API tools faq
paste
Advertisement
Guest User

Untitled

a guest
Feb 5th, 2014
85
0
Never
Add comment
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
text 10.41 KB | None | 0 0
raw download clone embed print report
  1. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
  2. <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-US">
  3. <head>
  4. <meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
  5. <link rel="stylesheet" href="/stylesheets/export.css"/>
  6. <link rel="shortcut icon" type="image/x-icon" href="/favicon.ico"/>
  7. <link rel="canonical" href="http://www.stackprinter.com/export?service=math.stackexchange&amp;question=633582&amp;printer=false&amp;linktohome=true"/>
  8. <title>If $(A-\lambda{I})$ is $\lambda$-similar to $(B-\lambda{I})$ then $A$ is similar to $B$</title>
  9. <script type="text/javascript" src="/javascripts/jquery-1.4.2.min.js"></script>
  10. <script type="text/javascript" src="/javascripts/main.js"></script>
  11. <script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML">
  12.  
  13. MathJax.Hub.Config({"HTML-CSS": { preferredFont: "TeX", availableFonts: ["STIX","TeX"] },
  14. tex2jax: { inlineMath: [ ["$", "$"], ["\\\\(","\\\\)"] ], displayMath: [ ["$$","$$"], ["\\[", "\\]"] ], processEscapes: true, ignoreClass: "tex2jax_ignore|dno" },
  15. TeX: { noUndefined: { attributes: { mathcolor: "red", mathbackground: "#FFEEEE", mathsize: "90%" } } },
  16. messageStyle: "none"
  17. });
  18. MathJax.Hub.Startup.onload();
  19. </script>
  20. <script type="text/javascript">
  21. MathJax.Hub.Queue(["Delay",MathJax.Callback,700],Print)
  22. </script>
  23. </script>
  24. </head>
  25. <body>
  26. <div id="home">
  27. <a href="/"><img title="Back to home" width="20px" height="20px" src="/images/icon_home.png" style="border:0"/></a>
  28. <a href="http://www.stackprinter.com/export?format=HTML&amp;service=math.stackexchange&amp;printer=false&amp;question=633582"><img title="Link to this printed question" width="20px" height="20px" alt="share" src="/images/Share.gif" style="border:0"/></a>
  29. </div>
  30. <div id="question-block">
  31. <div id="question-title">
  32. <img alt="Mathematics" src="http://cdn.sstatic.net/math/img/apple-touch-icon.png"/>If $(A-\lambda{I})$ is $\lambda$-similar to $(B-\lambda{I})$ then $A$ is similar to $B$<br/>
  33. </div>
  34. <div class="question-details">
  35. [+4] [2]
  36. user35660
  37. </div>
  38. <div class="question-details">
  39. [2014-01-10 10:44:23]
  40. </div>
  41. <div class="question-details">
  42. [
  43. linear-algebra
  44. matrices
  45. polynomials
  46. ]
  47. </div>
  48. <div class="question-details">
  49. [ http://math.stackexchange.com/questions/633582/if-a-lambdai-is-lambda-similar-to-b-lambdai-then-a-is-similar ]
  50. </div>
  51. <div id="question">
  52. <p>When reading the topic about primary and rational canonical form of matrices I stuck myself on this theorem:</p>
  53. <blockquote>
  54. <p>The matrices $A,B\in K^{n\times n}$ are similar if and only if their characteristic $\lambda$-matrices $(A-\lambda{I})$ and $(B-\lambda{I})$ are $\lambda$-equivalent; more precisely if $P(\lambda), Q(\lambda) \in K[\lambda]^{n\times n}$ are invertible such that $(B-\lambda{I})=Q(\lambda)(A-\lambda{I})P(\lambda)$ then $P(\lambda)=P$ and $Q(\lambda)=Q$ are constant matrices satisfying $Q=P^{-1}$ and $B=Q A P$.</p>
  55. </blockquote>
  56. <p><a href="http://thales.doa.fmph.uniba.sk/zlatos/la/LAG_A4.pdf" rel="nofollow">
  57. Theorem 21.5.1 on page 448
  58. </a>
  59. <sup style="font-size:9px">[1]</sup> (in Slovak).</p>
  60. <ul>
  61. <li>The main point in the proof of this theorem seems to be the assertion: If $P(\lambda), Q(\lambda)$ are invertible and $(B-\lambda{I})=Q(\lambda)(A-\lambda{I})P(\lambda)$ then $P(\lambda)=P$ and $Q(\lambda)=Q$ are constant matrices. What is the proof of this assertion?</li>
  62. </ul>
  63.  
