Null space of a rotation matrix If we have a rotation matrix of the kind:
link to the rotation matrix
how do i compute the null space of this matrix?
I know that to obtain the null space we need to write the matrix in echelon form, but in this case we have an orthonormal matrix, so my question is: how do we ge the null space of a rotation matrix knowing that it is an orthonormal matrix? Is there a different, and perhaps faster and easier way to do it?
Thank's in advance.
 A: Since it is a rotation matrix, its null space is $\{0\}$; $0$ is the only vector mapped into $0$ after rotating it around $0$.
A: For the $3 \times 3$ rotation matrix
$$A = 
\begin{pmatrix} 
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0 \\
0 & 0 & 1
\end{pmatrix},
$$
we compute its determinant. As $det(A) = 1 \neq 0,$ the matrix $A$ is invertible.
This implies $rank(A)=3,$ so the null space only contains the zero vector.
Alternatively, we proceed via trigonometry. Namely, the the null space condition
$$N(A) = 
\bigg\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} : \; 
\begin{pmatrix} 
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\bigg\}.$$
implies
$$\begin{pmatrix} cos(\theta) x - sin(\theta) y \\ sin(\theta) x + cos(\theta) y \\ z \end{pmatrix} = 
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$
Hence
$$cos(\theta) x = sin(\theta) y,$$
$$sin(\theta) x = -cos(\theta) y,$$
$$z = 0,$$
and the latter equation, $z = 0,$ suggests we focus on the preceding equations, which arise in the study of $2 \times 2$ rotation matrices.
Now, multiply the first equation by $cos(\theta)$ and the second equation by $sin(\theta),$ then add the modified equations to obtain
$$x = 1x = (cos^{2}(\theta) + sin^{2}(\theta)) x = (sin(\theta)cos(\theta) - sin(\theta)cos(\theta))y = 0y = 0.$$ Similarly, $y=0.$ Therefore the null space only contains the zero vector, i.e., $$N(A) = \bigg\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\bigg\}.$$
