Lemma 1. Let \(T \in \operatorname{End}(V)\) for \(V \in \mathrm{FinVect}_\mathbb{C}\). Then there exists an upper-triangular matrix representation of \(T\) with respect to some basis of \(V\).

Proof. By induction. Since we're over \(\mathbb{C}\), each operator has an eigenvalue. Clearly this holds if \(n = \dim V = 1\). If \(u \in \mathrm{im}(T - \lambda I)\) then \(Tu = (T - \lambda I) u + \lambda u\) so \(U = \mathrm{im}(T - \lambda I)\) is \(T\)-invariant. Then by the induction hypothesis, \(T|_U\) has an upper-triangular matrix representation with respect to some basis, say \(u_1, \dots, u_m\). Extend this to a basis \(u_1, \dots, u_m, v_1, \dots v_n\) of \(V\). For each \(1 \leq j \leq n, Tv_j = (T - \lambda I)v_j + \lambda v_j \implies Tv_j\) is of the form \(au + \lambda v_j\) for \(a \in \mathbb{C}\) and \(u \in U\), so \(Tv_j \in \mathrm{spam}(u_1, \dots, u_m, v_1, \dots, v_j)\) for each \(j\). Since the matrix of \(T\) with respect to \(u_1, \dots, u_m\) is already upper-triangular, we have \(u_j \in \mathrm{span}(u_1, \dots u_j\) for each \(1 \leq j \leq m\). Combining these two statements, we see \(T\) is upper-triangular with respect to \(u_1, \dots, u_m, v_1, \dots, v_n\), as desired. \(_\square\)

Lemma 2. Let \(T \in \mathrm{End}(V)\) for \(V\) a real inner-product space. Then whenever \(T\) is self-adjoint, \(T\) has an eigenvalue.

Proof. Let \(\dim V = n\). Then for any \(0 \neq v \in V\), there exist \(a_0, \dots, a_n\), not all zero, such that \[(\sum_{0 \leq i \leq n} a_i T^i)v = 0.\] Write \(p(x) = \sum_{0 \leq i \leq n} a_i x^i\) as a member of \(\mathbb{R}]x\); \(p(x)\) admits a unique irreducible factorization into quadratic and linear factors of the form \[p(x) = \prod q_i(x) \cdot \prod l_i(x).\] We see that if, for each \(i\), \(q_i(T)\) is invertible, then we are done, since one of the linear factors gives us our desired eigenvalue. Proceeding: fix \(i\); since \(q_i\) is irreducible, then writing \(q_i = x^2 + a_1 x + a_0\) gives \(a_1^2 < 4a_0\) since the quadratic formula would give us a further factorization into linear factors otherwise. Now, applying the Cauchy-Schwarz inequality and completing the square: \[\langle q_i(T)v,v \rangle = \langle T^2 v, v \rangle + a_1 \langle Tv, v \rangle + a_0 \langle v, v \rangle\] \[ = \langle Tv, Tv \rangle + a_1 \langle Tv, v \rangle + a_0 || v||^2 \geq ||Tv||^2 - |a_i| ||Tv||||v|| + a_0 ||v||^2\] \[(||Tv|| - \frac{|a_1||v||}{2})^2 + (a_0 - \frac{a_1^2}{4})||v||^2 > 0\]

so that \(q_i(T)v \neq 0\) whenever \(v \neq 0\), so that \(q_i(T)\) has trivial kernel and is hence invertible, which completes the proof. \(_\square\)

Theorem 1. (The complex spectral theorem.) Let \(V\) be a finite-dimensional \(\mathbb{C}\) inner-product space, and \(T\) an endomorphism of \(V\). Then \(V\) has an orthonormal basis of eigenvectors of \(T\) if and only if \(T\) is normal, i.e. commutes with its adjoint.

Proof. \((\Leftarrow)\) Applying Gram-Schmidt to the basis \(u_1, \dots, u_m\) (whose existence is guaranteed by Lemma 1) to which \(T\) has an upper-triangular matrix yields an orthonormal basis \(b_1, \dots, b_n\) such that for each \(m \leq n\), \(\mathrm{span}(b_1, \dots, b_m) = \mathrm{span}(u_1, \dots, u_m)\) is \(T\)-invariant and which therefore also gives an upper-triangular matrix for \(T\), say

\[M = (a_{ij}).\] Now, by the Pythagorean theorem, \(||Tb_1||^2 = |a_{11}^2|^2\) and \(||T^*b_1||^2 = ||Tb_1||^2 = \sum_{i \leq n} |a_{1,i}|^2\) so \(\sum_{1 < i \leq n} |a_{1,i}|^2 = 0 \implies a_{12} = 0, \dots, a_{1n} = 0\). Iterating this, we see all the nondiagonal entries of \(M\) are zero, and hence each \(b_i\) is an eigenvector of \(T\), as desired.

\((\Rightarrow)\). Let \(e_1, \dots, e_n\) generate \(V\). Then \(T\) has a diagonal matrix with respect to this basis; the result is immediate. \(_\square\)

Theorem 2. (The real spectral theorem.) Let \(V\) be a finite-dimensional \(\mathbb{R}\) inner-product space, and \(T\) an endomorphism of \(V\). Then \(V\) has an orthonormal basis of eigenvectors of \(T\) if and only if \(T\) is self-adjoint.

Proof. \((\Leftarrow)\) \(T\) self-adjoint \(\implies (\dim V = 1 \implies \text{ result})\). Inducting, by Lemma 2, \(T\) has an eigenvalue with eigenvector (say of norm \(1\)) \(u\). Consider U = \(u^{\perp}\). \(\forall v \in U, \langle u, Tv \rangle = \langle \lambda u, v \rangle = 0 \implies Tv \in U \implies T|_U\) is an operator on \(U\), which must have dimension less than that of \(V\). By the induction hypothesis, \(T|_U\) self-adjoint \(\implies \exists v_1, \dots, v_m\) all eigenvectors of \(T|_U\), which generate \(U\). Each \(v_i\) is also an eigenvector of \(T\), so adjoining \(u\) to this list completes the proof. (Clearly, \(T|_U\) is self-adjoint, as \(\langle T|_U v, w \rangle = \langle Tv, w \rangle = \langle v, Tw \rangle = \langle v, T|_U w \rangle\), for every \(v,w \in U\).