Unitary similarity transformation I have a matrix:
$ A= \dfrac{i}{3} \begin{bmatrix} 1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix} $
Could someone explain me how to find a corresponding diagonal matrix for a diagonalizable matrix or linear map by unitary similarity transformation, please?(in the simplest way, if it's possible)

@Amzoti: Could you tell me where is my mistake: $A= \frac{i}{3} \begin{bmatrix} 1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix} = \begin{bmatrix} \frac{i}{3}&\frac{-2i}{3}&\frac{i}{3}\\\frac{-2i}{3}&\frac{i}{3}&\frac{i}{3}\\\frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}\end{bmatrix} \\ A_{\lambda} = \begin{bmatrix} \frac{i}{3}-\lambda&\frac{-2i}{3}&\frac{i}{3}\\ \frac{-2i}{3}&\frac{i}{3}-\lambda&\frac{i}{3}\\ \frac{i}{3}&\frac{i}{3}&\frac{-2i}{3}-\lambda\end{bmatrix}$
After transformations($C_{1}-C_{2}$ and $W_{2}+W_{1}$), I have:
$A_{\lambda} = \begin{bmatrix} i-\frac{\lambda}{3}&\frac{-2i}{3}&\frac{i}{3}\\ 0&\frac{-i-\lambda}{3}&\frac{2i}{3}\\ 0&\frac{i}{3}&\frac{-2i-\lambda}{3}\end{bmatrix}$
So then I wanted to develop it with the first column:
$(i-\frac{\lambda}{3})(-1)^{2} \begin{bmatrix} \frac{-1-\lambda}{3}&\frac{2i}{3}\\\frac{i}{3}&\frac{-2i-\lambda}{3} \end{bmatrix}$
And the CP: $\lambda(i-\frac{\lambda}{3})(i+\frac{\lambda}{3})=0$
