Exercise 3.3.4

Question

A square matrix $(n 	imes n)$ is called skew-symmetric (or antisymmetric) if $A^\mathsf{T} = A$. Prove that if $A$ is skew-symmetric and $n$ is odd, then $\det A = 0$. Is this true for even $n$?

Jing Huang · Accepted Answer

\begin{proof}
$\det A = \det A^\mathsf{T} = \det (-A) = (-1)^n \det A$ by using the the properties of determinant and skew-symmetric matrices. If $n$ is odd, $(-1)^n = -1$, we have $\det A = -\det A$, thus $\det A = 0$.  \par
If $n$ is even, we just have $\det A = \det A$ so this conclusion generally is not true.    
\end{proof}

Exercise 3.3.4

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Exercise 3.3.4

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