Given a characteristic polynomial, make a conclusion about invariant subspaces I'm attempting to solve the following problem.

Find a real matrix whose characteristic polynomial is $X^4+1$. Then prove that for any such matrix $A$ there is no proper $A$-invariant subspace of $\mathbb{R}^4$.

Any $4\times 4$ rotation matrix will suffice for the first part. I believe the second part is actually false. It should be possible to find a $2$-dimensional subspace of $\mathbb{R}^4$ that is invariant under rotation. So I'm assuming the intended question is actually to show that there is no $3$-dimensional $A$-invariant subspace of $\mathbb{R}^4$. This seems true to me, but I'm struggling showing it, since I have to do so for any matrix with the given characteristic polynomial. Presumably, there are non-rotation matrices with that characteristic polynomial, which defeats my intuition about what's happening geometrically. Are there important tools for solving this type of problem or is it actually just a straightforward "suppose there is a $3$-dimensional $A$-invariant subspace...contradiction" type argument?
 A: The matrix
$$
\frac{1}{\sqrt 2}\begin{pmatrix}1 & 1 & 0 & 0\\-1 & 1 & 0 & 0\\0 & 0 & -1 & 1\\0 & 0 & -1 & -1\end{pmatrix}
$$
has the given characteristic polynomial but has obviously the invariant subspace $\mathbb R^2\times\{(0,0)\}$.
A: Since $x^4+1=(x^2-\sqrt 2x+1)(x^2+\sqrt 2x+1)$, such a matrix $A$ is conjugate to
$$\begin{pmatrix}
0 & -1 & 0 & 0\\
1 & \sqrt 2 & 0 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & 1 & -\sqrt 2
\end{pmatrix}.$$
We might as well pick a basis so that this is equal to $A$. Then $A$ preserves the proper subspaces
$$V=\operatorname{span}\left\{\begin{pmatrix}
1\\
0\\
0\\
0
\end{pmatrix},\begin{pmatrix}
0\\
1\\
0\\
0
\end{pmatrix}\right\}\qquad\text{and}\qquad
W=\operatorname{span}\left\{\begin{pmatrix}
0\\
0\\
1\\
0
\end{pmatrix},\begin{pmatrix}
0\\
0\\
0\\
1
\end{pmatrix}\right\}.$$
On the other hand, if $A$ preserves a $1$-dimensional subspace, then it has an eigenvalue, but $x^4+1$ has no real roots. Suppose $A$ preserves a $3$-dimensional subspace $U$. Then we have a restriction which is a linear map $U\to U$. But any map between odd-dimensional real vector spaces has an eigenvector since an odd-degree polynomial has an odd root, so $A$ cannot preserve any $3$-dimensional subspace.
