Suppose $E=\{X\in M_4(\mathbb{C}): X^T=-X\}$ and that there exists $a\in SL(4,\mathbb{C})$ such that for all $X\in E$ $$ aXa^T=X $$ I want to show that it follows that $a=\pm I$. This can be done by finding a basis for $E$ and working with matrix multiplication directly, but this is not an elegant argument. Is there perhaps a better one? Is it possible to show that $a\in Z(SL(4,\mathbb{C}))$ maybe?
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If you take $X$ to be the zero matrix then $X \in E$ and $a$ can be anything you want. – Wintermute Mar 31 '13 at 16:42
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@mtiano, the equality needs to hold for all $X\in E$, not just for $X=0$. – Jimmy R Mar 31 '13 at 16:44
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@JimmyR: In this case, you have to change the order of the quantifiers: There exists $a$, such that for all $E$... – gerw Mar 31 '13 at 16:59
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@gerw, sure, done it. – Jimmy R Mar 31 '13 at 17:02
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Consider $X$ of the form $vw^T - wv^T$, where $u,v$ are vectors.
Take $v$ as a right singular vector of $A$ such that $Av=su$. Then \begin{align*} &A(vw^T - wv^T)A^T = vw^T - wv^T\quad \forall w\in\mathbb{C}^n\\ \Rightarrow&su(Aw)^T - s(Aw)u^T = vw^T - wv^T\quad \forall w\in\mathbb{C}^n. \end{align*} As $w$ is arbitrary, it follows that $u,v$ are parallel to each other. Since $u$ and $v$ are (left or right) singular vectors, in turn we must have $u=\pm v$. Thus $A$ is unitarily diagonalizable.
Now suppose $v$ and $w$ are two linearly independent eigenvectors of $A$ corresponding to the eigenvalues $\lambda$ and $\mu$ respectively. Then \begin{align*} &A(vw^T - wv^T)A^T = vw^T - wv^T\\ \Rightarrow&\lambda\mu (vw^T - wv^T) = vw^T - wv^T. \end{align*} Therefore $\lambda\mu=1$ for any two eigenvalues $\lambda,\,\mu$ of $A$. Hence $A=\pm I$.
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