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Find a basis for $U=\{A\in\mathbb{M}_{22}\mid A^T=-A\}$.

$\mathbb{M}_{22}$ denotes the set of all $2 \times 2$ matrices. This question appeared on an examination I wrote yesterday. Does a basis even exist? I can't think of any matrices in which $A^T=-A$ except for the zero matrix. If this is the case, I was taught (if I recall correctly), that $0$ can not be in a basis because it is linearly dependent.

$\begin{bmatrix}a&b\\b&c\end{bmatrix}\to \begin{bmatrix}-a&-b\\-b&-c \end{bmatrix}\implies a=b=c=0?$

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  • $\begingroup$ $$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$ $\endgroup$ – Daniel Fischer Aug 10 '13 at 21:20
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Hint: The diagonal entries must be $0$ and the other entries must be opposite. (Why?) Can you think of a non-zero matrix with this form? What relationship will other elements of $U$ have to this matrix?


It is true that the only element of $U$ having the form $$\begin{bmatrix}a&b\\b&c\end{bmatrix}$$ is the zero matrix, but I see no reason to assume that we're dealing with symmetric matrices only.

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Well, let $$ A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} \in \mathbb{M}_{22}. $$ Then $$ A + A^T = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix} + \begin{pmatrix} A_{11} & A_{21} \\ A_{12} & A_{22} \end{pmatrix} = \begin{pmatrix} 2 A_{11} & A_{12}+A_{21} \\ A_{21}+A_{12} & 2A_{22}\end{pmatrix}, $$ so as long as you're working over a field of characteristic $\neq 2$, $A^T = -A$ if and only if $$ A_{11} = 0, \quad A_{22} = 0, \quad A_{21} = -A_{12}, $$ so that $$ U = \left\{\left. \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix} \; \right| \;a \in F \right\} = \operatorname{span}\left\{ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\}. $$ So, what can you conclude?

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  • $\begingroup$ A basis, by definition, is a linearly independent spanning set... $\endgroup$ – Branimir Ćaćić Aug 10 '13 at 21:37

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