin\varphi  \\
              \sin\varphi& \cos \varphi
             \end{smallmatrix}\right)   \
\ \  \text{and } \
I= \left(\begin{smallmatrix}
               1& 0  \\
              0& 1
             \end{smallmatrix}\right).
\end{equation}

\begin{defi}\label{defIndex} Let  $K_1, K_2, \ldots, K_\ell$ be positive constants and $J$ the matrix given in \eqref{rotace}.  We define the index
${\rm ind}: \mathbb{R}^{2\ell} \mapsto \{0,1,\cdots, \ell\}$  in the following way.  For  $x = (x_1,x_2, \ldots, x_{2\ell - 1}, x_{2\ell} )^T $, we put
\begin{itemize}
\item
  ${\rm ind}(x) = 0$,\  if \   $  \|(x_{2i-1}, x_{2i})\|_2 < K_i$   for all $i = 1,2, \ldots, \ell$;

\item
 otherwise,    \  ${\rm ind}(x) = $ the maximal index $j$   such that $  \|(x_{2j-1}, x_{2j})\|_2 \geq  K_j$.

\end{itemize}
 \end{defi}

First we  show two simple claims.

\begin{claim}\label{claim1} Given $q \in \mathbb N,$  denote $\mathcal{B} = \{ (0, 0)^T, (3q, 0)^T,   (0,3q)^T, (-3q, 0)^T,   (0,-3q)^T \}\subset \mathbb R^2$.   Then for each $z \in \mathbb R^2$ there exists $b \in \mathcal{B}$ such that
$$
\|z - b\|_2 \leq 6q \qquad \text{or} \qquad \|z - b\|_2 \leq \|z \|_2 - q. \
$$
\end{claim}
\begin{proof}  Because of the  symmetry of the digit set  $\mathcal{B}$,  it is enough to consider $z=(z_1,z_2)^T \in \mathbb R^2$ with $z_1\geq z_2\geq 1$.

\medskip

If $z_1\geq 3q$, put $b =(3q, 0)^T$. Then $\|z - b\|_2 = \sqrt{(z_1 - 3q)^2 + z_2^2}$. The inequality  $ \|z - b\|_2 \leq \|z \|_2 - q$  we want to show is equivalent to  $4q + \sqrt{z_1^2+z_2^2} < 3z_1$. The last inequality can be easily checked, since $4q < (3-\sqrt{2})z_1$  and  $\sqrt{z_1^2+z_2^2} \leq  \sqrt{2}z_1$.
\medskip

If $z_1<3q$, put $b =(0,  0)^T$. Then  $\|z - b\|_2 = \sqrt{z_1^2 + z_2^2}\leq \sqrt{(3q)^2 + (3q)^2} \leq 6q$.
\end{proof}
\begin{claim}\label{claim2} Let a constant $c_1\geq 0$ and $ R= \left(\begin{smallmatrix}
               \cos \varphi& - \sin\varphi  \\
              \sin\varphi& \cos \varphi
             \end{smallmatrix}\right)   \in \mathbb R^{2\times 2}$ be given.
             Then there exist  a constant $c_2>0$ and $\mathcal{B}  \subset \mathbb Z^2$ with $\# \mathcal{B} = 5$ such that  for any $u, v \in \mathbb R^2$, $ \|v\|_2 \leq c_1,$ there exists $b \in \mathcal{B}$ such that
$$\|Ru + v - b\|_2 < c_2 -\tfrac12 \qquad \text{or}\qquad \|Ru + v - b\|_2 < \|u\|_2 - 1.$$
\end{claim}
\begin{proof} Find  $q \in \mathbb N, q\geq c_1+1$.  For this  $q$,  Claim \ref{claim1} gives us  $\mathcal{B}$. Denote $c_2 = 6q+\tfrac12$.
Any rotation $R$ preserves the euclidean norm, i.e.,  $\|Ru\|_2  = \|u \|_2$.   According to Claim \ref{claim1}, for each vector $z =Ru + v$ we find $b \in \mathcal{B}$ such that
$$\|Ru + v - b\|_2 < 6q = c_2-\tfrac12  \quad \text{or}\quad \|Ru + v - b\|_2 < \|Ru + v \|_2 - q \leq  \|u \|_2 + \|v \|_2 - q \leq   \|u \|_2 + \underbrace {c_1 - q}_{\leq -1}\,.$$

\end{proof}

\begin{lem}\label{unit}  Let $J$ be the matrix given in \eqref{rotace}. There exist a finite digit set $\mathcal{D} \subset \mathbb Z^{2\ell}$ and constants $K_1,\dots,K_\ell$, such that the function ${\rm ind}$ defined by $K_1,\dots,K_\ell$ has the following property.

For each $x \in  \mathbb R^{2\ell}$ there exists a digit  $ d \in \mathcal{D}$  such that for all  $\varepsilon  \in  \mathcal{E}  $,   the vector $y= Jx-d+ \varepsilon$  satisfies
\medskip

\begin{itemize}

\item $ {\rm ind}(y) \leq  {\rm ind}(x) $;
\medskip

\item   If    \ $ {\rm ind}(y) = {\rm ind}(x) = j $, then $  \|(y_{2j-1},y_{2j})\|_2 \leq   \|(x_{2j-1}, x_{2j})\|_2 -\tfrac12$.

\end{itemize}
\end{lem}

\medskip

\begin{proof}

Consider the matrix   $J \in \mathbb R^{2\ell \times 2\ell}$  given in \eqref{rotace}. Using $\ell$  times Claim \ref{claim2} , we will construct $\ell$ constants $K_1, \ldots, K_\ell$ to determine  the function ${\rm ind}$:

by $K_\ell$  and $\mathcal{B}_\ell$   we denote the constant $c_2$ and the digit set found by Claim \ref{claim2} for $c_1 = 0$;

by $K_{\ell-1}$  and $\mathcal{B}_{\ell-1}$   we denote the constant $c_2$ and the digit set found by Claim \ref{claim2}  for $c_1 = K_\ell$;

by $K_{\ell-2}$  and $\mathcal{B}_{\ell-2}$   we denote the constant $c_2$ and the digit set found by Claim \ref{claim2}  for $c_1 = K_{\ell-1}$;

\noindent etc.

