Let $A,B$ be $n\times n$ matrix
If $(A^2)(B^2) = (B^2)(A^2)$ , is $AB=BA$?
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It is not true in general that $A^2B^2=B^2A^2$ implies $AB=BA$. For example let $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$
Clive Newstead
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Can I know what motivated you to find the counterexample? – tarit goswami Oct 11 '18 at 14:35
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2@taritgoswami: These matrices are both canonical examples of nonzero matrices whose squares are zero matrices, which means the condition $A^2B^2=B^2A^2$ is satisfied trivially. Matrices like this are great for counterexamples. – Clive Newstead Oct 11 '18 at 14:36
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No, just take the matrix (hope no miscalculations)
$ A = \bigg(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\bigg) $ and $ B = \bigg(\begin{array}{ll} 0 & 0 \\1 & 0 \end{array}\bigg) $
Ulrich
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Someone has already answered in same way. Can you explain it more clearly to make it an different answer ? – tarit goswami Oct 11 '18 at 14:45