Let \(P\) be an \(n\times n\) real matrix such that \[ x^TPy=-y^TPx \] for all column vectors \(x,y\in\RR^n\). Prove that \(P\) is skew-symmetric.
Solution
Let \[ P=(p_{ij})_{1\le i,j\le n} \] and let \(e_1,\ldots,e_n\) be the standard basis of \(\RR^n\).
1. For every \(i,j\),\[ p_{ij}=-p_{ji}. \]
Apply the given identity with \[ x=e_i, \qquad y=e_j. \] Then \[ e_i^TPe_j=-e_j^TPe_i. \] The left side is \(p_{ij}\) and the right side is \(-p_{ji}\), giving the claim.
2. The matrix \(P\) is skew-symmetric:\[ \boxed{P^T=-P.} \]
The \((i,j)\)-entry of \(P^T\) is \(p_{ji}\). By step 1, \[ p_{ji}=-p_{ij} \] for every \(i,j\), so \(P^T=-P\).
3. Q.E.D.
Step 2 is precisely the required conclusion.