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My problem:

Prove that for any matrix $A$ and for any polynomial $p(x)$ we have that $\sigma (p(A)) = p(\sigma (A))$, where $\sigma$ is the spectrum of a matrix. Prove that $A$ is an invertible matrix if no root of $p(x)$ is an eigenvalue.

My solution:

For example, if I have the polynomial $\;$ $p(x) = x^2 + x$ $\;$ and the matrix $$ A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} $$

then

$$ p(A)= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}^{2}+ {\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}}= {\begin{pmatrix} 2 & 0 \\ 0 & 2 \\ \end{pmatrix}} $$ $$ $$

We can see that the spectrum of $p(A)$ contains the eigenvalues $\lambda_{1,2} = 2$

The spectrum of the matrix $A$ contains eigenvalues $\lambda_{1,2} = 1$ and if we substitude it into $p(x)$, then $p(\lambda) = p(1) = 1^2 + 1 = 2$

The spectrum of $A$ is the set of its eigenvalues and the eigenvalues are the roots of the characteristic polynomial, but I am not sure how to prove $\sigma (p(A)) = p(\sigma (A))$ generally. The second proof I think I understand. Can anyone help me?

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    You need to work in an algebraically closed field, otherwise you might not be able to deal with cases like $p=x^2,\, A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. – 2017-02-15
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    Yes, I need to allow the complex eigenvalues. But what's next please? – 2017-02-15
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    @Leif do you know that every matrix can be upper triangularized? Or, do you know about Jordan form? – 2017-02-15
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    Actually not about that Jordan form, next semester. But I think I have found the solution. – 2017-02-15

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