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My goal is to show the following equivalence : $\forall X \in M_{n,1}(R), {}^tXAX=0 \iff$ A is antisymmetric

The indirect application is easy but the direct one is more difficult

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  • $\begingroup$ $A$ is an $n\times n$ matrix and $X$ is a vector? What have you tried? What happens if you plug in unit vectors for $X$? $\endgroup$ Jan 19 at 11:22
  • $\begingroup$ I managed to do it with scalar product $\endgroup$
    – Julien
    Jan 19 at 11:29
  • $\begingroup$ since you are not in characteristic 2, use the decomposition $A=\frac{1}{2}\big(A+A^T\big) + \frac{1}{2}\big(A-A^T\big)$, then use that to show $A+ A^T=\mathbf 0$ $\endgroup$ Jan 19 at 18:59
  • $\begingroup$ Please improve your question, after referring to How to ask a good question on math.se. $\endgroup$
    – amWhy
    Jan 24 at 21:26
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Hints.

  1. It suffices to prove that prove the statement when $A$ is symmetric.
  2. When $A$ is symmetric, by the polarisation identity, the condition that $x^TAx=0$ for all $x$ implies that $u^TAv=0$ for all vectors $u$ and $v$.
  3. Now you may pick a clever choice $u$ (that depends on $A$ and $v$) to prove that $Av=0$ for any arbitrary $v$. Alternatively, you may pick some appropriate $u$ and $v$ to show that $a_{ij}=0$ for all $i$ and $j$.
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If $e_1,e_2,\dots, e_n$ are vectors of the standard basis (columns of identity matrix), then $e_j^TAe_k$ means the $ a_{jk}$ entry of matrix $A$. Hence the main diagonal of $A$ is made of zeros.

Now take vector $v$ with two components $1$ on $j$ and $k$ position and the rest are $0$.
(for example $[1 \ 0 \ 1 \ 0]^T$ for 4 dimensional matrices, where $1$ is on $1$ and $3$ position).

It's easy to check that in this case $v^TAv$ leads to the sum $a_{jk}+a_{kj}+a_{kk}+a_{jj}=0$.

Hence $a_{jk}+a_{kj}=0$.

Consequently symmetric entries of the matrix are with opposite signs.

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