Comments on: Cayley-Hamilton Theorem and Jordan Canonical Form http://www.bigredbits.com/archives/215 Theory, Distributed Systems, and Other Random Bits Sat, 23 Apr 2011 06:32:20 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: renatoppl http://www.bigredbits.com/archives/215/comment-page-1#comment-360 renatoppl Fri, 19 Feb 2010 19:06:39 +0000 http://www.bigredbits.com/?p=215#comment-360 Thanks for the comment. I think it is actually the same proof, but you are hiding the Jordan Canonical Form in the concept of generalized eigenvectors. Thanks for the comment. I think it is actually the same proof, but you are hiding the Jordan Canonical Form in the concept of generalized eigenvectors.

]]> By: Jorge Miranda http://www.bigredbits.com/archives/215/comment-page-1#comment-359 Jorge Miranda Fri, 19 Feb 2010 18:56:44 +0000 http://www.bigredbits.com/?p=215#comment-359 I think it's even easier than that using the Jordan Canonical Form. The idea is that $p_A(A)=(A-\lambda_1I)...(A-\lambda_nI)$. If we compute $p_A(A)v$ for any $v$ that is a generalized eigenvector $p_A(A)v=0$ because we have enough factors in $p_A(A)$ to use that $(A-\lambda_iI)^kv=0$. Since the generalized eigenvectors are a basis, we conclude that $p_A(A)v=0$ for any $v$, and that means that $p_A(A)=0$ as we wanted to prove. Still, neat proof :D I think it’s even easier than that using the Jordan Canonical Form.

The idea is that $p_A(A)=(A-\lambda_1I)…(A-\lambda_nI)$. If we compute $p_A(A)v$ for any $v$ that is a generalized eigenvector $p_A(A)v=0$ because we have enough factors in $p_A(A)$ to use that $(A-\lambda_iI)^kv=0$.

Since the generalized eigenvectors are a basis, we conclude that $p_A(A)v=0$ for any $v$, and that means that $p_A(A)=0$ as we wanted to prove.

Still, neat proof :D

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