Question about a linear transformation 
For a linear transformation $A\in M_{n\times n}(\mathbb{R})$ can we prove there is a subspace $U$ where $\dim_{\mathbb{R}}(U)=n-2$ such that $A(U)\leq U$?

Starting in $\mathbb{R}^2$ we know that the $0$-space has this property but lines through the origin ($1$-spaces) do not necessarily have this property. The same if true for $\mathbb{R}^3$: For any matrix $A\in\mathbb{R}^3$ we can find some $1$-space such that the image of that $1$ space is either itself or the $0$-space but we cant necessarily find a $2$-space.
 A: This follows directly from the existence of real Schur form. In general, if $L:\mathbb R^n\to\mathbb R^n$ is a linear operator, there exists an orthonormal basis of $\mathbb R^n$ with respect to which the matrix representation of $L$ is in the form of
$$
\pmatrix{R_1&\ast&\ast&\cdots&\ast\\ &R_2& \ast&\cdots&\ast\\ &&\ddots&\ddots&\vdots\\ &&&\ddots&\ast\\ &&&&R_k},
$$
where each $R_i$ is either a real $1\times1$ matrix (i.e. a real scalar) or a real multiple of some $2\times2$ rotation matrix. The $R_i$s can also be arranged in a way that the $1\times1$ sub-blocks precede the $2\times2$ sub-blocks on the block diagonal.
In particular, if we take $L:x\mapsto Ax$, then there exists a real orthogonal matrix $Q$ such that $QAQ^T$ is in real Schur form. It follows that the first $n-2$ columns of $Q$ form an invariant subspace of $A$.
But surely, the use of real Schur form is an overkill. Since $A^T$ either has at least two real eigenvalues or a non-real spectrum, it can be shown (in the same spirit of the existential proof of real Schur form) that
$$
QA^TQ^T=\pmatrix{M&\ast\\ 0&\ast}
$$
for some real $2\times2$ matrix $M$ that is either upper triangular or a real scalar multiple of a rotation matrix. Then the last $n-2$ columns of $Q$ form an invariant subspace of $A$.
