$A$ is an $n \times n$ real matrix.
prove that
$$A=-A^T \iff AA^T=-A^2$$.
Thanks.
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1One implication is immediate. Where are you stuck on the other direction? – Pedro Nov 09 '13 at 11:36
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Actually I just don't know how to start.Whatever I try,It will require that A is inventible. – Yuxuan Dong Nov 09 '13 at 15:16
1 Answers
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Here's a proof for the nontrivial direction, right to left in the question. Like any real square matrix, $A$ can be represented as the sum $A=S+G$ of a symmetric matrix $S=(A+A^T)/2$ and an antisymmetric matrix $G=(A-A^T)/2$. The given equation $AA^T=-A^2$ can be rewritten as $(S+G)S=0$ or $S^2+GS=0$. Our task is to prove that $S=0$. Suppose not. By the spectral theorem, $S$ has a non-zero eigenvalue $\lambda$; let $v$ be a column eigenvector. So $v\neq0$ and $Sv=\lambda v$. Therefore $S^2v=\lambda^2v$ and, since $S^2+GS=0$, we get $$ \lambda Gv=G(\lambda v)=GSv=-S^2v=-\lambda^2v. $$ As $\lambda\neq0$, we infer $Gv=-\lambda v$. Using this, we compute $$ \lambda v^Tv=-v^TGv=+v^TG^Tv=(Gv)^Tv=-\lambda v^Tv, $$ where the second equality uses that $G$ is antisymmetric. So $2\lambda v^Tv=0$. But this is absurd, since $\lambda$ and $v^Tv$ (and 2) are non-zero.
Andreas Blass
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To Andreas Blass: that is just so cute! Very well done indeed, sir! Way cool, +1! – Robert Lewis Nov 10 '13 at 09:04
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I've never know the spectral theorem util now.This answer showed me what it is and how to use it.Thanks. – Yuxuan Dong Nov 12 '13 at 14:16