Question: Suppose $ A= \left( \begin{array}{ccc}
1 & 2 & 2 \\
2 & 0 & 3 \\
2 & 3 & 0 \end{array} \right) $

How to show there exists a unitary matrix $C$ such that $C^{-1}AC=B$, where $B\_{21} =B\_{31} =0$.

I appreciate some hints or solutions. Thank you.

Answer: 
The matrix is real symmetric, so has an orthonormal basis of eigenvectors. With $C$ the matrix whose columns describe that basis, $C^{-1}AC$ is a diagonal matrix; this is more than was asked for.
**Hint:** It suffices to select a unitary matrix $C$ whose first column is an eigenvector of $A$.