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Putnam Problem 1990-A5
math.stackexchange.com/questions/628154
Five Theorems in Matrix Analysis, with Applications
Bits
Yes for $n ≤ 2$; no for $n > 2$.
For $n = 2$: The characteristic polynomials of $AB$ and $BA$ are the same (for any $n$). Therefore, if $AB$ is nilpotent then $BA$ is also nilpotent. So, we necessarily have $BABA = (BA)^2 = 0$.
For $n≥3$. Let $A = \begin{pmatrix} 0&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}$ and $B = \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}$. Then \[ABAB = (AB)^2 = \begin{pmatrix} 0&0&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}^2 = 0\]but \[BABA = (BA)^2 = \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}^2 = \begin{pmatrix}0&0&1\\0&0&0\\0&0&0 \end{pmatrix} \neq 0.\]
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