(a) Let \(v \in V\) and \(v_{1},\ldots,v_{n} \in \mathbb{C}\) such that \(v = (v_{1},v_{2},\ldots,v_{n})\). Let \(\{e_{1},\ldots,e_{n}\}\) denote the standard basis. Then \(v_{i}e_{i} = (0,\ldots,0,v_{i},0,\ldots,0)\) where \(v_{i}\) is in the \(i\)th coordinate. Hence, \(v = \sum_{i=1}^n v_{i}e_{i}\) and the coordinate vector \([v]_{B}\) with respect to the standard basis is \(v\). (b) For all \(j \in \{1,\ldots,n \}\), the \(j\)th column of \([T]_{B}\) is \([Te_{j}]_B\). By definition of \(T\), \([Te_{j}]_B = [Ae_{j}]_B\) and from above, \([Ae_{j}]_B = Ae_{j} = (a_{1j},a_{2j},\ldots,a_{nj})^T\). Hence, each column of \([T]_{B}\) is the corresponding column of \(A\) so \([T]_B=A\).

(a) Denote the elements of \(B\) as \(\{b_{1},b_{2},\ldots,b_{n}\}\). Then \(\langle Fb_{i}, Fb_{j} \rangle = \langle (0,\ldots,b_{i},0,\ldots,0), (0,\ldots,b_{j},0,\ldots,0) \rangle = \delta_{ij}\). Hence, if \(v = \sum_{i=1}^{n} v_{i}b_{i}, w = \sum_{i=1}^{n} w_{i}b_{i} \in V\), then \(\langle Fv, Fw \rangle = \sum_{i=1}^n v_{i}\overline{w_{i}} = \langle v, w \rangle\). (b) Let \(C = \{ c_{1},\ldots,c_{n} \}\) be an orthonormal basis of \(V\). Then, from (a), \begin{align*} \delta_{ij} = \langle c_{i},c_{j} \rangle \iff \langle Fc_{i}, Fc_{j} \rangle = \delta_{ij} ~\forall i,j \in \{1,\ldots,n \}.\\ \end{align*} Hence the set \(F(C)\) is orthonormal if and only if \(C\) is orthonormal. Isomorphisms between vector spaces take bases to bases so \(F(C)\) is an orthonormal basis if and only if \(C\) is an orthonormal basis.

Let \(T \in \text{End}(V)\). Previously we proved that \(([T^*]_{B}) = ([T]_{B})^*\) and that \(T \mapsto [T]_{B}\) is an isomorphism from \(\text{End}(V)\) to \(M_n(\mathbb{C})\). Since isomorphisms preserve inverses, \([T^{-1}]_{B} = ([T]_{B})^{-1}\). Thus, \begin{align*} T \text{ is unitary } &\Leftrightarrow \langle Tv, Tw \rangle = \langle v, w \rangle \quad\forall v,w \in V \\ &\Leftrightarrow \langle T^*Tv-v, w \rangle = 0 \quad\forall v,w \in V \\ &\Leftrightarrow T^*Tv = v \quad\forall v \in V\\ &\Leftrightarrow T^* = T^{-1} \\ &\Leftrightarrow [T^*]_{B} = [T^{-1}]_{B} \text{ by the isomorphism } T \mapsto [T]_{B}; \\ &\Leftrightarrow ([T]_{B})^* = ([T]_{B})^{-1} \\ &\Leftrightarrow [T]_{B} \text{ is unitary}. \end{align*}

Let \(T\) be self-adjoint. Then, from previous work, \([T]_{B} = [T^*]_{B} = ([T]_{B})^*\). By the Spectral Theorem, the eigenvalues of \(([T]_{B})^*\) are real. Further, the proof of the Spectral Theorem constructs an orthonormal basis which is the eigenvectors of \([T]_{B}\). By Exercise B and the fact isomorphisms take bases to bases, the images of the eigenvectors of \([T]_B\) under the isomorphism \([v]_{B} \mapsto v\) form an orthonormal basis in \(V\). Now, we show that this orthonormal basis consists of the eigenvectors of \(T\). Let \(v \in V\) such that \([v]_{B}\) is an eigenvector of \([T]_B\) with corresponding eigenvalue \(\lambda\). Then, \[ [Tv]_B = [T]_{B}[v]_{B} =\lambda[v]_{B} = [\lambda v]_{B} \implies Tv = \lambda v. \] Conversely, assume that \(V\) has an orthonormal basis of eigenvectors of \(T\) and the eigenvalues are real. Let \(\Lambda_{i}\) be the \(i\)th eigenvector of \(T\) and \(\lambda_{i}\) the corresponding eigenvalue. Let \(v = \sum_{i=1}^n a_{i}\Lambda_{i},w = \sum_{j=1}^n b_{j}\Lambda_{j} \in V\). Then, \begin{align*} \langle Tv, w \rangle &= \left\langle T\sum_{i=1}^n a_{i}\Lambda_{i},\sum_{j=1}^n b_{j}\Lambda_{j} \right\rangle \\ &= \left\langle \sum_{i=1}^n a_{i}\lambda_{i}\Lambda_{i},\sum_{j=1}^n b_{j}\Lambda_{j} \right\rangle \text{ since } T\Lambda_{i} = T\lambda_{i} \quad \forall i \in \{1,\ldots,n \}\\ &= \sum_{i=1}^n a_{i}\lambda_{i} \left\langle \Lambda_{i},\sum_{j=1}^n b_{j}\Lambda_{j} \right\rangle \\ &= \sum_{i=1}^n a_{i}\lambda_{i}\left( \sum_{j=1}^n \overline{b_{j}} \left\langle \Lambda_{i},\Lambda_{j} \right\rangle \right) \\ &= \sum_{i=1}^n a_{i}\lambda_{i} \overline{b_{i}} \left\langle \Lambda_{i},\Lambda_{i} \right\rangle \text{ since } \Lambda_{i} \text{form an orthonormal basis};\\ &= \sum_{i=1}^n \left\langle a_{i}\Lambda_{i},\lambda_{i} b_{i} \Lambda_{i} \right\rangle \text{ since } \lambda_{i} \in \mathbb{R} \quad \forall i \in \{1,\ldots,n \};\\ &= \langle v, Tw \rangle. \end{align*} Since \(T^*\) is the unique linear operator such that \(\langle Tv, w \rangle = \langle v, Tw \rangle\) for all \(v,w \in V\) we conclude that \(T\) is self-adjoint.