Discussion
Let \(A\) be an \(n \times n\) real matrix. Consider the following statements.
P: If \(A\) is skew-symmetric, then \(X^{T}AX = 0\), for every \(n \times 1\) real matrix \(X\).
Q: If \(A\) is skew-symmetric and orthogonal, then \(n\) is even.
Then
Answer: C
Both P and Q are correct.
P: \(X^{T}AX\) is a \(1 \times 1\) matrix, so it equals its own transpose. Using \(A^{T} = -A\):
\[ X^{T}AX = (X^{T}AX)^{T} = X^{T}A^{T}X = -X^{T}AX \]
so \(X^{T}AX = 0\) for every real \(X\).
Q: take determinants of \(A^{T} = -A\): \(\det A = \det(-A) = (-1)^{n}\det A\). An orthogonal matrix has \(\det A = \pm 1 \neq 0\), so \((-1)^{n} = 1\) and \(n\) is even. The matrix \(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\) shows the case \(n = 2\) does occur. Every odd-order skew-symmetric matrix is singular, so it cannot be orthogonal.
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