Let \(V\) be an \(n\)-dimensional inner product space over the field \(F\), and let \(L(V, V)\) be the space of linear operators on \(V\). Show that there is a unique inner product on \(L(V, V)\) with the property that \(||T_{\alpha, \beta}||^2 = ||\alpha||^2 ||\beta||^2\) for all \(\alpha, \beta\) in \(V\). (\(T_{\alpha, \beta}\) is the operator defined in Exercise 10.) Find an isomorphism between \(L(V, V)\) with this inner product and the space of \(n \times n\) matrices over \(F\), with the inner product \((A \mid B) = \operatorname{tr}(AB^*)\).
Solution
Define \[ (S\mid T)_L=\operatorname{tr}(ST^*) \] for \(S,T\in L(V,V)\).
1. This formula defines an inner product on \(L(V,V)\).
Choose an orthonormal basis of \(V\). If \(S=(s_{ij})\) and \(T=(t_{ij})\) in that basis, then \[ \operatorname{tr}(ST^*) =\sum_{i,j}s_{ij}\overline{t_{ij}}. \] Thus \((\cdot\mid\cdot)_L\) is the standard inner product on the \(n^2\) matrix entries. In particular it is sesquilinear, conjugate-symmetric, and \[ (T\mid T)_L=\sum_{i,j}|t_{ij}|^2>0 \] for every nonzero \(T\).
2. Rank-one operators have the required norm.
Exercise 10 gives \[ T_{\alpha,\beta}^*=T_{\beta,\alpha}. \] Hence \[ T_{\alpha,\beta}T_{\alpha,\beta}^* =T_{\alpha,\beta}T_{\beta,\alpha} =\|\beta\|^2T_{\alpha,\alpha}. \] Taking traces and again using Exercise 10, \[ \begin{aligned} \|T_{\alpha,\beta}\|_L^2 &=\operatorname{tr}(T_{\alpha,\beta}T_{\alpha,\beta}^*)\\ &=\|\beta\|^2\operatorname{tr}(T_{\alpha,\alpha})\\ &=\|\beta\|^2\|\alpha\|^2. \end{aligned} \] Thus this inner product has the stated property.
3. The norm condition determines the inner product uniquely.
Suppose \([\cdot\mid\cdot]\) is any inner product on \(L(V,V)\) satisfying \[ [T_{\alpha,\beta}\mid T_{\alpha,\beta}] =\|\alpha\|^2\|\beta\|^2 \] for all \(\alpha,\beta\).
Fix \(\alpha\). The map \(\beta\mapsto T_{\alpha,\beta}\) is conjugate-linear in \(\beta\). Polarizing the displayed norm identity in the \(\beta\) variable therefore determines \[ [T_{\alpha,\beta}\mid T_{\alpha,\delta}] =\|\alpha\|^2(\delta\mid\beta). \] Now fix \(\beta,\delta\). The map \(\alpha\mapsto T_{\alpha,\beta}\) is linear. Polarizing the last identity in the \(\alpha\) variable gives \[ [T_{\alpha,\beta}\mid T_{\gamma,\delta}] =(\alpha\mid\gamma)(\delta\mid\beta). \] Hence the inner product is uniquely determined on every pair of rank-one operators.
Rank-one operators span \(L(V,V)\): if \(e_1,\ldots,e_n\) is an orthonormal basis, then \(T_{e_i,e_j}\) is the matrix unit sending \(e_j\) to \(e_i\) and all other \(e_k\) to zero. Sesquilinearity therefore determines the inner product uniquely on all of \(L(V,V)\).
4. Matrix representation in an orthonormal basis is an isometric isomorphism.
Fix an orthonormal basis \(\mathfrak B\) of \(V\) and define \[ \Phi:L(V,V)\longrightarrow M_n(F), \qquad \Phi(T)=[T]_{\mathfrak B}. \] This is a linear isomorphism. Since \(\mathfrak B\) is orthonormal, \[ [T^*]_{\mathfrak B}=[T]_{\mathfrak B}^*. \] Therefore \[ \begin{aligned} (S\mid T)_L &=\operatorname{tr}(ST^*)\\ &=\operatorname{tr}\bigl([S]_{\mathfrak B}[T]_{\mathfrak B}^*\bigr)\\ &=(\Phi(S)\mid\Phi(T))_{M_n(F)}. \end{aligned} \] Thus \(\Phi\) is the required inner-product-space isomorphism.