Prove that every real symmetric matrix \(A \in M_n(\mathbb{R})\) has real eigenvalues and can be orthogonally diagonalized: there exists an orthogonal matrix \(Q \in O(n)\) such that \(Q^T A Q = \operatorname{diag}(\lambda_1, \dots, \lambda_n)\) (The Real Spectral Theorem).
Solution
Let \(A=A^T\in M_n(\mathbb R)\). Regard \(A\) as a complex matrix. If \(Av=\lambda v\) with \(v\ne0\), then \[ v^*Av=\lambda v^*v. \] Because \(A^*=A\), the scalar \(v^*Av\) is real: \[ \overline{v^*Av}=v^*A^*v=v^*Av. \] Since \(v^*v>0\), it follows that \(\lambda=\overline\lambda\), hence every eigenvalue is real.
We prove orthogonal diagonalizability by induction on \(n\). The case \(n=1\) is immediate. For \(n>1\), choose a unit eigenvector \(u_1\in\mathbb R^n\) with \(Au_1=\lambda_1u_1\). Let \[ W=u_1^\perp. \] For \(w\in W\), \[ \langle Aw,u_1\rangle =\langle w,A u_1\rangle =\lambda_1\langle w,u_1\rangle=0, \] so \(W\) is \(A\)-invariant. The restriction \(A|_W\) is again symmetric. By induction, \(W\) has an orthonormal basis \(u_2,\dots,u_n\) of eigenvectors of \(A\).
Thus \(u_1,\dots,u_n\) is an orthonormal eigenbasis of \(\mathbb R^n\). If \[ Q=[u_1\ \cdots\ u_n], \] then \(Q^TQ=I\) and \[ Q^TAQ=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \] Hence every real symmetric matrix has real eigenvalues and is orthogonally diagonalizable.