# characterization of an antisymmetric matrix [closed]

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

• $A$ is an $n\times n$ matrix and $X$ is a vector? What have you tried? What happens if you plug in unit vectors for $X$? Jan 19 at 11:22
• I managed to do it with scalar product Jan 19 at 11:29
• 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 $A+ A^T=\mathbf 0$ Jan 19 at 18:59
• Please improve your question, after referring to How to ask a good question on math.se. Jan 24 at 21:26

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 $$j$$.

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.