What can you say about the eigenvalues of a real symmetric matrix?
Solution
Let \(A\in M_n(\RR)\) satisfy \(A^t=A\). Regard an eigenpair \(Av=\lambda v\) in \(\CC^n\). Then \[ \lambda\langle v,v\rangle =\langle Av,v\rangle =\langle v,Av\rangle =\overline\lambda\langle v,v\rangle. \] Since \(v\ne0\), this gives \(\lambda=\overline\lambda\), so every eigenvalue is real.
The real spectral theorem strengthens this: there is an orthogonal matrix \(Q\) and real numbers \(\lambda_1,\ldots,\lambda_n\) such that \[ Q^tAQ=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \] Thus a real symmetric matrix has only real eigenvalues, is diagonalizable over \(\RR\), and admits an orthonormal basis of eigenvectors. Eigenspaces belonging to distinct eigenvalues are orthogonal.