\medskip

The  digit set $\mathcal{D}$  is defined by
\begin{equation}\label{abecedaZB}
d\in {\mathcal{D}} \qquad \text{ if and only if } \qquad
d=\left(\begin{smallmatrix}
b^{(1)}\\\vspace*{-2mm}
b^{(2)} \\\vspace*{-1mm}
\vdots\\[2mm]
b^{(\ell)}
\end{smallmatrix}\right) \in \mathbb{Z}^{2\ell}, \ \
\text{where} \ \ b^{(k)}  \in \mathcal{B}_k, \text{ for }  k=1, 2, \ldots, \ell.
\end{equation}

We show that the function ${\rm ind}$ defined by the constants    $K_1, \ldots, K_\ell$  and the digit set $\mathcal{D}$ have the property declared in the statement of the lemma.

Let $x \in \mathbb R^{2\ell}$.
Then $$\left(\!\!\!\begin{array}{l}
(Jx)_{2i-1}\\%\\\vspace*{-2mm}
(Jx)_{2i}
\end{array}\!\!\!\right) = R \left(\!\!\!\begin{array}{l}
x_{2i-1}\\%\\\vspace*{-2mm}
x_{2i}
\end{array}\!\!\!\right) + \left(\!\!\!\begin{array}{l}
x_{2i+1}\\%\\\vspace*{-2mm}
x_{2i+2}
\end{array}\!\!\!\right), \ \ \ \text{ for $i = 1,2, \ldots, \ell$, where we put}\ \    \left(\!\!\!\begin{array}{l}
x_{2\ell+1}\\%\\\vspace*{-2mm}
x_{2\ell+2}
\end{array}\!\!\!\right)  =  \left(\!\!\!\begin{array}{l}
0\\%\\\vspace*{-2mm}
0
\end{array}\!\!\!\right).$$
To this  $x$, we define $d \in \mathcal{D}$ by determining  all its components $b^{(i)}$.

\begin{itemize}
\item
If  $i <   {\rm ind}(x)$,  then we choose $b^{(i)} \in \mathcal{B}_i$ randomly.

\item

If $i\geq   {\rm ind}(x)$, then $ \|(x_{2i+1}, x_{2i+2})\|_2 < K_{i+1}$ and  $b^{(i)} \in \mathcal{B}_i$ is determined by Claim \ref{claim2}   for $ u= (x_{2i-1}, x_{2i})^T$  and   $ v= (x_{2i+1}, x_{2i+2})^T$. Recall that  we used Claim \ref{claim2}  to find $K_{i}: = c_2$   for  $c_1 := K_{i+1}$.
Therefore,
\begin{equation}\label{oba} \left\|\left(\!\!\!\begin{array}{l}
(Jx)_{2i-1}\\%\\\vspace*{-2mm}
(Jx)_{2i}
\end{array}\!\!\!\right)  - b^{(i)} \right\|_2 <  K_{i} -\tfrac12 \quad \text{or} \quad  \left\|\left(\!\!\!\begin{array}{l}
(Jx)_{2i-1}\\%\\\vspace*{-2mm}
(Jx)_{2i}
\end{array}\!\!\!\right)  - b^{(i)} \right\|_2 <  \left\|\left(\!\!\!\begin{array}{l}
x_{2i-1}\\%\\\vspace*{-2mm}
x_{2i}
\end{array}\!\!\!\right) \right\|_2 - 1.
\end{equation}

\end{itemize}
For the digit $d$ we have described above and an  $\varepsilon  \in \mathcal{E}$,  we focus on  $y =Jx -d+\varepsilon$.
 Let us realize that the inequality $\|\varepsilon\|_\infty < \tfrac13$ means  $|\varepsilon_j|\leq \tfrac13$ for each coordinate $j$ and  thus
$\|(\varepsilon_{2i-1}, \varepsilon_{2i})\|_2 \leq  \tfrac12$.

\medskip

For each $i> {\rm ind}(x)$,  we have  $ \|(x_{2i-1}, x_{2i})\|_2 < K_{i}$,   and  thus  both inequalities in \eqref{oba}  imply
$$
 \left\|\left(\!\!\!\begin{array}{l}
(Jx)_{2i-1}\\%\\\vspace*{-2mm}
(Jx)_{2i}
\end{array}\!\!\!\right)  - b^{(i)} +     \left(\!\!\!\begin{array}{l}
\varepsilon_{2i-1}\\%\\\vspace*{-2mm}
\varepsilon_{2i}
\end{array}\!\!\!\right)  \right\|_2 < K_{i}.
$$
Consequently,  ${\rm ind}(y) \leq  {\rm ind}(x)$.

\medskip

 If  ${\rm ind}(y) =  {\rm ind}(x) =:j$,   then necessarily
$\|(y_{2j-1}, y_{2j})\|_2 =  \left\|\left(\!\!\!\begin{array}{l}
(Jx)_{2j-1}\\%\\\vspace*{-2mm}
(Jx)_{2j}
\end{array}\!\!\!\right)  - b^{(j)} +   \left(\!\!\!\begin{array}{l}
\varepsilon_{2j-1}\\%\\\vspace*{-2mm}
\varepsilon_{2j}
\end{array}\!\!\!\right)  \right\|_2 > K_{j}$  and the left most inequality in  \eqref{oba} cannot hold true. The  validity of the right most
inequality implies $\|(y_{2i-1}, y_{2i})\|_2  \leq \|(x_{2i-1}, x_{2i})\|_2 - \tfrac12$.

\end{proof}

In case of real $\lambda$, that is, $\lambda = \pm1,$ the approach is analogous, only easier. The digit set
$\{(a_1,\dots,a_m)^T : a_i\in\{-1,0,1\}\}$ now suffices for the analogy Lemma~\ref{unit}.

\begin{coro}\label{unitCoro} Let $J=J_m(\lambda)$ with $|\lambda|=1$. Then there exists $\mathcal D\subset\mathbb Z^m$ finite with the following property. For each $x \in \mathbb R^{m}$ there exists  a sequence  $(d_{i+1})_{i\in \mathbb N}$ of digits from $\mathcal{D} $  such that  for any  sequence  $(\varepsilon_{i+1})_{i\in \mathbb N}$ from  the perturbation set $\mathcal{E}$, the sequence   defined recursively

\medskip

\centerline{$x^{(0)} = x$  \ \ and \ \ \ $x^{(n+1)}:= Jx^{(n)} - d_{n+1}+\varepsilon_{n+1}$ for $n \in \mathbb N$}

\medskip

\noindent  satisfies  ${\rm ind}(x^{(N)})  = 0$ for all sufficiently large $N \in \mathbb N$.   In particular,  there exists a constant $C$ such that $\|x^{(N)}\|_\infty  < C$  for all sufficiently large $N\in \mathbb N$.

