[Math] Singularity of a skew symmetric matrix

linear algebramatrices

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

Thanks in advance 🙂

Best Answer

(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...

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