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?

  • 1
    $\begingroup$ What happens if you replace $B$ with $A^{-1}$ in the hint? $\endgroup$ – 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?

| cite | improve this answer | |
  • 4
    $\begingroup$ An odd sized skew-symmetric matrix cannot be invertible. $\endgroup$ – EuYu Apr 21 '13 at 12:43
  • $\begingroup$ $det(A) = 0$ if the dimension is odd $\endgroup$ – user67133 Apr 21 '13 at 12:43
  • $\begingroup$ Odd skew symmetric matrices aren't invertible. $\endgroup$ – Kaish Apr 21 '13 at 12:43
  • 1
    $\begingroup$ @Antoine $(A^T) = B = -A$, how do you get that? $\endgroup$ – Kaish Apr 21 '13 at 13:31
  • 1
    $\begingroup$ @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. $\endgroup$ – Cameron Buie Apr 21 '13 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.