Let \(V\) be a vector space and \(( \mid )\) an inner product on \(V\).
Show that \((0 \mid \beta) = 0\) for all \(\beta\) in \(V\).
Show that if \((\alpha \mid \beta) = 0\) for all \(\beta\) in \(V\), then \(\alpha = 0\).
Solution
1. The zero vector pairs to zero with every vector.
For any \(\beta\in V\), additivity in the first variable gives \[ (0\mid\beta)=(0+0\mid\beta)=(0\mid\beta)+(0\mid\beta). \] Subtracting \((0\mid\beta)\) from both sides yields \[ \boxed{(0\mid\beta)=0}. \]
2. If \(\alpha\) is orthogonal to every vector, then \(\alpha=0\).
Apply the hypothesis with \(\beta=\alpha\). Then \[ (\alpha\mid\alpha)=0. \] Positive definiteness of an inner product says that this happens only for \(\alpha=0\) [@HK71].