  64. <div id="question-links">
  65. [1] http://thales.doa.fmph.uniba.sk/zlatos/la/LAG_A4.pdf<br/>
  66. </div>
  67. </div>
  68. <div class="answers">
  69. <div class="answer-details">
  70. [+2]
  71. [2014-01-10 12:25:37]
  72. Vincent
  73. [<img height="17px" width="17px" src="/images/blackflag.png"/>ACCEPTED]
  74. </div>
  75. <div class="answer">
  76. <p>The only proof I know is non-trivial and involves division of matrices. For a reference see: Felix Gantmacher - The theory of matrices, Volume 1, Chapter VI, Theorem 6, p. 145.</p>
  77. <p>If you are familiar with division of matrices, the proof is very readable, but a bit too long and technical to copy down here.</p>
  78. <p>ps. I am not yet allowed to post this as a comment, but I believe it is too important not to mention.</p>
  79.  
  80. <br/>
  81. </div>
  82. <div class="answer-pagenumber">1</div>
  83.  
  84. <div class="answer-details">
  85. [+2]
  86. [2014-01-13 09:32:27]
  87. user35660
  88. </div>
  89. <div class="answer">
  90. <p>Inspired by @Vincent's hint and Gantmacher's book: </p>
  91. <ul>
  92. <li><p>Matrix polynomial $A(\lambda)$ of degree n is a polynomial of the form:
  93. \begin{equation*}
  94. A(\lambda)=A_n{\lambda}^n+A_{n-1}{\lambda}^{n-1}+\ldots+A_0
  95. \end{equation*}
  96. where $A_i$ are constant matrices.
  97. Matrix polynomial is said to be regular if $A_n$ is regular.
  98. (Each matrix $A(\lambda)\in K[\lambda]^{n\times n}$ is a matrix polynomial.)</p></li>
  99. <li><p>Assertion: Degree of product of two matrix polynomials is less or equal than sum od degrees of factors. If at least one of factors is regular, degree of product equal to sum of degrees of factors.
  100. (Easy to verify.)</p></li>
  101. <li><p>Division of matrix polynomials from left or right respectively.
  102. $$ \begin{array}{rcccl}
  103. A(\lambda) &amp; : &amp; B(\lambda) &amp; = &amp; C(\lambda) \\
  104. R(\lambda) &amp; &amp; &amp; &amp;
  105. \end{array}
  106. %
  107. \left(\begin{array}{rcccl}
  108. dividend &amp; : &amp; divisor &amp; = &amp; quotient \\
  109. remainder &amp; &amp; &amp; &amp;
  110. \end{array}\right) $$
  111. $$ \begin{array}{ccccl}
  112. (A_n{\lambda}^n+\ldots+A_0) &amp; : &amp; (B_m{\lambda}^m+\ldots+B_0) &amp; = &amp; C_{n-m}{\lambda}^{n-m}+\ldots+C_0 \\
  113. R_{m-1}{\lambda}^{m-1}+\ldots+R_0 &amp; &amp; &amp; &amp;
  114. \end{array} $$
  115. It si the same process as division of polynomials, just the quotient coefficients for example $C_{n-m}=A_n:B_m$ is $B_m^{-1}A_n$ or $A_nB_m^{-1}$ respectively, and product of two matrices $C_{n-m}\times B_m$ is $B_mC_{n-m}$ or $C_{n-m}B_m$ (when evaluating remainder).
  116. Conclusion: Whenever $B(\lambda)$ is regular matrix polynomial we have:
  117. $$A(\lambda)=B(\lambda)C(\lambda)+R(\lambda) \text{(division from left)}$$
  118. $$A(\lambda)=C(\lambda)B(\lambda)+R(\lambda) \text{(division from right)}$$
  119. where degree of $R(\lambda)$ is less than degree of $B(\lambda.)$</p></li>
  120. </ul>
  121. <p>Let $Q(\lambda)$ and $P(\lambda)$ be invertible matrices. It gives existence of inverse matrices $M(\lambda)$ and $N(\lambda)$. (I.e. $M(\lambda)Q(\lambda)=Q(\lambda)M(\lambda)=I$, the same for $P(\lambda)$ and $N(\lambda)$)</p>
  122. <p>Equality
  123. $$
  124. (B-\lambda{I})=Q(\lambda)(A-\lambda{I})P(\lambda)
