Let $T: \mathbb{R}^n \to \mathbb{R}^n$ be a linear map, with $n > 1$. Prove that there is a subspace $M \subset \mathbb{R}^n$, with $\dim M = 2$ such that $T(M) \subset M$.
This question is from my list of eigenvalues and eigenvectors. I don't want a solution to the question, my problem is that I really don't know how to approach the question. I think some hints should enough.
Using hints: Is $\lambda = a + bi \in \mathbb{C}^n$ is a non-real eigenvalue, so $$Ah = \lambda h \Longrightarrow A(h_1 + ih_2) = (a + bi)(h_1 + ih_2) \Longrightarrow Ah_1 + iAh_2 = ah_1 - bh_2 + i(ah_2 + bh_1),$$ from where $$Ah_1 = ah_1 - bh_2;\quad Ah_2 = bh_1 + ah_2.$$ Thus, $Ah_1, Ah_2 \in Span(h_1,h_2)$. Therefore, $Span(h_1,h_2)$ is two-dimensional invariant provided that $h_1, h_2$ are LI. How to ensure that?
Now, suppose that $T$ has no complex eigenvalues. If at least two real eigenvalues are distinct, is straighforward. But what about if no?
You are almost there with your solution using the hints.
Note that if $T(v)=\lambda v$, by taking complex conjugates, $T(v^*)=\lambda^* v^*$. Then $\lambda $ and $\lambda^* $ are unequal and therefore $v$ and $v^* $ are independent.
We have $T(v)=\lambda v$. Similarly, $T$ acting on $\mathbb{R}^n/(v)$ will have a real eigenvalue (which may of course also be $\lambda$). If the corresponding eigenvector is $(v)w$ then the requisite 2-dimensional subspace is generated by $v$ and $w$.