[Math] Trying to proof that skew-symmetric matrix $x^TAx=0$

linear algebramatrices

Skew-symmetric matrix is defined as:
$$ A^T=-A $$
Where matrix $A$ is real antisymmetric matrix.

Proof that $$x^TAx=0, \quad x \in \mathbb{R}^n$$

My progress so far

$$
x^TAx=0
$$
because $A^T=-A$
$$
(x^TAx)^T=-x^TAx
$$
because $(ABC)^T=(A(BC))^T=(BC)^TA^T=C^TB^TA^T$ ja $(A^T)^T=A$
$$
x^TA^Tx=-x^TAx
$$
$$
x^TA^Tx+x^TAx=0
$$

If someone can spot what i am doing wrong here that would be much appreciated.

Thanks,

Tuki

Best Answer

I think your difficulty lies in the fact that you never used (2) below, which you can do "for free" because of (1), since then $x^TAx = (x^TAx)^T$; that is, $1 \times 1$ matrices are symmetric. If you throw this fact in the mix, the demonstration is straightforward:

$x^TAx \in \Bbb R, \tag 1$

so

$x^TAx = (x^TAx)^T = x^TA^T(x^T)^T = x^TA^Tx; \tag 2$

but since

$A^T = -A, \tag 3$

(2) becomes

$x^TAx = -x^TAx, \tag 4$

which implies

$x^TAx = 0. \tag 5$

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