\end{coro}

\begin{comment}
\begin{coro}\label{unitCoro} Let $J$ be the matrix given in \eqref{rotace}  and  $\mathcal{D} \subset \mathbb Z^{2\ell}$  be the digit set  as in Lemma \ref{unit}.  Then for each $x \in \mathbb R^{2\ell}$ there exists  a sequence  $(d_{i+1})_{i\in \mathbb N}$ of digits from $\mathcal{D} $  such that  for any  sequence  $(\varepsilon_{i+1})_{i\in \mathbb N}$ from  $\mathcal{E}$ the sequence   defined recursively

\medskip

\centerline{$x^{(0)} = x$  \ \ and \ \ \ $x^{(n+1)}:= Jx^{(n)} - d_{n+1}+\varepsilon_{n+1}$ for $n \in \mathbb N$}

\medskip

\noindent  satisfies  ${\rm ind}(x^{(N)})  = 0$ for all sufficiently large $N \in \mathbb N$.   In particular,  there exists a constant $C$ such that $\|x^{(N)}\|_\infty  < C$  for all sufficiently large $N\in \mathbb N$.

\end{coro}
In case of real $\lambda$, that is, $\lambda = \pm1,$ the approach is analogous, only easier. Roughly speaking, instead of blocks of dimension $2$, we have blocks of dimension $1$. The digit set
$\{(a_1,\dots,a_d)^T : a_i\in\{-1,0,1\}\}$ now suffices for the analogy Lemma~\ref{unit}. In the analogy, the function {\rm ind} returns the index of the last row that is greater that $\tfrac12$ in absolute value.
\end{comment}

\subsection{A real Jordan block to an eigenvalue  $\lambda$ strictly inside the unit circle }
\begin{lem}\label{contract} Let $J\in \mathbb R^{m\times m}$ be the real Jordan block to $\lambda$ of modulus $<1$ and     $\mathcal{E}$ be the perturbation set.   Then there exist a norm $\|.\|_c$ of  $\mathbb R^{m}$ and  a constant $\gamma$ such that

\medskip
 for any  $x \in \mathbb R^m$  and  any $\varepsilon \in \mathcal{E}$, the vector $y=Jx+\varepsilon$ satisfies:

\medskip

\begin{itemize}

\item  If \ $ \| x\|_c   \leq \gamma$, then $\|y\|_c \leq \gamma$;
\medskip

\item   If  \ $ \| x\|_c   \geq \gamma$, then $\|y\|_c \leq \|x\|_c - \tfrac12$.

\end{itemize}
\end{lem}

\begin{proof}  Let us choose $\beta$ such that $1> \beta > |\lambda|$.  By  Theorem 3 from \cite{IsaacsonKeller}, there exists a norm $ \| x\|_c $  of  $\mathbb R^m$  such that    $\|Jz\|_c \leq \beta \|z\|_c$  \  for all $z \in \mathbb R^m$. Then
\medskip

\centerline{$\|Jx+ \varepsilon\|_c \leq \|Jx\|_c + \|\varepsilon\|_c\leq \beta \|x\|_c   +E$ \ \  where $E=\max\{\|\varepsilon\|_c : \varepsilon \in \mathcal{E}\}$.}

\medskip

 \noindent Now,  it is enough to check that

 $ \| x\|_c   \leq   \frac{1}{1-\beta}\bigl(\tfrac12+E\bigr) $,  implies  $ \beta \|x\|_c   +E \leq   \frac{1}{1-\beta}\bigl(\tfrac12+E\bigr)$

\noindent and

 $ \| x\|_c   \geq   \frac{1}{1-\beta}\bigl(\tfrac12+E\bigr) $,  implies  $ \beta \|x\|_c   +E \leq \|x\|_c   - \tfrac12 $.
\medskip

\noindent Therefore, we put $\gamma = \frac{1}{1-\beta}\bigl(\tfrac12+E\bigr)$.

\end{proof}

\begin{coro}\label{contractCoro} Let $J\in \mathbb R^{m\times m}$ be the real Jordan block to $\lambda$ of modulus $<1$. Then there exists a constant $\gamma$  and a norm $ \| x\|_c $  of  $\mathbb R^m$  such that  for each $x \in \mathbb R^{m}$  and   for any  sequence  $(\varepsilon_{i+1})_{i\in\mathbb N}$ from  $\mathcal{E}$, the sequence   defined recursively
$$x^{(0)} = x  \ \ and \ \ \ x^{(n+1)}:= Jx^{(n)} +\varepsilon_{n+1} \quad\text{ for }\quad n \in \mathbb N$$
satisfies   $\|x^{(N)}\|_c  < \gamma$  for all sufficiently large $N \in \mathbb N$.

\end{coro}

\subsection{A real Jordan block to an eigenvalue  $\lambda$ strictly outside the unit circle }
\begin{lem}\label{expand}   Let $J\in \mathbb R^{m\times m}$ be the real Jordan block to $\lambda$ of modulus $>1$ and     $\mathcal{E}$ be the perturbation set.   Then there exists a digit set $\mathcal{D} \subset \mathbb Z^m$ with the property:

\medskip
for each $x \in  \mathbb R^{m}$    with  $\|x\|_\infty \leq 1$  there exists a digit  $ d \in \mathcal{D}$  such that
 for all  $\varepsilon  \in  \mathcal{E}  $,   the vector $y= Jx-d+ \varepsilon$  satisfies  $\|y\|_\infty\leq 1$.

\end{lem}
\begin{proof} Since   $\|Jx\|_\infty \leq (2|\lambda| + 1)\|x\|_\infty $ for each $x\in  \mathbb R^{m}$, we put
$\mathcal{D} = \{ d\in \mathbb Z^m : \|d\|_\infty \leq  2|\lambda| + 2 + E\}$,  where $E=\max\{\|\varepsilon\|_\infty : \varepsilon \in \mathcal{E}\}$.

\end{proof}

\begin{coro}\label{expandCoro}  Let $J\in \mathbb R^{m\times m}$ be the real Jordan block to $\lambda$ of modulus $>1$,  and $\mathcal{D}$   and   $\mathcal{E}$  be  as in Lemma \ref{expand}. Then for each $x \in \mathbb R^{m}$ there exists  a sequence  $(d_{i+1})_{i\in \mathbb N}$ of digits from $\mathcal{D} $  such that  for any  sequence  $(\varepsilon_{i+1})_{i\in \mathbb N}$ from  $\mathcal{E}$ the sequence   defined recursively

\medskip

\centerline{$x^{(0)} = x$  \ \ and \ \ \ $x^{(n+1)}:= Jx^{(n)} - d_{n+1}+\varepsilon_{n+1}$  \ \ for $n \in \mathbb N$,}

\medskip

\noindent  satisfies  $\|x^{(N)}\|_\infty  < 1$  for all sufficiently large $N \in \mathbb N$.

