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Let $A \in R^{m\times k}, B\in R^{k \times m}$ be non-square matrices.

A vector $v \in R^m$ is said to be a generalized eigenvector of $AB$ corresponding to $\lambda \neq 0$ if $(AB- \lambda I)^m v = 0$.

I know that if $v$ is an eigenvector of $AB$, then $Bv$ is an eigenvector of $BA$.

However, if $v$ be any generalized eigenvector of $AB$, i.e., $ (AB- \lambda I)^m v = 0$ then can I say that $Bv$ is a corresponding generalized eigenvector of $BA$ ? That is, is it true that $(BA-\lambda I)^k Bv = 0$ ?

By expanding the binomial series,and pre-multiplying with B, I think am getting the result. However, since generalized eigenvectors include ordinary eigenvectors, am worried if my purported proof is correct.

me10240
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  • Is there a reason for assuming $\lambda\neq0$? – Arthur Sep 24 '19 at 06:08
  • Yes, it ensures Bv is not 0. – me10240 Sep 24 '19 at 07:09
  • Is there a hole in my proof ? I basically want to use this further and show that the algebraic multiplicity of a nonzero eigenvalue is the same for AB, BA. – me10240 Sep 24 '19 at 07:11
  • I can't tell whether there is a hole in your proof, because you haven't shown me your proof. You've only told me the general idea behind it. But the result is correct. – Arthur Sep 24 '19 at 07:32

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Let $n = \max(k, m)$. We have $$ (BA - \lambda I)^nBv = \left(\sum_{i = 0}^n \binom ni\lambda^{n-i}(BA)^i\right)Bv\\ = \left(\lambda^nI + \sum_{i = 1}^n\binom ni \lambda^{n-i}B(AB)^{i-1}A\right)Bv\\ = \left(\lambda^n B + \sum_{i = 1}^n\binom ni \lambda^{n-i}B(AB)^{i-1}AB\right)v\\ = B\left(\lambda^nI + \sum_{i = 1}^n\binom ni \lambda^{n-i}(AB)^{i}\right)v\\ = B(AB-\lambda I)^nv = B0 = 0 $$ So yes, indeed, if $v$ is a generalized eigenvector of $AB$ with generalized eigenvalue $\lambda$ and generalized eigenvector $v$, and $Bv\neq 0$, then $Bv$ is a generalized eigenvector of $BA$.

Arthur
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  • I think this is fine if $k > m$, but not enough if case $k < m$, for we do not know if $v$ is then a generaized eigenvector. But if you instead start with $(BA - \lambda I)^m Bv$, that will work for both cases using the property that the nullspaces are nested for increasing powers and finally become constant once the exponent equals the dimension of the space. – me10240 Sep 26 '19 at 06:24
  • @me10240 That's true. I should've just used $\max(k, m)$ as the exponent from the start. – Arthur Sep 26 '19 at 06:42
  • @me10240 Could you please give me a hint for this question https://math.stackexchange.com/questions/4035136/show-that-dim-v-dim-w?noredirect=1#comment8332207_4035136 – Priya Feb 23 '21 at 08:52
  • @Arthur We can prove that $AB$ and $BA$ have the same eigenvalues but they are order or size is different $AB_{m\times m}$ and $BA_{k\times k}$, so the Jordan normal form of them should be different, am I right? Does that mean, they have the same eigenvalues but multiplicities are different? Then generalized eigenvectors are also different, am I right? – Priya Feb 24 '21 at 01:08
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    @Priya They do have different Jordan forms, yes. That's not the question here, though. It's been a while since I wrote this, but I think what I'm saying here is that any non-zero eigenvalue has the same algebraic multiplicity in $AB$ as in $BA$. So the diagonal off the Jordan matrix only differs by the number of zeroes. – Arthur Feb 24 '21 at 06:24
  • @Arthur I am trying to find an answer to another question https://math.stackexchange.com/questions/4035136/show-that-dim-v-dim-w Here I want to show that the equality of the dimension of generalized eigenspaces for $BA$ and $AB$. – Priya Feb 25 '21 at 01:08