Is there a unitary diagonalization matrix for $A$?

Question

We have the matrix $$\begin{pmatrix}1 & i \\ 1+i & -1\end{pmatrix}$$ To check if there is a unitary diagonalization matrix to we have to check if the eigenvectors of $A$ are orthogonal?

The characteric polynomial is equal to $$p(\lambda )=\lambda^2-1$$ So the eigenvalus are $\lambda_{1,2}=\pm \sqrt{i}$.

For $\lambda =\sqrt{i}$ we get the eigenvector $v_1=\begin{pmatrix}1 \\ i(1-\sqrt{i})\end{pmatrix}$ and for $\lambda =-\sqrt{i}$ we get the eigenvector $v_2=\begin{pmatrix}1 \\ i(1+\sqrt{i})\end{pmatrix}$.

Is that everything correct so far?

It holds that $$v_1\cdot v_2=\begin{pmatrix}1 \\ i(1-\sqrt{i})\end{pmatrix}\cdot \begin{pmatrix}1 \\ i(1+\sqrt{i})\end{pmatrix}=1+i^2(1-\sqrt{i})(1+\sqrt{i})=1-1+i=i\neq 0$$ Is this correct? Does this mean that there is no unitary matrix that diagonalizes $A$ ?

Answer 1

It is not a good idea to write $\sqrt i$, since that's ambiguous. The number $i$ has two square roots: $\pm\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)$.

And the easiest way to determina whether or not a square matrix is unitarily diagonalizable is to apply the spectral theorem. Since $A$ doesn't commute withits conjugate transpose, it is not unitarily diagonalizable.

Answer 2

If you want to check whether the matrix is unitary diagonalizable the easy and fast way, without much calculations, check whether it is a normal matrix:

$$A=\begin{pmatrix}1&i\\ 1+i&-1\end{pmatrix}\implies A^*=\overline{A^t}=\begin{pmatrix}1&1-i\\ -i&-1\end{pmatrix}$$

and then

$$AA^*=\begin{pmatrix}2&1-2i\\ 1+2i&3\end{pmatrix}\;,\;\;A^*A=\begin{pmatrix}3&-1+2i\\ 1-2i&2\end{pmatrix}\neq AA^*$$

so the matrix isn't normal and is thus not unitary diagonalizable (this is the spectral theorem)