I have a symmetric $k$ by $k$ matrix B with eigenvalues $\lambda_1,\dots \lambda_k$. I found a result in an old linear models book that says "there exists an orthogonal matrix Q such that Q'BQ= D" where D = diag($\lambda_1,\dots \lambda_k$).
My question is whether or not there is a more efficient way to determine Q than to solve the entire set of equations?
UPDATE
In the simplest case, *B*$=-\frac{\theta^2}{4+2\theta^2}\cdot$ \begin{pmatrix} 5+5\theta^2+\theta^4 & \sqrt{1+\theta^2} \\ \sqrt{1+\theta^2} & 1\end{pmatrix}