  125. $$
  126. can be rewritten as
  127. $$
  128. M(\lambda)(B-\lambda{I})=(A-\lambda{I})P(\lambda). \;\;\;\;\;(1)
  129. $$
  130. Let us divide $M(\lambda)$ from left side by $(A-\lambda{I})$. It means
  131. $$
  132. M(\lambda)=(A-\lambda{I})R(\lambda)+M. \;\;\;\;\;(2)
  133. $$
  134. Let us divide $P(\lambda)$ from right side by $(B-\lambda{I})$. It means
  135. $$
  136. P(\lambda)=S(\lambda)(B-\lambda{I})+P. \;\;\;\;\;(3)
  137. $$
  138. Matrices $P,M$ are constant matrices.
  139. After plugging (2) and (3) into (1) and a quick modification we obtain:
  140. $$
  141. (A-\lambda{I})[R(\lambda)-S(\lambda)](B-\lambda{I})=(A-\lambda{I})P-M(B-\lambda{I}) \;\;\;\;\;(4)
  142. $$
  143. Matrices $(A-\lambda{I}),(B-\lambda{I})$ are regular matrix polynomials of 1-st degree. Degree of the left side of (4) is at least 2, provided $[R(\lambda)-S(\lambda)]$ isn't constant zero matrix. (based on Assertion) But the degree of the right side is at most 1.
  144. It means $[R(\lambda)-S(\lambda)]=\mathbb{O}$ and consequently:
  145. $$
  146. (A-\lambda{I})P=M(B-\lambda{I})
  147. $$
  148. To finish the proof, it's enough to show regularity of $M$.</p>
  149. <p>Let us divide $Q(\lambda)$ by $(B-\lambda{I})$ from left side.
  150. $$
  151. Q(\lambda)=(B-\lambda{I})U(\lambda)+Q \;\;\;\;\;(5)
  152. $$
  153. Let us plug $Q(\lambda)$ from (5) to $M(\lambda)Q(\lambda)=I$.
  154. We obtain:</p>
  155. <p>$$
  156. M(\lambda)((B-\lambda{I})U(\lambda)+Q)=I
  157. $$
  158. $$
  159. M(\lambda)(B-\lambda{I})U(\lambda)+M(\lambda)Q=I
  160. $$
  161. Now substitute $M(\lambda)(B-\lambda{I})$ in the first summand by $(A-\lambda{I})P(\lambda)$ according to(1) and substitute $M(\lambda)$ in the second summand by $(A-\lambda{I})R(\lambda)+M$ according to (2).
  162. We obtain:
  163. $$
  164. (A-\lambda{I})P(\lambda)U(\lambda)+((A-\lambda{I})R(\lambda)+M)Q=I
  165. $$
  166. $$
  167. (A-\lambda{I})[P(\lambda)U(\lambda)+R(\lambda)Q]+MQ=I \;\;\;\;\;(6)
  168. $$
  169. Expression in the brackets in (6) equals $\mathbb{O}$. Otherwise the degree of the first summand in (6) is greater than $0$ (based on Assertion and the regularity of $(A-\lambda{I})$). Degree of the second summand and of the right side is $0$. Contradiction.
  170. It gives:
  171. $$
  172. MQ=I
  173. $$
  174. Regularity of $M$ is proved.</p>
  175.  
  176. <br/>
  177. </div>
  178. <div class="answer-pagenumber">2</div>
  179.  
  180. </div>
  181. </div>
  182. <script type="text/javascript">
  183. var gaJsHost = (("https:" == document.location.protocol) ? "https://ssl." : "http://www.");
  184. document.write(unescape("%3Cscript src='" + gaJsHost + "google-analytics.com/ga.js' type='text/javascript'%3E%3C/script%3E"));
  185. </script>
  186. <script type="text/javascript">
  187. try {
  188. var pageTracker = _gat._getTracker("UA-4276204-5");
  189. pageTracker._trackPageview();
  190. } catch(err) {}
  191. </script>
  192. <script src="http://static.getclicky.com/js" type="text/javascript"></script>
  193. <script type="text/javascript">clicky.init(250663);</script>
  194. <noscript><p><img alt="Clicky" width="1" height="1" src="http://in.getclicky.com/250663ns.gif" /></p></noscript>
  195. </body>
  196. </html>
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement
Public Pastes
Advertisement
We use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy.  OK, I Understand
Not a member of Pastebin yet?
Sign Up, it unlocks many cool features!