Let \(V\) be a finite-dimensional complex inner product space, and let \(T\) be a linear operator on \(V\). Prove that \(T\) is self-adjoint if and only if \((T\alpha \mid \alpha)\) is real for every \(\alpha\) in \(V\).
Solution
1. If \(T\) is self-adjoint, then \((T\alpha\mid\alpha)\) is real.
Assume \(T=T^*\). Then \[ (T\alpha\mid\alpha) =(\alpha\mid T^*\alpha) =(\alpha\mid T\alpha). \] By conjugate symmetry, \[ (\alpha\mid T\alpha)=\overline{(T\alpha\mid\alpha)}. \] Thus \((T\alpha\mid\alpha)\) equals its complex conjugate and is therefore real.
2. Under the reality hypothesis, \(S=T-T^*\) has zero quadratic form.
Assume \((T\alpha\mid\alpha)\) is real for every \(\alpha\). Put \[ S=T-T^*. \] Then \(S^*=-S\). Moreover, \[ \begin{aligned} (S\alpha\mid\alpha) &=(T\alpha\mid\alpha)-(T^*\alpha\mid\alpha)\\ &=(T\alpha\mid\alpha)-(\alpha\mid T\alpha)\\ &=(T\alpha\mid\alpha)-\overline{(T\alpha\mid\alpha)}\\ &=0. \end{aligned} \] So \[ q(\alpha):=(S\alpha\mid\alpha)=0 \] for every \(\alpha\in V\).
3. The vanishing quadratic form forces every mixed value \((S\alpha\mid\beta)\) to vanish.
Fix \(\alpha,\beta\in V\) and write \[ a=(S\alpha\mid\beta). \] Since \(S^*=-S\), \[ (S\beta\mid\alpha) =(\beta\mid S^*\alpha) =-(\beta\mid S\alpha) =-\overline{a}. \]
Now expand \(q(\alpha+\beta)=0\). The diagonal terms vanish by step 2, so \[ 0=a-\overline a. \] Thus \(a\) is real.
Next expand \(q(\alpha+i\beta)=0\). Again the diagonal terms vanish, while the cross terms are \[ (S\alpha\mid i\beta)+i(S\beta\mid\alpha) =-ia-i\overline a. \] Hence \[ a+\overline a=0. \] Thus \(a\) is purely imaginary. Being both real and purely imaginary, \(a=0\). Therefore \[ (S\alpha\mid\beta)=0 \] for all \(\alpha,\beta\).
4. Therefore \(T\) is self-adjoint.
Fix \(\alpha\). Step 3 says that \(S\alpha\) is orthogonal to every vector in \(V\). Non-degeneracy of the inner product gives \(S\alpha=0\). Hence \(S=0\), so \[ T-T^*=0. \] Thus \(T=T^*\) [@HK71]. Combining with step 1 proves \[ \boxed{T\text{ is self-adjoint}\iff (T\alpha\mid\alpha)\in\mathbb R \text{ for every }\alpha.} \]