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What's the relation between a matrix $A$ that represents a skew-symmetric form, and a skew-symmetric matrix $B$?

The matrix $A$ is such that for all vectors $v,w$, we have $v^TAw=-w^TAv$.

The matrix $B$ is such that $B^T=-B$.

So their characterizations are different. Is there a relation between these two sets of matrices?

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  • $\begingroup$ What do you get when you take the transpose of $v^tAw$? $\endgroup$ – Gerry Myerson Aug 9 '13 at 13:02
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    $\begingroup$ On the one hand, (2) implies (1), since if $A^T = -A$, then $$v^T A w = (v^T A w)^T = w^T A^T v = - w^T A v.$$ On the other hand, (1) implies (2), since if $\{e_1,\dotsc,e_n\}$ is the standard ordered basis of $F^n$, then $$ A_{ij} = e_i^T A e_j = -e_j^T A e_i = -A_{ji}, $$ and hence $A = -A^T$. $\endgroup$ – Branimir Ćaćić Aug 9 '13 at 13:02
  • $\begingroup$ Oh, I absolutely forgot that $v^TAw$ is just a scalar, and so $v^TAw=(v^TAw)^T$ $\endgroup$ – PJ Miller Aug 9 '13 at 13:09
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There is no difference. Observe:

$$\vec{v}^T{A}\vec{w}=\vec{w}^T{A}^T\vec{v}$$

This is true for all vectors $\vec{v}$, $\vec{w}\in\mathbb{R}^n$ and $A\in\mathbb{R}^{n\times{n}}$.

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