\end{coro}

\subsection{Completion of proof of Proposition \ref{skoroHotovo}}

Now we combine properties of all types of the real Jordan  blocks.

\begin{proof}[Proof of Proposition \ref{skoroHotovo}]
Let $J_1, J_2, \ldots , J_s$ be the real Jordan blocks of the size $m_1, \ldots, m_s$ respectively,  such that
$J=J_1\oplus J_2 \oplus \cdots \oplus J_s$.

To each block $J_k \in \mathbb R^{m_k}$ corresponding to  $\lambda$ with $|\lambda|\geq 1$ we find a constant $C_k$ and  a finite digit set $\mathcal{D}_k\subset \mathbb Z^{m_k}$  with the properties described in   Corollaries \ref{expandCoro} and \ref{unitCoro}.

 To  each block $J_k$ corresponding to  $\lambda$ with $|\lambda|< 1$, we  assign the constant $\gamma_k$  found in Corollary \ref{contractCoro}  for the norm $\|.\|_c$.    As two norms on a finite dimensional space are equivalent, there exists a constant  $C_k$ such that $\|x\|_c < \gamma_k$ implies  $\|x\|_\infty < C_k$.   We put $\mathcal{D}_k=\{0\} \subset \mathbb R^{m_k}$.

We use the digit sets $\mathcal{D}_k$  for construction of a new digit set $\widetilde{\mathcal{D}}\subset   \mathbb Z^m$
\begin{equation}\label{abeceda}
\tilde{d}\in \widetilde{\mathcal{D}} \qquad \text{ if and only if } \qquad
\tilde{d}=\left(\begin{smallmatrix}
d^{(1)}\\\vspace*{-2mm}
d^{(2)} \\\vspace*{-1mm}
\vdots\\[2mm]
d^{(s)}
\end{smallmatrix}\right) \in \mathbb{Z}^{m}, \ \
\text{where} \ \ d^{(k)}  \in \mathcal{D}_k, \text{ for }  k=1, 2, \ldots, s.
\end{equation}
Put $C =\max_k{C_k}$. In this notation, we get the following.

\medskip
\noindent {\bf Claim A:}
 for each      $x \in \mathbb R^m$  with   $\|\pi_e(x)\|_\infty \leq 1$  there exists a sequence  $(\tilde{d}_i)_{i\in \mathbb N}$ of digits from $\widetilde{\mathcal{D}} $  such that  for any  sequence  $(\varepsilon_i)_{i\in \mathbb N}$ from  $\mathcal{E}$ the sequence   defined recursively
\begin{equation}\label{jesteIntegerAlphabet}
\text{$x^{(0)} = x$  \ \ and \ \ \ $x^{(n+1)}:= Jx^{(n)} - \tilde{d}_{n+1}+\varepsilon_{n+1}$  \ \ for $n \in \mathbb N$,}
\end{equation}
\noindent  satisfies  $\|x^{(N)}\|_\infty  < C$  for all sufficiently large $N \in \mathbb N$.

\medskip

Since $L$ is close to  $ \mathbb{Z}^{m}$,   we can assign  to  each $\tilde{d} \in \widetilde{\mathcal{D}}$ a lattice point  $ d\in L$ such that $d \in \tilde{d} +\mathcal{E}$. We obtain a new digit set  $\mathcal{D} \subset L$ and the size of  $\mathcal{D}$ does not exceed the size of   $\widetilde{\mathcal{D}}$.   We rewrite Equation \eqref{jesteIntegerAlphabet} for the specific choice of the sequence   $(\varepsilon_i)_{i\in \mathbb N}$, namely for $\varepsilon_i : = {\tilde{d}}_i - d_i \in \mathcal{E}$.   We get a modification of the previous claim

\medskip
\noindent {\bf Claim B:}
 for each      $x \in \mathbb R^m$  with   $\|\pi_e(x)\|_\infty \leq 1$  there exists a sequence  $({d}_i)_{i\in \mathbb N}$ of digits from ${\mathcal{D}} $  such that   the sequence   defined recursively
$$
\text{$x^{(0)} = x$  \ \ and \ \ \ $x^{(n+1)}:= Jx^{(n)} - {d_{n+1}}$  \ \ for $n \in \mathbb N$,}
$$
\noindent  satisfies  $\|x^{(N)}\|_\infty  < C$  for all sufficiently large $N \in \mathbb N$.

\medskip
Multiplying the equality  $x^{(n+1)}= Jx^{(n)} - {d_{n+1}}$  by $J^{-n -1}$  and summing up for $n= 0,1,\ldots, N-1$ we get
$$\sum_{n=0}^{N-1} J^{-n-1}x^{(n+1)} = \sum_{n=0}^{N-1} J^{-n}x^{(n)} - \sum_{n=0}^{N-1} J^{-n-1}d_{n+1}   \quad \text{and thus}   \quad \sum_{n=1}^{N} J^{-n}x^{(n)} = \sum_{n=0}^{N-1} J^{-n}x^{(n)} - \sum_{n=1}^{N} J^{-n}d_{n}\,.$$
It implies $ J^{-N}x^{(N)} =  x^{(0)} - \sum_{n=1}^{N} J^{-n}d_{n}$  with     $\|x^{(N)}\|_\infty  < C$, as required in the statement of the proposition.

\end{proof}

\section{${\rm Fin}_{\mathcal{D}}(M)$ \ \  versus  \ \  $\bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m}$}
In general, we have that ${\rm Fin}_{\mathcal D}(M)\subseteq\bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m}$. In this section we describe when the inclusion turns into an equality.
The inverse to a matrix $M \in \mathbb{Z}^{m\times m}$ with the determinant  $\Delta \neq 0$  belongs to  $\frac{1}{\Delta}\mathbb{Z}^{m\times m}$. Therefore,
the definition of the set  ${\rm Fin}_{\mathcal{D}}(M)$  directly implies the following properties.

\begin{lem}\label{list}  Let $M \in \mathbb{Z}^{m\times m}$ with  $\det M = \Delta \neq 0$ and $\mathcal{D} \subset \mathbb{Z}^m$, $\mathcal{D}$ finite.  Then
\begin{enumerate}
\item  $M^k\Bigl({\rm Fin}_{\mathcal{D}}(M)\Bigr) =  {\rm Fin}_{\mathcal{D}}(M)$ for each $k \in \mathbb Z$\,.

