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$A$ is an $n \times k$ matrix and $B$ is an $k \times n$ matrix.

If $v_1, ..., v_l$ are linearly independent eigenvectors of $BA$ corresponding to a single nonzero eigenvalue $c$, then $Av_1, ..., Av_l$ are linearly independent eigenvectors of $AB$ corresponding to $c$.

I need help proving this.

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    $\begingroup$ Be careful with notation: you use the letter $k$ to represent two different things. $\endgroup$ – Alex Provost May 25 '13 at 18:48
  • $\begingroup$ What have you tried? Can you prove some part of the question? (Notice there are two statements to prove) $\endgroup$ – Najib Idrissi May 25 '13 at 19:12
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Hint: What happens if you left-multiply the equation $BAv_i=kv_i$ with $A$?

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