# Proof of the linear independence of the generalized eigenvectors of a square matrix

I'm currently stuck on this problem:

Let $V$ be a finite dimensional vector space. If $S: V\rightarrow V$ and $T: V\rightarrow V$ are linear maps and $ST=TS$, prove every eigenvalue of $ST$ is a product of an eigenvalue of $S$ with an eigenvalue of $T$.

One thing that I do know - as a consequence of the above - is that there is a basis of simultaneous eigenvectors. Do I somehow use this fact?

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This is true only over algebraically closed fields. E.g. if $T=S=$ matrix of rotation by $+90^\circ$, then $T$ and $S$ don't have real eigenvalues, but $TS=-Id$ does have. –  Berci Apr 25 '13 at 10:30
@Berci Right. Good example. –  julien Apr 25 '13 at 10:31
Your "consequence" is false. This would imply that $S$ and $T$ are simultaneuously diagonalizable. It is true, nevertheless, if $S$ and $T$ commute and are both diagonalizable. But for the result you want, all you need is that they be simultaneously triagularizable. And this is true for any commuting $S$, $T$. –  julien Apr 25 '13 at 11:13

There is only a basis of simultaneous eigenvectors if $S$ and $T$ are both diagonalizable, but in that case the answer is yes - try multiplying some element of the basis of simultaneous eigenvectors by $ST$ and seeing what happens. This will give you a set of eigenvalues of $ST$, and you need to see why this set must in fact be all the eigenvalues.

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Thanks, I've shown that the product of the eigenvalues for the simultaneous eigenvectors (when used in $S$ and $T$) are eigenvalues for $ST$. Is it true that every eigenvector of $ST$ must be one of the simultaneous eigenvectors? This would show that the set you mentioned would consist of all the eigenvalues. –  Mel Apr 25 '13 at 10:34
You should have found $n$ eigenvalues for $ST$ (where $n$ is the dimension of $V$), possibly with some multiplicity; one for each of your linearly independent eigenvectors. A map $V\to V$ can't have more than $\dim{V}$ eigenvalues. –  Matt Pressland Apr 25 '13 at 10:37
Thinking about it more, you should only know that there is a basis of simultaneous eigenvectors if $S$ and $T$ are both diagonalizable, which isn't an assumption you have. However, Julien's statement that the two are simultaneously triangularizable is enough. (I would also recommend accepting that answer rather than this, because it addresses that issue properly). –  Matt Pressland Apr 25 '13 at 10:51

You need the scalar field to be algebraically closed (see Berci's example in he comments above). I am pretty sure it is $\mathbb{C}$ in your case, but the argument works exactly the same for a general algebraically closed field $K$.

Fact: any $S\in L(V)$ is triangularizable. That is, there exists a basis of $V$ with respect to which the matrix of $T$ is upper triangular.

Generalization: any pair of commuting $S,T\in L(V)$ is simultaneously upper triangularizable.

Application: take a basis of simultaneous upper triangularization for $S$ and $T$. Then the matrix of $ST=TS$ is also upper triangular. Its diagonal is the product of the diagonals of $S$ and $T$. In particular, every eigenvalue of $ST$ is the product of an eigenvalue of $S$ by an eigenvalue of $T$.

Proof of the fact: induction on $n$, the dimension of $V$. The key point is that there always exists at least an eigenvalue $\lambda$, as the characteristic polynomial splits over $K$. For $n=1$, the assertion is trivial. Assume it holds up to $n-1\geq 1$. Then take an eigenvalue $\lambda$. Take a basis of the eigenspace $V_\lambda:=\mbox{Ker}(S-\lambda I)$, and complete it into a basis of $V$. If $V_\lambda$ was already equal to$V$, this means $S$ is scalar and we are done. If not, then the matrix of $S$ with respect to this basis is a $2\times 2$ block matrix. The upper left block is $\lambda I$, and the lower left block is $0$. Apply the induction hypothesis to the lower right block. QED.

Proof of the generalization: you can do that by induction on $n\dim V$ again. On a one-dimensional space, it is trivial. Assume this holds for any dimension up to $n-1\geq$. Take $S,T$ commuting. Since $K$ is algebraically closed, the characteristic polynomial of $S$ has at least one root $\lambda$. This is an eigenvalue of $S$. On the eigenspace $V_\lambda:=\mbox{Ker}(S-\lambda I)$, $Sx=\lambda x$. Therefore $STx=TSx=\lambda Tx$, i.e. $Tx\in V_\lambda$ for every $x\in V_\lambda$. In other words, $V_\lambda$ is invariant under $T$. Using the fact, you get a basis of $V_\lambda$ in which $T$ is upper triangular and, of course, $S$ is diagonal. If $V_\lambda=V$, we are done. If not, take a basis of $V_\lambda$, and complete it into a basis of $V$. Then the matrices of $S$ and $T$ are $2\times 2$ block matrices. As $V_\lambda$ is invariant under $S$ and $T$, they are block upper triangular. We don't care about the upper right blocks. By writing down the corresponding $2\times 2$ block product, the condition $ST=TS$ implies that the lower right blocks are still commuting. Apply the induction hypothesis to them. QED.

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Can you sketch how the Generalization is shown? –  Berci Apr 25 '13 at 10:41
@Berci I was actually writing this down when you commented. I hope it is clear enough. –  julien Apr 25 '13 at 10:57
Great, thank you. –  Berci Apr 25 '13 at 15:00