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.

  • 1
    $\begingroup$ You can easily prove the following: if $A$ is a square matrix such that $Av = \lambda v$, and $f$ is a polynomial, then $f(A)v = f(\lambda) v$. $\endgroup$ – Crostul Oct 6 '14 at 7:24
  • $\begingroup$ Oh, that's nice. @Crostul En, will there be any problem about multiplicity of eigenvalues? $\endgroup$ – gaoxinge Oct 6 '14 at 7:37
  • 1
    $\begingroup$ My argument shows only that $\{ f(\lambda) : \lambda \mbox{ eigenvalue of } A\} \subseteq \{ \lambda : \lambda \mbox{ eigenvalue of } f(A) \}$. I have no much ideas on how proving the other inclusion: this is the reason why I wrote a comment instead of a complete answer. $\endgroup$ – Crostul Oct 6 '14 at 7:51
  • $\begingroup$ His argument actually shows that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $v$ is an eigenvector of $f(A)$ with eigenvalue $f(\lambda)$. Doesn't that do it? $\endgroup$ – Rijul Saini Oct 6 '14 at 8:04
  • $\begingroup$ @RijulSaini I think you are right. $\endgroup$ – gaoxinge Oct 6 '14 at 8:30

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 $A$.

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