Finding a diagonal matrix $B$ and a unitary matrix $C$ that satisfy $B=C^{-1}AC$.

Question

The matrix $A$ is given as $$A=\frac{1}{9} \begin{bmatrix} 4+3i & 4i & -6-2i \\ -4i & 4-3i & -2-6i \\ 6+2i & -2-6i & 1 \end{bmatrix}$$ Find a diagonal matrix $B$ and a unitary matrix $C$ that satisfy $B=C^{-1}AC$.

Could anyone help me deal with this + explain the algorithm to find $B$ and $C$ in a little bit more detailed way? I didn't understand everything well by reading the book.

I've already found the eigenvalues $\lambda_1=1, \ \ \lambda_2=i, \ \ \lambda_3=-i$ and the eigenvectors $v_1=(-2i \ \ -2 \ \ 1)^T, \ \ v_2=(i \ \ -1/2 \ \ 1)^T, \ \ v_3=(-i/2 \ \ 1 \ \ 1)^T$.

Answer 1

Now, you divide each of those vectors by its norm, thereby getting$$w_1=\frac13(-2i\ \ -2\ \ 1)^T,\ w_2=\frac13(2i\ \ -1\ \ 2)^T\text{, and }w_3=\frac13(-i\ \ 2\ \ 2)^T.$$And then you take$$C=\frac13\begin{bmatrix}-2i&2i&-i\\-2&-1&2\\1&2&2\end{bmatrix},$$whose columns are the vectors $w_1$, $w_2$, and $w_3$. And you're done.

Answer 2

You already have the answer. Since you know the eigenvalues are $1,i,-i$, then a diagonal matrix $B$ is $$ \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i\end{array}\right] $$