Let \(V\) be an inner product space and \(\beta, \gamma\) fixed vectors in \(V\). Show that \(T\alpha = (\alpha \mid \beta)\gamma\) defines a linear operator on \(V\). Show that \(T\) has an adjoint, and describe \(T^*\) explicitly.
Now suppose \(V\) is \(C^n\) with the standard inner product, \(\beta = (y_1, \ldots, y_n)\), and \(\gamma = (x_1, \ldots, x_n)\). What is the \(j, k\) entry of the matrix of \(T\) in the standard ordered basis? What is the rank of this matrix?
Solution
Define \[ T\alpha=(\alpha\mid\beta)\gamma. \]
1. The map \(T\) is linear.
Hoffman–Kunze use the convention that the inner product is linear in its first argument [@HK71]. Thus, for scalars \(a,b\) and vectors \(\alpha_1,\alpha_2\), \[ \begin{aligned} T(a\alpha_1+b\alpha_2) &=(a\alpha_1+b\alpha_2\mid\beta)\gamma\\ &=a(\alpha_1\mid\beta)\gamma+b(\alpha_2\mid\beta)\gamma\\ &=aT\alpha_1+bT\alpha_2. \end{aligned} \]
2. The adjoint is another rank-one operator:\[ \boxed{T^*\delta=(\delta\mid\gamma)\beta.} \]
For arbitrary \(\alpha,\delta\in V\), \[ \begin{aligned} (T\alpha\mid\delta) &=((\alpha\mid\beta)\gamma\mid\delta)\\ &=(\alpha\mid\beta)(\gamma\mid\delta). \end{aligned} \] On the other hand, conjugate-linearity in the second argument gives \[ \begin{aligned} (\alpha\mid(\delta\mid\gamma)\beta) &=\overline{(\delta\mid\gamma)}(\alpha\mid\beta)\\ &=(\gamma\mid\delta)(\alpha\mid\beta). \end{aligned} \] The two expressions agree for every \(\alpha,\delta\), so the displayed formula is the adjoint.
3. In the standard basis of \(\mathbb C^n\), the \((j,k)\) entry is\[ \boxed{x_j\overline{y_k}.} \]
Let \(e_k\) be the \(k\)th standard basis vector. Since \[ (e_k\mid\beta)=\overline{y_k}, \] we have \[ Te_k=\overline{y_k}\gamma =\begin{pmatrix} x_1\overline{y_k}\\ \vdots\\ x_n\overline{y_k} \end{pmatrix}. \] The \(k\)th column of the matrix of \(T\) is therefore the displayed vector. Hence \[ [T]=xy^*, \] where \(x=(x_j)^t\) and \(y=(y_j)^t\).
4. The rank is zero exactly when one of the defining vectors is zero, and otherwise it is one.
If \(\beta=0\) or \(\gamma=0\), then \(T=0\) and the rank is \(0\). Suppose both are nonzero. Every value of \(T\) is a scalar multiple of \(\gamma\), so \(\operatorname{range}T\subseteq\operatorname{span}\{\gamma\}\). But \[ T\beta=(\beta\mid\beta)\gamma\ne0, \] so \(\gamma\) lies in the range. Therefore \[ \operatorname{range}T=\operatorname{span}\{\gamma\}, \] and \[ \boxed{\operatorname{rank}T=1.} \]