Let \(V\) be a finite-dimensional inner product space. For each \(\alpha, \beta\) in \(V\), let \(T_{\alpha, \beta}\) be the linear operator on \(V\) defined by \(T_{\alpha, \beta}(\gamma) = (\gamma \mid \beta)\alpha\). Show that
\(T_{\alpha, \beta}^* = T_{\beta, \alpha}\).
\(\operatorname{trace}(T_{\alpha, \beta}) = (\alpha \mid \beta)\).
\(T_{\alpha,\beta}T_{\gamma,\delta} = T_{\alpha,(\beta \mid \gamma)\delta}\).
Under what conditions is \(T_{\alpha, \beta}\) self-adjoint?
Solution
1. The adjoint is \(T_{\beta,\alpha}\).
For arbitrary \(\gamma,\eta\in V\), \[ \begin{aligned} (T_{\alpha,\beta}\gamma\mid\eta) &=((\gamma\mid\beta)\alpha\mid\eta)\\ &=(\gamma\mid\beta)(\alpha\mid\eta), \end{aligned} \] while \[ \begin{aligned} (\gamma\mid T_{\beta,\alpha}\eta) &=(\gamma\mid(\eta\mid\alpha)\beta)\\ &=\overline{(\eta\mid\alpha)}(\gamma\mid\beta)\\ &=(\alpha\mid\eta)(\gamma\mid\beta). \end{aligned} \] Thus the two inner products agree for every \(\gamma,\eta\), so \[ \boxed{T_{\alpha,\beta}^*=T_{\beta,\alpha}}. \]
2. The trace is \((\alpha\mid\beta)\).
Choose any orthonormal basis \(e_1,\ldots,e_n\) of \(V\). The diagonal entry of \(T_{\alpha,\beta}\) in position \(i\) is \[ (T_{\alpha,\beta}e_i\mid e_i) =(e_i\mid\beta)(\alpha\mid e_i). \] Hence \[ \operatorname{tr}(T_{\alpha,\beta}) =\sum_{i=1}^n(e_i\mid\beta)(\alpha\mid e_i). \] Since \[ \beta=\sum_i(\beta\mid e_i)e_i, \] conjugate symmetry gives \((e_i\mid\beta)=\overline{(\beta\mid e_i)}\), and therefore \[ (\alpha\mid\beta) =\sum_i(\alpha\mid e_i)(e_i\mid\beta). \] Thus \[ \boxed{\operatorname{tr}(T_{\alpha,\beta})=(\alpha\mid\beta)}. \]
3. The product formula is\[ T_{\alpha,\beta}T_{\gamma,\delta} =T_{\alpha,(\beta\mid\gamma)\delta}. \]
For every \(\eta\in V\), \[ T_{\alpha,\beta}T_{\gamma,\delta}(\eta) =(\eta\mid\delta)(\gamma\mid\beta)\alpha. \] On the other hand, \[ \begin{aligned} T_{\alpha,(\beta\mid\gamma)\delta}(\eta) &=(\eta\mid(\beta\mid\gamma)\delta)\alpha\\ &=\overline{(\beta\mid\gamma)}(\eta\mid\delta)\alpha\\ &=(\gamma\mid\beta)(\eta\mid\delta)\alpha. \end{aligned} \] The two operators therefore agree.
4. Characterize when \(T_{\alpha,\beta}\) is self-adjoint.
By step 1, self-adjointness is equivalent to \[ T_{\alpha,\beta}=T_{\beta,\alpha}. \] If either \(\alpha=0\) or \(\beta=0\), then \(T_{\alpha,\beta}=0\), so the operator is self-adjoint.
Assume now that \(\alpha,\beta\ne0\) and that the two rank-one operators are equal. Their ranges are respectively \(\operatorname{span}\{\alpha\}\) and \(\operatorname{span}\{\beta\}\), hence these lines coincide. Write \[ \alpha=c\beta \] with \(c\ne0\). Then \[ T_{\alpha,\beta}=cT_{\beta,\beta}, \] whereas, because the second slot is conjugate-linear, \[ T_{\beta,\alpha} =T_{\beta,c\beta} =\overline c\,T_{\beta,\beta}. \] Since \(T_{\beta,\beta}\ne0\), equality holds if and only if \(c=\overline c\), i.e. \(c\in\mathbb R\).
Conversely, if \(\alpha=c\beta\) with \(c\in\mathbb R\), the two operators are equal. Thus \[ \boxed{\text{$T_{\alpha,\beta}$ is self-adjoint iff one vector is zero, or $\alpha=c\beta$ for some $c\in\mathbb R$.}} \]