# Cayley Hamilton Theorem Question

• Oct 17th 2010, 01:28 PM
lordslytherin
Cayley Hamilton Theorem Question
Hello, a tad stuck on an algebra assignment.

The Cayley Hamilton theorem implies that, for any nxn matrix A over a field K, there is a polynomial p(x) with coefficents in K and degree n in x such that p(A)=0.
By considering the matrices A^i for ,0 less than or eqaul to (i) less than or equal to (n^2), prove, without using the cayley hamilton theorem, that there is a non-zero polynomial p(x) of degree at most (n^2) in x, such that p(A)=0.

Thankyou
• Oct 17th 2010, 01:42 PM
Also sprach Zarathustra
Look at the n^2 matrices:

I,A,A^2,...,A^(n^2).

Remember that the vector space M of matrices nxn is n^2, but you have n^2+1 matrices, so...?
• Oct 17th 2010, 01:53 PM
lordslytherin
Quote:

Originally Posted by Also sprach Zarathustra
Look at the n^2 matrices:

I,A,A^2,...,A^(n^2).

Done this, but still stuck.
• Oct 17th 2010, 02:09 PM
lordslytherin
Quote:

Originally Posted by Also sprach Zarathustra
but you have n^2+1 matrices

why?
• Oct 17th 2010, 02:11 PM
Also sprach Zarathustra
Quote:

Originally Posted by lordslytherin
why?

I,A,A^2,...,A^(n^2) <==>A^0,A,A^2,...,A^(n^2).
• Oct 17th 2010, 02:14 PM
lordslytherin
Quote:

Originally Posted by Also sprach Zarathustra
I,A,A^2,...,A^(n^2) <==>A^0,A,A^2,...,A^(n^2).

Thought as much, when i read it at n^3, rather than (n^2)+1. Still at a loss to do after that though.
• Oct 17th 2010, 02:23 PM
Also sprach Zarathustra
I will continue from my first comment.

...hence, n^2+1 matrices are linear dependence, in other words:

there exist t_0,t_1,...t_{n^2+1} scalars that not all of them is zero, which for them:

t_{n^2+1}A^{n^2+1}+...+t_1A+t_0I=0

or in other writing...

A is a root of f(x)=t_{n^2+1}x^{n^2+1}+...+t_1x+t_0I.
• Oct 17th 2010, 02:29 PM
lordslytherin
Thanks alot Also sprach zarathustra, most helpful.