# Prove that two operators have a common eigenvector

Suppose $S,T:\mathbb C^3\to\mathbb C^3$ are linear operators. The degree of the minimal polynomial of each of the operators is at most 2. Show that they share a common eigenvector.

I tried to exploit the condition on the degree. I obtained that there are two possible Jordan canonical forms for $S,T$. The first possibility is that the JCF is diagonal of the form $(a,a,b)$. The second possibility is that the JCF has one block of size 2 with eigenvalue $a$ and 1 block of size 1 with eigenvalue $a$. For the other operator the possibilities for the JCF are the same except that the eigenvalues may be different. But I don't know how to proceed from this point.

• Another possibility for the JCF is the diagonal matrix of the form $(a,a,a)$, which I forgot to mention. – user437309 Jul 17 '18 at 21:37

## 2 Answers

Your work on the JCFs is correct:

• If the minimal polynomial is $(X-a)^2$ then the only eigenvalue is $a$ and the largest block size is $2$, so the other block is of size $1$

• If the minimal polynomial is $(X-a)(X-b)$ then there are two eigenvalues $a$ and $b$ and since the minimal polynomial has no multiple roots the matrix is diagonalisable.

Now note that in each case, there is an eigenvalue whose associated eigenspace has dimension $2$. The dimension is indeed the number of Jordan blocks for a given eigenvalue.

All you need now is that in dimension $3$, two planes must intersect non-trivially.

Edit: one case was forgotten, since the degree of the minimal polynomial is at most $2$, it could be $1$. In that case, the minimal polynomial is $X-a$ for some $a$, meaning that $S$ (or $T$) is a multiple of the identity and the whole of $\Bbb C^3$ is their eigenspace. Of course, the intersection of the eigenspaces is again non-trivial in that case.

• Wait, can't the minimal polynomial be $X-a$? – user437309 Jul 17 '18 at 20:44
• @user437309 Absolutely, I will edit. – Arnaud Mortier Jul 17 '18 at 21:21

We can mark $T_\lambda,S_\mu$ s.t $dim T_\lambda=2,dim S_\mu=2$ the eigenspaces coressponding to the 2 jordan blocks and $dim (T_\lambda\cap S_\mu)=dim(T_\lambda)+dim(S_\mu)-dim(T_\lambda+S_\mu)\geq2+2-3=1$

Therefore, $T_\lambda\cap S_\mu\neq\ \{\theta\}$ and there are shared eigenvectors.