# How to do a unitary diagonalization of a normal matrix?

It is easy to diagonalize a normal matrix such that $D = P^{-1} A P$ by simply putting all the orthogonal eigenvectors as columns for $P$. But I spent hours trying a unitary diagonalization of the following Hermitian (and therefore Normal) matrix: $$A = \begin{bmatrix} 0 & i & 1 \\ -i & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ such that $D = U^*AU$. I know that by definition every normal matrix is unitarily diagonalizable. The eigenvalues of this matrix are $\{ 0, -\sqrt{2}, \sqrt{2} \}$.

What did not work but was my most promising try, was to scale down the eigenvectors by their norm so the matrix $P$ became orthonormal. The result does not give me the diagonal matrix with the desired eigenvalues though. Also, Google search did not yield a single nicely explained way to do a unitary transform of a normal matrix. The only document that I believe to try to explain it is here, although it does not show clearly how to construct $U$.

• Note that the Matlab expression U' returns the conjugate transpose of U for matrices U with complex entries and that U.' returns the nonconjugate transpose. I'm guessing that your promising try was correct, but your verification failed. – DCarter Nov 7 '16 at 15:33
• Nearly! My try was not correct since I didn't normalize properly. – Flaudre Nov 8 '16 at 3:17

The eigenvalues of $A$ are $0, \sqrt{2}, -\sqrt{2}$. These eigenvalues correspond to the eigenvectors $$\begin{bmatrix} 0\\ i\\ 1 \end{bmatrix},\quad \begin{bmatrix} \sqrt{2}\\ -i\\ 1 \end{bmatrix},\quad \begin{bmatrix} -\sqrt{2}\\ -i\\ 1 \end{bmatrix},$$ respectively. You will observe that the eigenvectors are orthogonal with respect to the standard inner product on $\mathbb{C}^n$. Normalizing the eigenvectors gives the unitary matrix $$U = \begin{bmatrix} 0 & 1/\sqrt{2} & -1/\sqrt{2}\\ i/\sqrt{2} & -i/2 & -i/2\\ 1/\sqrt{2} & 1/2 & 1/2 \end{bmatrix}$$ that diagonalizes $A$ to $D = \operatorname{diag}(0,\sqrt{2},-\sqrt{2})$.
• Absolutely. I failed to orthonormalize properly with respect to the standard inner product in $\mathbb{C}^n$ as you showed. Thanks! – Flaudre Nov 8 '16 at 3:22
• Is any diagonalizable matrix can be in the form $U D {U}^{H}$ or there are special requirements for that? – Royi Aug 25 '17 at 11:15
• Only the normal matrices are unitarily diagonalizable. A matrix $A$ is normal if $A^*A = AA^*$. – K. Miller Aug 25 '17 at 13:15