Singularity of a skew symmetric matrix

I have to prove the following:

If A is an nxn skew symmetric matrix, and n is odd, show A is singular.

I have some kind of idea on how to solve it, but I'm still missing some steps.

Here are my steps.

skew symmetric:

$A^T=-A$

$Det(A^T) = Det(-A)$

Then here I have to show that $det(A^T)=det(A)$ Since you switch rows and columns when taking the transpose, the determinant will clearly be the same, but I don't know how to properly show this.

Once I have shown $det(A^T)=det(A)$ I'll get:

$det(A)=det(-A)$

And from here on I also don't know how to continue.

To prove that A is nonsingular for odd n I know I have to show that det(A) for odd n is zero, but I don't know how

• Don't you actually know that for any square matrix $\;\det A^t=\det A\;$ ? – DonAntonio Dec 8 '16 at 20:37

(Assuming you're probably working over $\mathbb R$ or $\mathbb C$ or some subfield.)

The step I think you're missing is that $\det(-A)=\det(-I)\det(A)=(-1)^n\det(A)$.

So you're looking at the equation $\det(A)=(-1)^n\det(A)$ for odd $n$.

I think you can see what happens here if $\det(A)$ is nonzero...

• Thanks, that was indeed something I didn't see :) – Amaluena Dec 8 '16 at 20:47

You need to know (or prove) that $\;\det A^t=\det A\;$ in any case, and also that if $\;k\;$ is any scalar, then $\;\det(kA)=k^n\det A\;,\;\;n=$ the matrix's order , so now:

$$\det A=\det A^t=\det(-A)=(-1)^n\det A=-\det A \;\;\text{(because \;n\; is odd...)}$$

and we're done... as long as we're working on a field of characteristic$\;\neq2\;$