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Let $V$ be a vector space over K and $B: V × V \to K$ a bilinear form on $V$.

$\beta$ is a basis of $V$ and $G$ the gram matrix of $B$ with respect to $\beta$.

How to proof: $B$ is skew-symmetric $\Leftrightarrow$ $G=-G^T$?

For $\Rightarrow$ I used:

Let $B$ be skew-symmetric and $\beta=(v_1,...,v_n)$ a basis of $V$. Then it's $B(x,y)=-B(y,x)$.

So $g_{ij}=-g_{ji}$ and $G=-G^T$.

For $\Leftarrow$ I used:

Let $G=(g_{ij}$) be a skew-symmetric gram matrix.

For $v=\lambda_1v_1+...+\lambda_nv_n$ and $w=\mu_1v_1+...\mu_nv_n$ with $\lambda_1,...\lambda_n,\mu_1,...\mu_n \in K$ put $\langle v,w\rangle:=(\lambda_1,...,\lambda_n)A\begin{pmatrix} \mu_1\\.\\\mu_n \end{pmatrix}$.

Since $G$ is skew-symmetric, $\langle v_i,v_j\rangle=-g_{ji}$ and

$\langle v,w\rangle = \sum_{i,j=1}^{n}{\lambda_i\mu_jg_{ij}}=-\sum_{i,j=1}^{n}{\mu_j\lambda_ig_{ij}} = -\langle w,v\rangle$

So $B$ is skew-symmetric.

Does this work or is there another way to show this.

Humertun
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    The argument is correct and the end of the second part could also be done as follows: $$B(x,y) = x^T A y =- x^T A^Ty = -(x^T A^Ty )^T = - y^T A x =- B(y,x)$$ where $x,y$ are column matrices. – Hugo Jun 13 '18 at 11:29
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    Where I am using that for $1\times 1$ matrices, also known as numbers, the identity $a^T = a$ always holds. – Hugo Jun 13 '18 at 11:31

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