# Cayley Hamilton Theorem

• Feb 28th 2010, 06:27 PM
chiph588@
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 PM
Bruno 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 PM
Drexel28
Quote:

Originally Posted by chiph588@
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?

Cayley?Hamilton theorem - Wikipedia, the free encyclopedia
• Feb 28th 2010, 06:35 PM
chiph588@
Quote:

Originally Posted by Bruno 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!).

Oh haha! (Surprised)(Rofl)
• Feb 28th 2010, 06:38 PM
tonio
Quote:

Originally Posted by chiph588@
Oh haha! (Surprised)(Rofl)

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
• Sep 6th 2011, 08:32 AM
lesnik
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: