Could someone walk me through why saying for a matrix $\displaystyle A $

if $\displaystyle p(t) = det(tI-A) $,

then $\displaystyle p(A) = det(AI-A) = det(A-A) = 0 $

is an invalid proof of the Cayley Hamilton Theorem?

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- Feb 28th 2010, 06:27 PMchiph588@Cayley Hamilton Theorem
Could someone walk me through why saying for a matrix $\displaystyle A $

if $\displaystyle p(t) = det(tI-A) $,

then $\displaystyle p(A) = det(AI-A) = det(A-A) = 0 $

is an invalid proof of the Cayley Hamilton Theorem? - Feb 28th 2010, 06:32 PMBruno J.
Well if you look at the matrix $\displaystyle A-tI$ you see that $\displaystyle t$ occurs within the matrix; with this in mind, it certainly makes no sense to substitute a matrix for $\displaystyle t$ (you'd get matrices within a matrix, and only on the diagonal at that!).

- Feb 28th 2010, 06:33 PMDrexel28
- Feb 28th 2010, 06:35 PMchiph588@
- Feb 28th 2010, 06:38 PMtonio
- Sep 6th 2011, 08:32 AMlesnikRe: Cayley Hamilton Theorem
This topic has been bugging me for a while now. Since there's no official textbook for the linear algebra class that I had, all I (we) have is a printed latex document compiled from various notes taken by students a few years back. For a proof of this theorem I have a one-liner:

$\displaystyle (tI-A)adj(tI-A)=det(tI-A)I=p(t)I$, hence $\displaystyle p(A)=0$

As far as I can see, there's no equating between a scalar and a matrix and also no insertion of a matrix in the diagonal, since we get

$\displaystyle (AI-A)adj(AI-A)=p(A)I$

where $\displaystyle p(A)$ is a matrix polynomial, as defined here Matrix polynomial - Wikipedia, the free encyclopedia and we use this identity Adjugate matrix - Wikipedia, the free encyclopedia

My question is: could this be valid?