# If $A$ is an invertible skew-symmetric matrix, then prove $A^{-1}$ is also skew symmetric

Let $A$ be an invertible skew-symmetric $(2n \times 2n)$-matrix. Prove that $A^{-1}$ is also skew-symmetric. (You may assume that $(AB)^T = B^TA^T$).

I did this with a $2 \times 2$ matrix and got that it worked, but I don't know how to show it for a general $2n \times 2n$ matrix, as it is a little harder to calculate the inverse of that. Obviously the hint comes into play somehow but I can't see how.

I have the definition of a skew symmetric bileanr function to be $B(u,v) = - B(v,u)$, but again, I can't see how to put this into matrix form and use that.

Can someone give me some hints please?

• What happens if you replace $B$ with $A^{-1}$ in the hint? – Glen O Apr 21 '13 at 12:36

$(A^T)^{-1}=(A^{-1})^T$ and according to Wikipedia, a skew-symmetric matrix is a matrix that satisfies the condition $A^T=-A$. So $(A^{-1})^T=(A^T)^{-1}=(-A)^{-1}=-A^{-1}$ Why do you need $2n\times 2n$ condition?
• $det(A) = 0$ if the dimension is odd – user67133 Apr 21 '13 at 12:43
• @Antoine $(A^T) = B = -A$, how do you get that? – Kaish Apr 21 '13 at 13:31
• @Kaish: Since $A$ is skew-symmetric, then $A^T=-A$ (swapping the $u,v$ in the bilinear form corresponds to taking the transpose of a matrix), so of course $(A^T)^{-1}=(-A)^{-1}$. Making use of the hint with $B:=A^{-1}$, we can show (as Antoine did in the most recent comment) that $(A^{-1})^T=(A^T)^{-1}$. You should be able to show that for any invertible matrix $A$, $-A$ is also invertible, and $(-A)^{-1}=-(A^{-1}).$ At that point, you're done. – Cameron Buie Apr 21 '13 at 13:48