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I am asked the following:

Assume that $M$ is skew-symmetric and invertible. If $M$ is an $n \times n$ matrix, prove that $n$ is even.

The following is my attempt at the solution. I'm not too sure if it is right, or if I'm allowed to do this.

So, if a matrix is skew-symmetric, then,

$$A^T=-A$$

and by taking the determinant of both sides, $$\det(A^T)=\det(-A)$$

Using the property of transpose and the multi-linear property,

$$\det(A)=(-1)^n\det(A)$$

Here is where my problem arises. Are we allowed to assume that $n$ is even and then show that it is invertible? However, if we do this we get

$$\det(A) = \det(A)$$

and I'm not too sure what this equality means. Or where to go from here. Any help would be greatly appreciated.

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    $\begingroup$ Why are you trying to prove the converse of the claim? The converse is not true. If $n$ is even, $A$ may and may not be invertible. $\endgroup$ – darij grinberg May 26 at 10:07
  • $\begingroup$ What are you suggesting I do? I've been a bit stuck $\endgroup$ – Joshie Bread May 26 at 10:14
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    $\begingroup$ Nothing; you're done. You have to prove that $n$ is even if $A$ is invertible. You have shown that if $n$ is odd, then $A$ is not invertible. This is an equivalent statement. $\endgroup$ – darij grinberg May 26 at 10:21
  • $\begingroup$ Great! Thank you/. $\endgroup$ – Joshie Bread May 26 at 10:23
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Note that if $n$ is odd, you would get $\det (A) = - \det (A)$, so A has determinant $0$ and is not invertible. Some invertible matrices that are even by even are skew-symmetric; try constructing them yourself!

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  • $\begingroup$ Thank you. Is what I have done valid though for the proof? I am trying to prove that It is has an even $n$, Is it fair to assume that it is even? $\endgroup$ – Joshie Bread May 26 at 10:08
  • $\begingroup$ @JoshuaPeters Not quite; to prove $n$ is even you will need to prove that odd $n$ fails as in my answer. $\endgroup$ – auscrypt May 26 at 10:10
  • $\begingroup$ So a proof by contradiction? where we assume that $n$ is odd and invertible. Then we arrive at a conclusion where $2 det(M)=0$ which implies it is not invertible. So $n$ must be even. $\endgroup$ – Joshie Bread May 26 at 10:13
  • $\begingroup$ @JoshuaPeters Yes, that’s right. $\endgroup$ – auscrypt May 26 at 10:15
  • $\begingroup$ Is it fair to say that since the negation of the statement is false, this implies that if n is even, M is skew-symmetric and invertible. $\endgroup$ – Joshie Bread May 26 at 10:53

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