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If $A$ is a Jordan cell for an eigenvalue $\lambda_1$ of multiplicity $m_1$, $B$ is a Jordan cell for an eigenvalue $\lambda_2$ of multiplicity $m_2$, and $K$ is an $m_1\times m_2$ matrix such that $$ AK=KB, $$ then show that if $\lambda_1\neq\lambda_2$, $K=0$.

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  • $\begingroup$ The same question has been asked here (but doesn't have an accepted answer). $\endgroup$ – Omnomnomnom Nov 19 '14 at 2:25
  • $\begingroup$ What does $J(0,k)$ mean? $\endgroup$ – Dia McThrees Nov 19 '14 at 2:35
  • $\begingroup$ The Jordan cell for eigenvalue $0$ with size $k$ $\endgroup$ – Omnomnomnom Nov 19 '14 at 2:35
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Let $e_1, \ldots, e_{m_2}$ be the standard $1 \times m_2$ column vectors. Since $Ke_i$ is just the $i^\text{th}$ column of $K$ what you want is to prove that $Ke_i = 0$ for all $i$. You'll do this by induction on $i$.

Here's the base case. First I'm assuming that you take Jordan normal form to have 1's above the diagonal so that $Ae_1 = \lambda_1e_1$ and $Ae_i = \lambda_1e_i + e_{i - 1}$ when $i > 1$, and similarly for $B$. Now just multiply $AK = KB$ by $e_1$ to get $$AKe_1 = KBe_1 = \lambda_2Ke_1.$$ If $Kv \neq 0$ then this would say that $Kv$ was an eigenvector of $A$ with eigenvalue $\lambda_2$. But $\lambda_1$ is the only eigenvalue of $A$ and $\lambda_1 \neq \lambda_2$ so it can't be that $Ke_1 \neq 0$, it must be that $Ke_1 = 0$.

Now the inductive step I'm going to leave to you. You'll use the fact that $Be_i = \lambda_2e_i + e_{i - 1}$.

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  • $\begingroup$ Also for the more mathy of you that are reading this: $K$ is a homomorphism between $k[x]$-modules with non-isomorphic composition factors, so it must be zero. $\endgroup$ – Jim Nov 19 '14 at 4:34

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