If $P(x)$ is characteristic polynomial of $A$ then is $P(A) = 0$?

Question

I'm a student and I've just read the Characteristic polynomial on Wiki. I have a feeling that:

If $P(x)$ is characteristic polynomial of $A$ then is $P(A) = 0$

thanks to the Matrix calculator I've tested it for the matrix that is the wikis example and another matrix, and this was true for both of them. the matrices were $ \begin{bmatrix}2 & 1 \\-1 & 0 \end{bmatrix} $ and $ \begin{bmatrix}-1 & -1 \\1 & 1 \end{bmatrix}$.

I've thought about it. if the eigenvectors of the matrix span the space. so we can write every vector in the space as a linear combination of its eigenvectors. So for every $v$ in space $P(A)v = P(A)(\alpha_1e_1 + \alpha_2e_2 + ...)$ and it is easy to see that for every $e_i$ there is $P(A)e_i = 0$

BUT the problem is for the matrices like $ \begin{bmatrix}-1 & -1 \\1 & 1 \end{bmatrix}$ that dosent have $n$ distinct eigenvectors that span the space. I have no idea for this kind.

Any help would be appreciated.

Answer 1

Yes, that is correct. Actually, that's what the Cayley-Hamilton theorem says.

Answer 2

This is the Cayley-Hamilton theorem.