\item   ${\rm Fin}_{\mathcal{D}}(M) \subseteq  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m} $\,.

\item   If $ \mathbb{Z}^m \subset {\rm Fin}_{\mathcal{D}}(M)$,  then  ${\rm Fin}_{\mathcal{D}}(M)$ is closed under addition and subtraction.
\end{enumerate}
\end{lem}

Any  classical $b$-ary numeration system, with  base $b \in \mathbb N, b\geq 2$ and the canonical digit set $\mathcal{D} = \{0,1,\ldots, b-1\}$  can be considered as a matrix system with $M=b \in \mathbb Z^{1\times 1}$ and the determinant  $\Delta =b$.   The canonical digit set  allows to represent only non-negative numbers and obviously ${\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{b^k}\mathbb{N} $.

In the next auxiliary claim we denote by $e_i \in \mathbb Z^m$ the vector $(\delta_{i1}, \delta_{i2}, \ldots, \delta_{im})^T$, where $\delta_{ij}$ is the Kronecker symbol, i.e. $\delta_{ij} = 0$, if $i\neq j$ and $\delta_{ii} = 1$.

\begin{claim}\label{C1}  $  {\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m}$ \ \  if and only if \ \    for each $i \in \{ 1,2, \ldots, m\}$ there  exists  $\ell_i \in \mathbb N$ such that  $\frac{1}{\Delta} {e}_i \in  M^{-\ell_i}(\mathbb Z^m)\,.$
\end{claim}
\begin{proof}  $(\Leftarrow)$   Due to Item (2) of Lemma \ref{list}, it is enough to show that  $ \frac{1}{\Delta^k}\mathbb{Z}^{m} \subset {\rm Fin}_{\mathcal{D}}(M) $ for each $k \in \mathbb N$. We show it by induction on $k \in \mathbb N$.

The assumption  $\frac{1}{\Delta} {e}_i \in  M^{-\ell_i}(\mathbb Z^m)$ gives $\tfrac{1}{\Delta}\mathbb Z^m\subset \sum_{i=1}^mM^{-\ell_i}(\mathbb Z^m)$.  Multiplying  this inclusion by $\tfrac{1}{\Delta^k}$ and
 using the induction hypothesis we obtain
$$
\tfrac{1}{\Delta^{k+1}} \mathbb Z^m  \subset  \sum_{i=1}^mM^{-\ell_i}\bigl(\tfrac{1}{\Delta^{k}} \mathbb Z^m\bigr) \subset \sum_{i=1}^mM^{-\ell_i}\bigl({\rm Fin}_\mathcal{D}(M)\bigr) =  {\rm Fin}_\mathcal{D}(M)\,.
$$
The last equality follows from Items (1) and (3) of Lemma \ref{list}.

$(\Rightarrow)$   This implication is obvious.
\end{proof}

\begin{thm}\label{rovnost}  Let $M \in \mathbb{Z}^{m\times m}$, $\det M  = \Delta \neq 0$.   Let  $\mathcal{D} \subset \mathbb Z^m$ be a finite digit set such that   $\mathbb Z^m \subset
  {\rm Fin}_{\mathcal{D}}(M)$.    Then
$$  {\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{m}  \qquad \Longleftrightarrow
 \qquad  \exists \ell \in \mathbb N \text{  such that }  M^\ell = \Theta \mod \Delta\,.$$
In particular,  if  $\det M = \pm 1$, then $ \mathbb{Z}^{m} =  {\rm Fin}_{\mathcal{D}}(M)$\,.

\end{thm}
\begin{proof} By Claim \ref{C1} we have to study when $\frac{1}{\Delta}e_i \in M^{-\ell_i}(\mathbb Z^m)$ for some $\ell_i \in \mathbb N$.  It can be rewritten into  $M^{\ell_i} e_i \in \Delta \mathbb Z^m$, or equivalently, $ M^{\ell_i} e_i = 0\mod \Delta$.     We  denote the   $i^{th}$ column of a matrix $A$ by $A_{\bullet i}$.  Obviously,  $M^{\ell_i}e_i = (M^{\ell_i})_{\bullet i}$.

 Let us realize that  if $(M^{\ell_i})_{\bullet i}= 0 \mod \Delta$,  then   $(M^{\ell_i})_{\bullet i}=0 \mod \Delta$ for all $\ell\in \mathbb N, \ell>\ell_i$. Indeed,  $(M^{\ell_i})_{\bullet i} = M^{\ell -\ell_i} (M^{\ell_i})_{\bullet i} =  M^{\ell -\ell_i} 0 \mod \Delta $.  Therefore,   we can look for a common exponent $\ell$ for all columns of the matrix $M$.
\end{proof}

\bigskip
\begin{example} Let $M_1 = \left( \begin{array}{rr} 2&1\\2&2 \end{array} \right) $  and  $M_2 = \left( \begin{array}{rr} 1&2\\2&2 \end{array} \right) $.
Then $\det M_1 = 2$ and $\det M_2 = -2$.
$$M_1 = \left( \begin{array}{rr} 0&1\\0&0 \end{array} \right) \mod \Delta,   \qquad   M_1^2 = \Theta \mod \Delta \qquad \text{and thus \ } {\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{2^k}\mathbb{Z}^{2}   $$
for a suitable   $\mathcal{D} \subset \mathbb Z^2$.

On the other hand,  $$M_2 = \left( \begin{array}{rr} 1&0\\0&0 \end{array} \right) \mod \Delta,   \qquad   \text{and thus \ } \ \  M_2^\ell  \neq  \Theta \mod \Delta \  \text{ for all $\ell \in \mathbb N$\ }  \,. $$
In particular,  $\Bigr(\!\!\! \begin{array}{r} \tfrac12\\0 \end{array}\!\!\! \Bigl) \notin {\rm Fin}_{\mathcal{D}}(M_2)$ for any choice $\mathcal{D} \subset \mathbb Z^2$.

\end{example}

\section{Comments}
The main idea of this paper comes from~\cite{VavraIsrael} concerning number systems with algebraic base without any restrictions on its Galois conjugates. We do not give an explicit bound on the size of the digit set, even though our proof is a constructive one. For matrices without eigenvalues on the unit circle, the method used in \cite{FrPeSv},\cite{FrHePeSv} can be used to obtain explicit digit sets, as well as arithmetic algorithms.

