Cayley-Hamilton Theorem for matrices

Last_Singularity · #1 $Old$ November 5th, 2009, 08:50 PM

The Cayley Hamilton Theorem states that if $T$ is a linear operation on vector space $V$ and $f(t)$ is the characteristic polynomial of $T$ , then $f(T)$ is the zero transformation.

How do I extend this to matrices? As in, how do I show that: if $A$ is $n \times n$ and $f(t)$ is the characteristic of $A$ , then $f(A)$ is the zero matrix?

I attempted to prove it using $f(A) = det(A - AI) = det(O) = 0$ but this reasoning seems faulty...

Jose27 · #2 $Old$ November 5th, 2009, 09:31 PM

Remember that linear transformations between finite dimensional vector spaces are in 1-1 correspondence with matrices (and they do the same thing with the exception that the matrices act on $\mathbb{F} ^n$ ) so translate everything from matrices to lin.tranf. and see what $f(A)$ does in $\mathbb{F} ^n$