1. ## Cayley Hamilton Theorem

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

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

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

is an invalid proof of the Cayley Hamilton Theorem?

2. Well if you look at the matrix $A-tI$ you see that $t$ occurs within the matrix; with this in mind, it certainly makes no sense to substitute a matrix for $t$ (you'd get matrices within a matrix, and only on the diagonal at that!).

3. Originally Posted by chiph588@
Could someone walk me through why saying for a matrix $A$

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

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

is an invalid proof of the Cayley Hamilton Theorem?
Cayley?Hamilton theorem - Wikipedia, the free encyclopedia

4. Originally Posted by Bruno J.
Well if you look at the matrix $A-tI$ you see that $t$ occurs within the matrix; with this in mind, it certainly makes no sense to substitute a matrix for $t$ (you'd get matrices within a matrix, and only on the diagonal at that!).
Oh haha!

5. Originally Posted by chiph588@
Oh haha!

Great answer...and yes: of course that was an illegal "proof", since you'd first have to define polynomials with matrices as argument, which can be done btw, but first you have to define it properly, at least.

Tonio

6. ## Re: 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:

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

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

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

where $p(A)$ 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?