To find a digit set $\mathcal{D} \subset \mathbb Z^m$ of the minimal cardinality may be a hard problem. It is known that if $M$ is expanding, then we need a digit set of size at least  $|\det M|$ to be able to represent integer vectors in the form $\sum_{k=0}^nM^kd_k$. In our case it seems that an appropriate lower bound would be around the product of absolute values of the eigenvalues outside the unit circle.

In any case, the digit set needs to have at least one digit beside zero. Two digits might be already enough for some matrices of dimension two or more, as is illustrated by the following examples.

The first example is a case of a matrix having only one eigenvalue outside the unit circle. It can be seen from the proof of Proposition~\ref{skoroHotovo} that this dominant eigenvalue will determine the size of the digit set. We will, however, utilize the connection between matrix numeration systems and those with real base.

\begin{example}   Let $T = \left(\!\!\! \begin{array}{rrr} 0&0&-1\\1&0&1\\0&1&-1 \end{array} \!\!\!\right) $.
As $\det T =-1$, we have $  {\rm Fin}_{\mathcal{D}}(T)  = \mathbb Z^3$ for a suitable digit set guaranteed by Theorem \ref{main}. The characteristic polynomial of $T$ is  $x^3+x^2-x+1$. $T$ has one negative eigenvalue $\beta = -1,839\dots$ and a pair of complex conjugate eigenvalues in  modulus smaller than 1. As shown in \cite{KrStVa},  any element of  the ring  $\mathbb Z[\beta]=\{a+b\beta +c\beta^2: a,b,c \in \mathbb Z\}$    can be written as
\begin{equation}\label{tribo} a+b\beta +c\beta^2 = \sum_{i=k}^n a_i\beta^i,  \quad \text{where }  k, n \in \mathbb Z, k\leq n  \text{ and \ }  a_i \in \{0,1\} \,.\end{equation}

We use the isomorphism $\psi :  \mathbb Z[\beta]  \to \mathbb Z^3$ of two additive groups  given by   $\psi(a+b\beta +c\beta^2) = \left(\!\!\!  \begin{array}{r} a\\b\\c \end{array}\!\!\! \right)$.
Since $\beta  (a+b\beta +c\beta^2 ))=  -c +(a+c)\beta +(b-c)\beta^2$, we have
$$\psi(\beta (a+b\beta +c\beta^2 ) )=\left(\!\!\!  \begin{array}{c} -c\\ a+c \\b-c \end{array}\!\!\! \right) = \left(\!\!\! \begin{array}{rrr} 0&0&-1\\1&0&1\\0&1&-1 \end{array} \!\!\!\right)\left(\!\!\!  \begin{array}{r} a\\b\\c \end{array}\!\!\! \right)  =T\left(\!\!\!  \begin{array}{r} a\\b\\c \end{array}\!\!\! \right).$$
Applying this rule to  Equation \eqref{tribo} we deduce that   any element of $ \mathbb Z^3$  can be expressed as
$$\left(\!\!\!  \begin{array}{r} a\\b\\c \end{array}\!\!\! \right) =  \psi(a+b\beta +c\beta^2)=  \sum_{i\in I}\psi(\beta^i a_i) =  \sum_{i=k}^n T^i \psi( a_i) = \sum_{i=k}^n T^i\left(\!\!\!  \begin{array}{c} a_i\\0\\0 \end{array}\!\!\! \right)\,.$$
In other words, $  {\rm Fin}_{\mathcal{A}}(T)  = \mathbb Z^3$ already for the small digit set
$
\mathcal{A} =  \{ (0,0,0)^T, (1,0,0)^T \}
$.

\end{example}

\begin{comment}
The second example shows the relation between representation in a real base and a matrix base.
In particular, the representations with algebraic base $\beta$ are isomorphic to matrix based systems via the companion matrix. If $\beta$ does not have a Galois conjugate on the unit circle, then there is a simple argument for the existence of digit sets allowing finite representations of elements of $\mathbb Z[\beta,\beta^{-1}]$. The argument involves the parallel addition algorithm, see ...., and this approach was utilized in \cite{BMPV}. An analogous approach would be possible also for matrix bases, however, only under the assumption of non-existence of unimodular eigenvalues.
\end{comment}

The second example is a rotation matrix in $\mathbb R^2.$ Notice that in this case the link to real/complex base numeration systems is missing, for there is no eigenvalue of modulus $>1.$ Moreover, the actual minimal digit set is significantly smaller that the one ensured by the the proof of Proposition~\ref{skoroHotovo}.

\begin{example}
Consider $M = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$ with the digit set $\mathcal D=\{(0,0)^T,(1,0)^T\}.$

We have the identities
\[
M^{4k}=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad M^{4k+1} = \begin{pmatrix}0&1\\-1&0\end{pmatrix},\quad M^{4k+2}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix},\quad M^{4k+3}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
\]

Hence
\[
M^{4k}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix},
\quad M^{4k+1}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix},
\quad M^{4k+2}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\0\end{pmatrix},
\quad M^{4k+3}\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}.
\]
It is now easy to see how the representations are created. For instance, if $a,b>0,$ then
\[
\begin{pmatrix}a\\b\end{pmatrix} = \sum_{k=1}^a M^{4k}\begin{pmatrix}1\\0\end{pmatrix} + \sum_{k=1}^b M^{4k+3}\begin{pmatrix}1\\0\end{pmatrix}.
\]
\end{example}
\begin{comment}
\begin{example}\label{JordanUnit}  Consider
$$M =\left(\!\! \!\begin{array}{rr} 1&1\\0&1 \end{array} \!\!\!\right) \qquad   \text{and \ } \ \ \mathcal{D} =  \left\{\Bigr(\!\!\! \begin{array}{r} 0\\0 \end{array}\!\!\! \Bigl), \Bigr(\!\!\! \begin{array}{r} 0\\1 \end{array}\!\!\! \Bigl) ,  \Bigr(\!\!\! \!\begin{array}{r} 0\\ \text{-}1 \end{array}\!\!\! \Bigl) \right\}  \,. $$
We show that any vector $ \Bigr(\!\!\! \begin{array}{r} a\\b \end{array}\!\!\! \Bigl)  \in \mathbb Z^2$ can be written in the form
$$ \Bigr(\!\!\! \begin{array}{r} a\\b \end{array}\!\!\! \Bigl) = \sum_{k\in I}M^k\Bigr(\!\!\! \begin{array}{c} 0\\d_k \end{array}\!\!\! \Bigl),\quad  I\subset \mathbb Z,  \ I \text{ finite},  \ d_k \in \{1, \text{-}1\}.$$
Because of the symmetry of the digit set $\mathcal{D}$, we assume without loss of generality  that  $b \geq 0$.

