Eigenvalues and eigenvector of a matrix containing $\sin \theta$ and $\cos \theta$

Question

$$
A
= \begin{bmatrix}
0 & 1 & -\sin \theta \\
-1 & 0 & \cos \theta \\
-\sin \theta & \cos \theta & 0
\end{bmatrix}
$$
Given this matrix I want to calculate its eigenvalues and eigenvectors. What I don't understand is how this can be done because I find a characteristic polynomial of $\lambda^3 = 0$ which means that its eigenvalue is $0$ with algebraic multiplicity of $3$. When I perform the row operations to find the eigenvectors I can't because the top left element is $0$ so I can't eliminate the elements below it.

Answer 1

Since you have a $0$ there, exchange the first and the second rows, thereby getting$$\begin{bmatrix}-1&0&\cos\theta\\0&1&\sin\theta\\-\sin\theta&\cos\theta&0\end{bmatrix}.$$Now, proceed in the usual way. In the end, you will get that the eigenvectors with eigenvalue $a$ are those vectors of the form $\lambda(\cos\theta,\sin\theta,1)$.