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Let $A$ be a skew-symmetric $n\times n$-matrix over the real numbers. Show that $\det A$ is nonnegative.

I'm breaking this up into the even case and odd case (if $A$ is an $n\times n$ skew-symmetric matrix).

So when $n$ is odd, we have:

$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)\Rightarrow \det(A)=-\det(A)\Rightarrow \det(A) = 0$

So $\det(A)$ is non-negative when $n$ is odd.

When $n$ is even, we have:

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

But why can't $\det(A)$ be negative in this case?

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Since $A$ is skew-symmetric, if $\lambda$ is an eigenvalue of $A$ then $-\lambda$ is also an eigenvalue. Thus, in even dimension, the eigenvalues of $A$ come in pairs $\pm \lambda.$ Moreover, it is known that all the eigenvalues are imaginary. That is, the set of eigenvalues of $A$ is of the form $\{ \pm \lambda_1 i,\cdots,\pm \lambda_n i\}$ Thus,

$$\det(A)=\lambda_1 i\cdot (-\lambda_1 i)\cdots \lambda_n i\cdot (-\lambda_n i)=\lambda_1^2\cdots \lambda_n^2\ge 0.$$

Without the use of eigenvalues we can proceed as follows. The result is clear if the matrix is of order $2.$ If it has order $4$ is of the form:

$$\left(\begin{array}{cccc}0 & a_{12}& a_{13}& a_{14}\\ -a_{12}& 0 & a_{23}& a_ {24}\\ -a_{13}& -a_{23} & 0 & a_{34}\\ -a_{14}& -a_{24}& -a_{34} & 0\end{array}\right).$$ If the first row is $0$ then $\det=0$ and we are done. So, there is $a_{1i}\ne 0.$ Assume it is $a_{12}\ne 0.$ In other case change row $2$ with row $i$ and column $2$ with column $i.$ Now, we make zeros in the first two rows and columns and get the matrix

$$\left(\begin{array}{cccc}0 & a_{12}& 0& 0\\ -a_{12}& 0 & 0& 0\\ 0& 0 & 0 & a'_{34}\\ 0& 0& -a'_{34} & 0\end{array}\right).$$ All changes we have done doesn't change the determinant of the matrix. Now we get that the determinant is $$a_{12}^2(a'_{34})^2\ge 0.$$

We can do this argument with any symmetric matrix of order $n.$ We change columns and rows, if necessary, to get

$$\left(\begin{array}{ccc}0 & a_{12}& \cdots \\ -a_{12}& 0 & \cdots \\ \vdots & \vdots & A\end{array}\right)$$ with $a_{12}\ne 0.$ Now, we make zeros in the two first rows and columns to get

$$\left(\begin{array}{ccc}0 & a_{12}& 0 \\ -a_{12}& 0 & 0 \\ 0 & 0 & A'\end{array}\right).$$ We can show that $A'$ is skew-symmetric and that the determinant is $a_{12}^2\det(A').$ Now, $A'$ is a skew-symmetric matrix of order $n-2$ and we repeat the process.

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  • $\begingroup$ Is there a solution that doesn't use eignevalues? We haven't covered them yet. $\endgroup$ – Caleb Nov 26 '14 at 1:01
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See Lemma 5.6 (and the following discussion) in Keith Conrad's Bilinear Forms.

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