Finding generators for $A$-invariant subspace 
For the real matrix $A=\pmatrix{3 & -3 & 1 & 3\\
-1 & 4 & -1 & -2\\
3 & 0 & -1 & 2\\
-5 & 9 & -2 & -6}$ we have $A^4=4I$.

(1) Show that the minimum polynomial of $A$ divides $x^4 − 4$.
(2) Give generators for a $A$-invariant subspace $U \subset \mathbb{R}^4$ with $U\neq \{0\}$ and $U\neq \mathbb{R}^4$.

I solved (1) by noting that $A^4-4I=0$, this implies that the polynomial $x^4-4$ vanishes at $A$, so the minimal polynomial of $A$ divides $x^4-4$.
Now since $x^4-4=(x-\sqrt{2})(x+\sqrt{2})$ in $\mathbb{R}$ I think that means that $\pm\sqrt{2}$ are eigenvalues of $A$. Meaning that $Av=\pm\sqrt{2}v$ for $v$ in the eigenspaces of $\pm\sqrt{2}$ respectively.
I suppose to solve (2) I should calculate these eigenspaces, because they're invariant under $A$, but this seems like tedious amounts of work so I was wondering if there are other (more efficient) ways to find such $A$-invariant subspaces of $\mathbb{R}^4$.
Any help is greatly appreciated!
 A: Whether this is really less "tedious" is subjective, but here's a way to answer the question while avoiding square roots and instead only requires that we compute $A^2$. In particular, we find
$$
A^2 = \left[\begin{matrix}0 & 6 & -1 & -1\\0 & 1 & 0 & -1\\-4 & 9 & 0 & -5\\0 & -3 & 0 & -1\end{matrix}\right].
$$
First of all, note that the minimal polynomial is indeed $x^4 - 4I$; we can actually deduce that this is the case from the fact that $A$ has rational entries, which means that the only other candidates are $(x^2 \pm 2)$, which would imply that $A^2$ is a multiple of the identity matrix, which is not the case.
Note that $(A^2 - 2I)(A^2 + 2I) = 0$. So, the column-space of $A^2 + 2I$ must span the kernel of $A^2 - 2I$, which consists of the (direct) sum of the eigenspaces of $A$ associated with $\pm \sqrt{2}$. Note that this 2-dimensional subspace is indeed an invariant subspace of $A$. We know that this column-space must be $2$-dimensional.
Alternatively, we could simply have noted that $\operatorname{im}(A^2 + 2I)$ is an invariant subspace of $A$ (as is $\operatorname{im}(p(A))$ for any polynomial $p$), then deduced that $A^2 + 2I$ cannot be invertible since $(A^2 - 2I)$ is non-zero, which in turn means that $\operatorname{im}(A^2 - 2I)$ is a proper and non-trivial invariant subspace.
It is easy to see that the first two columns of $A^2 + 2I$ are linearly independent, which means that they form a basis for the column space of $A^2 + 2I$, which is the space we're interested in. Thus, we can take our vectors to be
$$
(2,0,-4,0),\quad(6,3,9,-3).
$$
