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I supose that if $t$ is an eigenvalue of $B$ and hence there is an eigenvector $x$ associated such that

$Bx=tx$

$MBx=M(tx)$

$AMx=A(Mx)=t(Mx)$

How can I ensure that $Mx$ is different from $0$ so that $t$ is an eigenvalue of $A$?

Feng
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Renato lorentz
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  • If $Mx=0$ for all EigenVectors $x$ , then $M=0$ , & we are given that $M$ is not Zero , hence there is atleast 1 EigenVector $x$ where $Mx$ is not Zero. That matching $t$ is the common EigenValue. – Prem Oct 22 '22 at 06:14
  • @Prem The problem is we don't know whether the sum of the eigenspaces of $A$ and/or $B$ covers all of $\mathbb{C}^n$ ($x$ here is not an eigenvalue of $M$), so we can't conclude that there exists such an $x$ that simply, but I very well could be missing something – Bruno B Oct 22 '22 at 06:18
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    @Prem Why all eigenvectors of $B$ do span the whole space $\mathbb{C}^n$? – Landau Oct 22 '22 at 06:20

2 Answers2

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Let $p$ be the characteristic polynomial of $B$.

Because $AM=MB$, we have $p(A)M=Mp(B)$. It follows from Cayley's theorem that $p(A)M=0$. We see $p(A)$ is not invertible since $M\neq 0$. I think this is enough to draw the conclusion.

Landau
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    (+1) Yes! Assume that $p(\lambda)=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n)$ where $\lambda_i$ are the eigenvalues of $B$. If $p(A)$ is not invertible, then $0=\det(p(A))=\det(A-\lambda_1I)\cdots \det(A-\lambda_nI)$, which implies that $\det(A-\lambda_iI)=0$ for some $i$, and this $\lambda_i$ is a common eigenvalue of $A$ and $B$. – Feng Oct 22 '22 at 06:48
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(IMPORTANT: this is currently only a partial answer that was too long to be a comment, the answer from Landau is much quicker, simpler, and actually proves the result in its entirety! However, I wanted to show something that followed the footsteps of what Renato and Prem tried, plus it at least covers more than just invertible $M$ which would be trivial due to being able to write $B = M^{-1}AM$.)

Eigenvectors alone might not get the job done, but I thought we could get the same kind of method to work, so we'll go a small step further:

Assume $M$ is not nilpotent, i.e. $\forall j \in \mathbb{N},\,\, M^j \neq 0$. (This is why the answer is only partial, but since it covers slightly more than just $M$ invertible I'll write it anyway).
Since we're in $\mathbb{C}$, the characteristic poynomial of $B$ $\chi_B$ can be written in the form: $$\chi_B(X) =: \prod_{k=0}^s \left(X - \lambda_k\right)^{m_k}$$ with $\sum_{k=0}^s m_k = n$ since $\deg(\chi_B) = n$.
Let $C_k := \operatorname{Ker}(B - \lambda_kI_n)^{m_k}$ be the set of generalized eigenvectors of $B$ for $\lambda_k$. It is a known result that: $$\mathbb{C}^n = \bigoplus_{k=0}^{s} C_k$$ in contrast to the usual eigenspaces whose sum might not always cover all of $\mathbb{C}^n$.

By our assumption on $M$, there must exist $k$ and $x \in C_k$ such that $y := M^{m_k} x \neq 0$ by the above statement.
We then obtain: $$\left(A - \lambda_k I_n \right)^{m_k}y = \left(AM - \lambda_k M\right)^{m_k}x = \left(MB - \lambda_k M\right)^{m_k}x = M^{m_k}\left(B - \lambda_k I_n\right)^{m_k}x = 0$$ Therefore, $y$ is a generalized eigenvector of $A$ for $\lambda_k$, and as such $\lambda_k$ is an eigenvalue of $A$ (and for at most the same multiplicity as for $B$), which concludes the proof for $M$ non-nilpotent (well, technically the proof only needs $M$ with a nilpotence index of over $\max_k{m_k} + 1$ but that's still not every $M$...).

Bruno B
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