I'm a TA with Advanced Algebra in school and teach the Jordan Form now. There are three questions about eigenvalues in this chapter:
Given matrix $A$, $B$ and polynome $f$, consider the eigenvalues' relation between
$A$ and $A^T$
$AB$ and $BA$
$A$ and $f(A)$
Except the last, we can solve them without the Jordan Form because of $|\lambda I-A|=|\lambda I-A^T|$ and $|\lambda I-AB|=|\lambda I-BA|$. Last is easily solved by Jordan Form, and the conclusion is
If $\lambda_1$, $\cdots$, $\lambda_n$ are $A$'s eigenvalues, then $f(\lambda_1)$, $\cdots$, $f(\lambda_n)$ are $f(A)$'s eigenvalues.
However I am not satisfied with the method of Jordan Form since
The first and second are just use the properties of determinant
If using the method of Jordan Form, we actually introduce the field of coefficient into its algebraically closed field. But I want to avoid the situation.
So are there any other approaches to the last question? Any advice is helpful. Thank you.
Let's try to understand what the statement is: the characteristic polynomial of $f(A)$ is the monic polynomial with roots $f(\lambda_i)$, where $\lambda_i$ are the roots of the characteristic polynomial of $A$.
Now, for a given polynomial $P$ with roots $\lambda_i$, $1\le i \le n$ and another polynomial $f$, the monic polynomial with roots $f(\lambda_i)$ has coefficients that are polynomial expressions in the coefficients of $P$ and $f$ ( this uses the fundamental theorem of symmetric functions).
Therefore, to show that $P_{f(A)}$ has the eigenvalues $f(\lambda_i)$ we need to prove $n$ polynomial identities involving the coefficients of $P_A(t)$, $P_A(t)$ and $f$, hence the entries of $A$ and the coefficients of $f$. These identities have coefficients in $\mathbb{Z}$. To show they are true it's enough to show they are true if the entries of $A$ and the coefficients of $f$ are from an infinite field. Let's choose that field $\mathbb{C}$. Fix the polynomial $f$. Now you need to show some identities in the entries of $A$. These identities are valid for $A$ in some dense open subset- the subset of matrices with distinct eigenvalues. Therefore, the identities are always true. Details below:
If $A$ is a diagonal matrix then the statement is true - easy.
If the statement is true for $A$ then it is true for any $A'$ conjugate to $A$ : $A' = T A T^{-1}$ - easy.
The set of complex matrices $A$ conjugate to a diagonal matrix with distinct eigenvalues $=$ set of matrices whose eigenvalues are distinct $=$ matrices $A$ so that the discriminant of $P_A(t)$ is nonzero - an open (Zariski) nonvoid set of complex matrices.
If a polynomial identity in the entries of $A$ holds for all $A$ in a non void open subset it holds for all $A$.