How to get the eigenvectors from eigenvalues in a rotation matrix?

Question

For the rotation matrix \begin{bmatrix} cos(t) & sin(t) & 0 \\ -sin(t) & cos(t) & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} I already got to the three corresponding eigenvalues [1, cos(t) + i sin(t), cos(t) - i sin(t)].

Now I don't know how to apply them to find each corresponding eigenvectors. I tried some examples I saw in YouTube, but they are only examples about matrices with real numbers, and in this case the eigenvectors involve complex numbers.

In a solutions book I found that the eigenvectors are:

For e-val1 = |0 0 1|

for e-val2 = |$\frac{1}{\sqrt{2}}$ $\frac{i}{\sqrt{2}}$ 0|

and for e-val3 = |$\frac{1}{\sqrt{2}}$ $\frac{-i}{\sqrt{2}}$ 0|

But I have no idea on how to get to those values. The examples I saw simply don't work.

Could someone please, if it's not trouble, tell me some tip, clue or explanation on how to reach those eigenvectors? Maybe I'm missing some trigonometric identity or something like that.

Answer 1

In general we solve the equation plugging in each eigenvalue. $$A v_1 = \lambda_1 v_1$$ For this example, setting up the first one looks like: $$\begin{bmatrix} cos \theta &\sin \theta & 0\\ - \sin \theta & cos \theta & 0\\ 0&0&1 \end{bmatrix} \begin{bmatrix}v_{1x} \\v_{1y} \\v_{1z} \end{bmatrix}= 1 \begin{bmatrix}v_{1x} \\v_{1y} \\v_{1z} \end{bmatrix}$$

$$ cos \theta v_{1x} +sin \theta v_{1y} = v_{1x}$$$$ -sin \theta v_{1x} +cos \theta v_{1y} = v_{1y} $$ $$ v_{1y} =v_{1z} $$ From this we see that $v_{1x}=v_{1y}=0$ and $v_{1z}$ is arbitrary. Normalizing leaves $v_{1z}=1$ or: $$v_{1}= \begin{bmatrix} 0\\0\\1 \end{bmatrix}$$ Hope this helps finding the other ones.

Answer 2

Solve $\det(A-\lambda I) = 0$ to get eigenvalues $\lambda$, if you are unsure if you obtained the [correct] eigenvalues, test them once again in the characteristic equation. It is just finding roots to a polynomial.

Once having all eigenvalues, solve for vector $v$ $$ Av = \lambda v \iff (A-\lambda I)v = 0, \forall \lambda$$ which is a basic matrix equation. For each $\lambda$ you will obtain a corresponding eigenvector $v$ (since all roots above are simple).

Complex numbers will not change the methodology, in fact it will only make it favorable since the fundamental theorem of algebra provides solutions to the polynomial.