Let $ A \in M_{n \times n}$. I know that $ A^T = \pm A $ and I need show that $ \ker(A) = \ker(A^2) $. I was wandering about diagonal form but a skew matrix need not be diagonalizable.
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what do you know about $\ker (A^TA)$? – daw Jun 28 '23 at 12:41
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for sure i can see that $ Ker(A) \subset Ker(A^TA) $ – olaf Jun 28 '23 at 12:44
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$\ker(A)=\ker(A^TA)=\ker(\pm A^2)=\ker(A^2)$ by the duplicate. – Dietrich Burde Jun 28 '23 at 12:45