Assume matrices $A,B\in\mathbb{C}^{n\times n}$ s.t.

  • $A,B$ nilpotent.
  • $A,B$ have the same minimal polynomial.
  • $\dim\mathcal{N}(A-\lambda I)=\dim\mathcal{N}(B-\lambda I)$, for all $\lambda\in\mathbb{C}$.

I want to prove that $A,B$ similar for $n=6$ and check if it also holds for $n=7$.

My attempt so far is (will cite theorems from Horn and Johnson's Matrix Analysis book):

Since $A,B$ nilpotent their characteristic polynomial is $p_A(t)=p_B(t)=t^n$ so they can only have the zero eigenvalue. We are also given that $\dim\mathcal{N}(A-\lambda I)=\dim\mathcal{N}(B-\lambda I)$, $\forall\lambda\in\mathbb{C}$ thus it will hold for their eigenvalue i.e. their eigenvalue will have the same geometric multiplicity in both cases.

I think the proof is complete here (since they will have the same Jordan Canonical Form) and we don't need the fact that they have the same minimal polynomial. Is this true?

But I will state some proven facts about the minimal polynomial. By Cor. 3.3.4 same minimal polynomial means same eigenvalues and by Thm. 3.3.6 the largest Jordan block that corresponds to each eigenvalue of them will be the same on the JCFs of $A$ and $B$. Thm. 3.3.6 says that

Let $A\in\mathbb{C}^{n\times n}$ with distinct eigenvalues $\lambda_1,\dots,\lambda_d$. The minimal polynomial is $$q_A(t)=\prod_{i=1}^d(t-\lambda_i)^{r_i}$$ in which $r_i$ is the size of the largest Jordan block of $A$ corresponding to the eigenvalue $\lambda_i$.

I also think that this proof works for any $n$ so also for $n=7$.


You have to check that both matrices have the same Jordan form, i.e., the Jordan blocks have the same size.

This can be proven for $n\le6$ given the assumptions.

For $n=7$ and degree of minimal polynomial $=3$, one can have matrices $A$ and $B$ with Jordan blocks of sizes $3,3,1$ and $3,2,2$. Hence they satisfy the assumptions and are not similar.

  • $\begingroup$ Can you give some insight on the proof for $n\leq6$ and can you pinpoint where is the problem that my proof is not general? $\endgroup$ – mgus Mar 13 '18 at 18:42
  • $\begingroup$ your proof only shows that all eigenvalues are zero, which is not enough. dimension of null space gives number of Jordan blocks. Degree of minimal polynomial gives largest Jordan block. Then you can show that for $n\le6$ this is enough to show that these matrices have the same Jordan normal form. $\endgroup$ – daw Mar 13 '18 at 19:14

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.