Diagonalize a complex matrix

Question

We have a matrix $A = \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \\ \end{array} \right)$

I have found the eigenvalues to be simple and equal to $$\lambda_1 = i$$ $$\lambda_2 = -i$$

Computing the eigenvectors we have:$\left( \begin{array}{ccc} -i & 1 \\ -1 & -i \\ \end{array} \right)s_1 = 0$ and :$\left( \begin{array}{ccc} i & 1 \\ -1 & i \\ \end{array} \right)s_2 = 0$

What are the eigenvectors?

Answer 1

Eigenvectors are solutions to the equation

$A v= \lambda v$.

Since you know both $A$ and $\lambda$, you can solve for $v$:

$$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} v_0\\v_1 \end{bmatrix}= \begin{bmatrix} i v_0\\i v_1 \end{bmatrix}$$

In other words, you need to solve the linear equations: $v_1=iv_0$ and $-v_0=iv_1$

The other eigenvector is just as easy.

Answer 2

Hint matrix A is rotation matrix where $\theta=90$ you can write $i=cis(90)$ now do simple caculation which is usually done that is $(A-(cis(90)I_2)x=0$ you will get the eigenvectors.