Suppose \(A^* = A\). It is then a fact that \(A\) is self-adjoint, and so for every \(\vector{v}\in V\) we have \[ \inner{A\vector{v}}{\vector{v}} = \inner{\vector{v}}{A^*\vector{v}} = \inner{\vector{v}}{A\vector{v}}. \]
Let \((\lambda, \vector v)\) be an eigenvalue of \(A\) with one of its corresponding eigenvectors, so \(A\vector{v} = \lambda{\vector v}\).
On one hand, \[\begin{align*} \inner{A\vector{v}}{\vector{v}} = \inner{\lambda\vector{v}}{\vector{v}} = \lambda \inner{\vector{v}}{\vector{v}} = \lambda \norm{\vector{v}}^2 ,\end{align*}\]
while on the other hand,
\[\begin{align*} \inner{A\vector{v}}{\vector{v}} = \inner{\vector{v}}{A^*\vector{v}} = \inner{\vector{v}}{A\vector{v}} = \inner{\vector{v}}{\lambda\vector{v}} = \overline{\lambda} \inner{\vector{v}}{\vector{v}} = \overline{\lambda} \norm{\vector{v}}^2 .\end{align*}\]
Equating these expressions, we find that \[ \lambda = \overline{\lambda} \implies \lambda \in \RR. \qed \]
We can make use of the following fact:
Theorem (Schur): Every square matrix \(A \in M_n(\CC)\) is unitarily similar to an upper triangular matrix, i.e. there exists a unitary matrix \(U\) such that \(A = UT\inverseof{U}\) where \(T\) is upper-triangular.
Applying this theorem yields \(A = UT\inverseof{U}\) and thus \(T = \inverseof{U} A U\). In particular, \(A \sim T\).
Noting that if \(U\) is unitary then \(\inverseof{U} = U^*\), we have
\[\begin{align*} T^* &= (\inverseof{U} A U)^* \\ &= U^* A^* (\inverseof{U})^* \\ &= U^* A^* U^{**} \\ &= \inverseof{U} A^* U \\ &= T ,\end{align*}\]
and so \(T^* = T\).
Since \(T\) is upper triangular, this forces \(T_{ij} = 0\) whenever \(i\neq j\) But this makes \(T\) diagonal, so \(A\) is similar to a diagonal matrix. \(\qed\)
Proof of Schur’s Theorem: We’ll proceed by induction on \(n = \dim_\CC(V)\), and showing that there is an orthonormal basis of \(V\) such that the matrix of \(A\) is upper triangular.
Lemma: If \(V\) is finite dimensional and \(\lambda\) is an eigenvalue of \(A\), then \(\overline{\lambda}\) is an eigenvalue of \(A^*\).
Proof: Since \(A^* - \bar\lambda I = (A - \lambda I)^*\) and \(\det(M^*) = \overline{\det M}\), \[ \operatorname{det}\left(A^{*} - \bar{\lambda} I\right) = \overline{\operatorname{det}(A-\lambda I)} = \bar 0 = 0. \qed \]
Since \(\CC\) is algebraically closed, every matrix \(A \in M_n(\CC)\) will have an eigenvalue, since its characteristic polynomial will have a root by the Fundamental Theorem of Algebra.
So let \(\lambda_1, \vector v_1\) be an eigenvalue/eigenvector pair of the adjoint \(A^*\).
Consider the space \(S = \spanof_\CC\theset{\vector v_1}\); then \(V = S \oplus S^\perp\). The claim is that the original \(A\) will restrict to an operator on \(S^\perp\), which has dimension \(n-1\). The inductive hypothesis will then apply to \(\restrictionof{A}{S^\perp}\).
Note that if this holds, there will be an orthonormal basis \(\mathcal{B}\) of \(S^\perp\) such that the matrix \[ \mathbf{A}' \definedas [\restrictionof{A}{S^\perp}]_{\mathcal{B}} \] will be upper triangular. We would then be able to obtain an orthonormal basis \(\mathcal{C} \definedas \mathcal{B} \union \theset{\vector{v_1}}\) of \(S \oplus S^\perp = V\), ordered with \(\vector v_1\) last.
Only \(S^\perp\) is \(A\dash\)invariant here; \(S\) is an eigenspace of \(A^*\), not of \(A\), so \(\restrictionof{A}{S}\) is not defined and the matrix is not block diagonal. What the invariance of \(S^\perp\) gives is that the first \(n-1\) columns have no \(\vector v_1\) component, i.e. a zero in the last row:
\[\begin{align*} [A]_{\mathcal{C}} &= \left[\begin{array}{cc} [\restrictionof{A}{S^\perp}]_{\mathcal{B}} & \vector u \\ 0 & c \end{array}\right] = \left[\begin{array}{cc} \mathbf{A}' & \vector u \\ 0 & c \end{array}\right] ,\end{align*}\]
where \(\vector u\) and the scalar \(c\) record \(A\vector v_1\) in the basis \(\mathcal C\), and are otherwise unconstrained. This is upper-triangular since \(\mathbf{A}'\) is upper-triangular.
Ordering \(\vector v_1\) first would instead put the unconstrained column on the left and give a lower triangular matrix.
To see that \(A\) does indeed restrict to an operator on \(S^\perp\), we need to show that \(A(S^\perp) \subseteq S^\perp\). So let \(\vector s \in S^\perp\); then \(\inner{\vector v_1}{\vector s} = 0\) by definition. Then \(A\vector s \in S^\perp\) since
\[\begin{align*} \inner{\vector v_1}{A\vector s} &= \inner{A^* \vector v_1}{\vector s} \\ &= \inner{\lambda_1 \vector v_1}{\vector s} \\ &= \lambda_1 \inner{\vector v_1}{\vector s} \\ &= 0 .\end{align*}\]
\(\qed\)
Solution
Let \(A\) be self-adjoint on a finite-dimensional complex inner-product space.
If \(Av=\lambda v\) with \(v\ne0\), then \[ \lambda\langle v,v\rangle =\langle Av,v\rangle =\langle v,Av\rangle =\overline\lambda\langle v,v\rangle. \] Hence \(\lambda=\overline\lambda\), so every eigenvalue is real.
For diagonalization, Schur’s theorem gives a unitary \(U\) such that \[ T=U^*AU \] is upper triangular. Since \(A^*=A\), \[ T^*=U^*A^*U=U^*AU=T. \] An upper-triangular self-adjoint matrix is diagonal: upper triangularity gives \(T_{ij}=0\) for \(i>j\), and \(T=T^*\) then gives \(T_{ij}=0\) also for \(i<j\). Thus \(T\) is diagonal and \[ A=UTU^* \] is unitarily diagonalizable.
For completeness, Schur’s theorem itself follows by induction on dimension. Choose an eigenvector \(v\) of \(A^*\); then \(v^\perp\) is \(A\)-invariant because \[ \langle v,Aw\rangle=\langle A^*v,w\rangle=0 \qquad(w\in v^\perp). \] Apply induction to \(A|_{v^\perp}\) and place \(v\) last in the resulting orthonormal basis.