\begin{itemize}

\item First,  assume that $b>0$. For parameters  $ n \in \mathbb Z$  and $s \in \mathbb N$,   which will be determined later, we look for a  representation in the form
$$ \Bigr(\!\!\! \begin{array}{r} a\\b \end{array}\!\!\! \Bigl) = \sum_{k= n+1}^{n+b+2s}\left(\!\! \!\begin{array}{rr} 1&k\\0&1 \end{array} \!\!\!\right)\Bigr(\!\!\! \begin{array}{c} 0\\d_k \end{array}\!\!\! \Bigl)\,,$$
$ \text{where\ } \ d_{k} = 1 \  \text{for\ } k =n+1, n+2,\ldots, n+b$, \   and $d_{k} =( -1)^k\text{\ for} \
k=n+b+1,n+b+2,\ldots,n+b+2s $.
The  described choice of digits implies
$$\sum_{k= n+1}^{n+b+2s} \left(\!\! \!\begin{array}{rr} 1&k\\0&1 \end{array} \!\!\!\right)\Bigr(\!\!\! \begin{array}{c} 0\\d_k \end{array}\!\!\! \Bigl) = \sum_{k= n+1}^{n+b+2s} \Bigr(\!\!\! \begin{array}{c} kd_k\\d_k \end{array}\!\!\! \Bigl) =  \sum_{k= n+1}^{n+b} \Bigr(\!\!\! \begin{array}{c} k\\1 \end{array}\!\!\! \Bigl)  +  \sum_{k= n+1 +b}^{n+b+2s} \Bigr(\!\!\! \begin{array}{c} (\text{-}1)^kk\\(\text{-}1)^k \end{array}\!\!\! \Bigl)  =  :  \Bigr(\!\!\! \begin{array}{c} A\\mathcal B \end{array}\!\!\! \Bigl)    $$
Obviously,
$$ B = b\quad \text{and} \quad A =   nb + \frac{b(b+1)}{2}  +s $$
Now, we find $s\geq 0$ such that $ \frac{b(b+1)}{2}  +s   = a \mod b$  \  and \ put   $n=   b^{-1}\bigl(a-\frac{b(b+1)}{2}  -s \bigr)$.
The choice of parameters $s$  and $n$ gives  $A=a$,  as desired.
\item If $b=0$  and $a \neq  0$, then the symmetry of the digit set allows us to consider $a> 0$. In this  case,
$$ \Bigr(\!\!\! \begin{array}{r} a\\0 \end{array}\!\!\! \Bigl) = \sum_{k= 1}^{2a} \left(\!\! \!\begin{array}{rr} 1&k\\0&1 \end{array} \!\!\!\right)\Bigr(\!\!\! \begin{array}{c} 0\\(\text{-}1)^k \end{array}\!\!\! \Bigl)\,,$$
\item If $a=b=0$, then
$$ \Bigr(\!\!\! \begin{array}{r} 0\\0 \end{array}\!\!\! \Bigl)  =  \left(\!\! \!\begin{array}{rr} 1&\text{-}1\\ 0&1 \end{array} \!\!\!\right)\Bigr(\!\!\! \begin{array}{r} 0\\ 1 \end{array}\!\!\! \Bigl)  + \left(\!\! \!\begin{array}{rr} 1&0\\ 0&1 \end{array} \!\!\!\right) \Bigr(\!\!\! \begin{array}{r} 0\\\text{-}1 \end{array}\!\!\! \Bigl)  + \left(\!\! \!\begin{array}{rr} 1&1\\ 0&1 \end{array} \!\!\!\right) \Bigr(\!\!\! \begin{array}{r} 0\\\text{-}1 \end{array}\!\!\! \Bigl)  + \left(\!\! \!\begin{array}{rr} 1&2\\ 0&1 \end{array} \!\!\!\right) \Bigr(\!\!\! \begin{array}{r} 0\\ 1 \end{array}\!\!\! \Bigl) $$
\end{itemize}
\end{example}
\end{comment}

\begin{comment}
\begin{example} Let $M = \left( \begin{array}{rr} 3&1\\5&1 \end{array} \right) $. The spectrum of $M$ is
$\sigma(M) = \{ 2+ \sqrt{6}, 2-\sqrt{6}\}$ and by Theorem  \ref{main} there exists $\mathcal{D} \subset \mathbb Z^2$ such   that  $ \mathbb Z^2\subset  {\rm Fin}_{\mathcal{D}}(M)$.      Moreover,   $\det M =\Delta = -2$  and thus
$$M = \left( \begin{array}{rr} 1&1\\1&1 \end{array} \right) \mod \Delta  \qquad  \text{and} \qquad  M^2 = \Theta \mod \Delta\,.$$
By Theorem \ref{rovnost},  ${\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{2^k}\mathbb{Z}^{2}  $.

\end{example}

Let us discuss how is difficult to decide whether ${\rm Fin}_{\mathcal{D}}(M) =  \bigcup\limits_{k\in \mathbb{N}} \frac{1}{\Delta^k}\mathbb{Z}^{2} $. By Theorem \ref{rovnost} we have to decide whether  the following condition is satisfied

\medskip

\centerline{{\bf Condition (T)} :  \ \
$  \exists \ell \in \mathbb N \text{  such that }  M^\ell = \Theta \mod \Delta\,.$}

\begin{prop}    Let $M \in \mathbb{Z}^{d\times d}$  and $|\det M|>1$. To decide Condition (T) one needs to perform at most $\sum_{p \in  P} \lceil \log_2(d^2p) \rceil $  matrix multiplications, where $P =\{ p\in \mathbb N: p \text{  prime},\  p\text{ divides}  \ \det M\}$.
\end{prop}
\begin{proof}
Let us recall some simple rules on modular arithmetic. For any $a \in \mathbb Z$,  $p,q, i \in \mathbb N$, $p,q, i >0$:

--  If  $p,q$ are coprime, then  $a = 0\mod pq$ if and only if  $a = 0\mod p$ and $a = 0\mod q$.

-- If $a = 0\mod p^i$, then  $a = 0\mod p$.

-- If $a = 0\mod p$, then  $a^i = 0\mod p^i$.

