# Diagonalizing a real normal matrix

Given the matrix $$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{pmatrix}$$, how would I find a real orthogonal matrix $$P$$ such that $$PAP^t$$ is a diagonal matrix?

I've found the eigenvalues $$0, \dfrac{9\pm\sqrt{105}}{2}$$, but I don't know how to proceed from here.

There exists a basis of eigenvectors since $$A$$ is symmetric. Form an orthonormal basis of eigenvectors. Form the matrix whose columns are the basis vectors, as your $$P$$.
• For each eigenvalue $\lambda$, find the kernel of $A-\lambda I$. Get three linearly independent eigenvectors, then use Gram-schmidt. – Chris Custer Feb 21 at 18:36