5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Another possibility for the JCF is the diagonal matrix of the form $(a,a,a)$, which I forgot to mention. $\endgroup$ – user437309 Jul 17 '18 at 21:37
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Wait, can't the minimal polynomial be $X-a$? $\endgroup$ – user437309 Jul 17 '18 at 20:44
  • $\begingroup$ @user437309 Absolutely, I will edit. $\endgroup$ – Arnaud Mortier Jul 17 '18 at 21:21
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.