\medskip
\noindent  Therefore Condition (T) is equivalent to Condition (T'):  for any prime number $p$ which divides $\det M$ there exists $\ell_p \in \mathbb N$ such that $M^{\ell_p} = \Theta \mod p$.
The  set $\{M^i \mod p :  i \in \mathbb N\}$ has at most $d^2 p$ elements.  Find $k$ such that $2^k \geq d^2 p$. Then  $M^{2^k} =\Theta \mod p$ if and only if $M^{\ell_p} = \Theta \mod$ for some $\ell_p \in \mathbb N$. To compute   $M^{2^k}$ one needs to multiply matrices  $M^{2^i}  M^{2^i}   = M^{2^{i+1}}$, for $i=0,1,\ldots, k-1$.     Therefore, for each prime $p$ dividing $\det M$ one needs to perform  $k \geq \log_2(d^2p)$ multiplications.
\end{proof}

\end{comment}

\section*{Acknowledgements}
E.P.  acknowledges support of the project CZ.02.1.01/0.0/0.0/16\_019/0000778. T.V. was supported by Charles University,  projects PRIMUS/20/SCI/002 and UNCE/SCI/022.

redpajama_set_name: RedPajamaArXiv


text:
Saskatchewan NDP Caucus
REALITY CHECK: On potash, Wall's record speaks louder than words
Upon news of the possible merger between Potash Corp. and Agrium, Brad Wall assembled journalists to apparently say for the record that he is concerned about the potash industry and jobs. But, if you actually check the record, you'll see that he and the Sask. Party have not actually protected either.
Sask. Party must fix chronic understaffing in hospitals and health centres: NDP
As the Sask. Party continues to ignore the growing crisis in access to health care in rural Saskatchewan, nurses at the Davidson Health Centre are speaking out about low staffing levels that are leaving them feeling "desperate" and unable to properly do their jobs.
Plagued by rough summer, Sask. Party shuffles cabinet but changes nothing
Since beginning this session, the government has been plagued by scandal and the real-life consequences Saskatchewan people are facing as a result of Sask. Party mismanagement but the negligence and lack of accountability in the Sask. Party cabinet have not been solved by today's shuffle.
"Today, the Premier wasted an opportunity to demonstrate the change the Sask. Party has promised," said NDP Leader Trent Wotherspoon. "A few titles have changed but the direction and tone that has seen the Sask. Party stumble from ethical scandal to ethical scandal and mismanaged crisis to mismanaged crisis has stayed the same."
Health regions waiting in the dark for health commissioner to be named
After touting the need for "transformational change" this spring, the Sask. Party government continues to create uncertainty for health regions, health care workers and communities by apparently reneging on its commitment to name a health commissioner to lead that change.
Lack of Sask. Party funding is not resolved by black-tie fund-raising
Saskatchewan has one of the worst records in the country when it comes to domestic violence and violence against women but, instead of taking real action and committing predictable and sustainable government funding to support victims and bring an end to these tragedies, the Sask. Party is providing only the funds raised at their black-tie fundraiser Sunday night.
"Raising much-needed money for a shelter like SOFIA House is obviously a good thing and all those who dip into their pockets and give what they can, should feel good about helping such an important cause," said NDP Justice, Policing and Corrections Critic Nicole Sarauer. "But, it comes across as much less genuine from the Sask. Party who, for nearly a decade, have chosen to withhold the government funding and support needed to make a more significant difference for these shelters as well the women and children who rely on them. This is especially true as we see the Sask. Party making deeper cuts to funding for some of Saskatchewan's most vulnerable."
Boyd out of cabinet, but scandals remain
With news that Bill Boyd is stepping down from cabinet, NDP Leader Trent Wotherspoon noted that the Premier should have shown leadership and held the now former Minister of Economy, the GTH, and SaskPower to account for his mismanagement and scandal with his files, which have cost Saskatchewan taxpayers billions of dollars.
"This scandal plagued Minister should have been kicked out of cabinet a long time ago," Wotherspoon said. "The fact that, even today, the Premier and the rest of the Sask. Party caucus still stand by Mr. Boyd and his record means that this scandal and waste lays at their feet still."


(Host) Vermont's three gubernatorial candidates have very different plans to provide affordable health care coverage to all Vermonters.
Their views range from expanding the private market approach to creating a single risk pool for all Vermonters that would be financed by a broad based tax.
(Kinzel) Health care reform remains a critical issue because the cost of medical care continues to rise much faster than the rate of inflation. Vermont hospitals recently won state approval to boost their rates, on average, by almost 10 percent.
Independent candidate Anthony Pollina is backing the biggest change to the way health care is currently provided in Vermont.
Pollina says the current system is broken and should be replaced with a single payer system that's modeled after the federal Medicare program. It would be financed by using state tax dollars.
(Pollina) "So when you look at health care, you have to start with the knowledge that what we are doing is not working and it is bankrupting businesses and it is undermining our state budget. We are not controlling costs at all, so it is clearly time to go down a different path."
(Kinzel) Republican incumbent Jim Douglas says Pollina's plan will drive up taxes and fail to control costs. He says the state's current program, known as Catamount Health Care, is a better way to provide coverage to uninsured Vermonters.
Douglas says one way to make Catamount more affordable is to create a new plan that offers cheaper rates in return for fewer benefits and higher deductibles.
(Douglas) "One idea that I'm looking at is something I'm tentatively calling Catamount Light – a plan that is less extensive, less costly, something that would provide some level of coverage to Vermonters who might not otherwise choose to or be able to afford another plan."
(Kinzel) Douglas also believes that Vermonters should be allowed to purchase health care insurance from out-of-state companies.
(Douglas) "In this 21st century economy, people can log onto to the Internet and see what premiums are like in other states. And in many cases they're a lot less. We buy auto insurance online. We buy life insurance online. I think it would be great to find a way to do that for health insurance, as well."
(Kinzel) Democratic candidate Gaye Symington opposes this plan because, she says, the insurance companies will cover only young and healthy Vermonters – leaving older and sicker individuals to rely on state programs. It's a move she says will drive up the cost of these programs.
(Symington) "I think that that would move Vermont away from the notion of a community risk pool. It would allow for more cherry picking. It would allow for more administrative hassles within the system."
(Kinzel) Instead, Symington wants to expand Catamount and she's calling for a major change in the way that health care providers are paid.
(Symington) "We can do that by reimbursing our health care providers for the steps of keeping people healthy. Right now we don't. Our reimbursements and incentives are geared towards procedures, tests and fixing folks when they're ill."
(Kinzel) Symington says it will be difficult for Vermont to achieve universal access without the strong help of the federal government and she says she looks forward to working with a Democratic president and Congress who are committed to make major changes in the health care system.