Is there any good strategy for computing null space of a matrix with entries $\cos x$ and $\sin x$?

Question

For example, say

$A= \left ( \begin{matrix} \cos x & -\sin x & 0 \\ \cos y \sin x & \cos x \cos y & -\sin y \\ \sin x \sin y & \sin y \cos x & \cos y \end{matrix} \right)$.

How do i conpute null space of $A-I$?

Since I don't know whether $\cos x$ , the 1-1 entry, is $0$ or not, I cannot simply take elementary operators on this matrix.

Is there any good strategy to find its null space?

(Note that $A$ is just an example. I'm asking how to compute null space of such matrices with entries $\cos x$, $\sin x$, $\cos y$, $\sin y$...)

Answer 1

Recall that the rank of a matrix is equal the the dimension of the space spanned by either its rows or its columns. So if you can prove the rows or columns are linearly independent, it must have full rank (i.e. trivial null space).

In this case, the dot product of any two distinct rows/columns is 0, and none of the rows/columns can be identically zero, so you're done.

In fact, the matrix is orthogonal, since all the rows/columns are in fact of length 1 as well. In some sense this isn't so surprising, because sines and cosines arise naturally in problems about rotations, which are represented by orthogonal matrices.

Answer 2

In your example:

$$\det A=\color{red}{\cos^2x}\cos^2y+\color{green}{\sin^2x}\sin^2y+\color{red}{\cos^2x}\sin^2y+\cos^2y\color{green}{\sin^2x}=\cos^2x+\sin^2x=1$$

Thus your matrix is regular $\,\iff \ker A=0\;$

In general you will need to resort to do some calculations as the above one, but I don't think there's a general formula for the general case.

Please note we used basic trigonometry above...