Unitary similarity transformation

Question

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)

Answer 1

@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: $\lambda(i-\frac{\lambda}{3})(i+\frac{\lambda}{3})=0$

Answer 2

$$|xI-A|=\left(\frac i3\right)^3\begin{vmatrix}x-1&2&-1\\2&x-1&-1\\-1&-1&x+2\end{vmatrix}=$$

$$=-\frac i{27}(x-1)^2(x+2)+4-2(x-1)-4(x+2)=$$

$$=-\frac i{27}\left(x^3-3x+\color{red}2-2x+\color{red}2+\color{red}4-4x-\color{red}8\right)=-\frac i{27}(x^3-9x)=$$

$$=-\frac i{27}x(x-3)(x+3)$$

and the above is close (at least) to what you got (and not what Amzoti got), so you better check stuff.