Proving existence of $T$-invariant subspace Let $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ be a linear transformation. I'm trying to prove that there exists a T-invariant subspace $W\subset \mathbb{R}^3$ so that $\dim W=2$.
How can I prove it? Any advice?

If you've not learned about minimal polynomials or Jordan normal form, here's a simple proof involving the Cayley-Hamilton Theorem. 
Let $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be a linear map and $A = [T]_{\text{std}}$ be the matrix representation of $T$ in the standard basis.
Take the set $\{v,Tv\}$. If this is linearly dependent for all $v \in \mathbb{R}^{n}$, then we can choose any $\{v_{1},v_{2}\} \subseteq \mathbb{R}^{n}$ such that $v_{1}$ and $v_{2}$ are linearly independent.
Then, $\operatorname{span}\{v_{1},v_{2}\}$ is the required $2$-dimensional subspace.
If $\{v,Tv\}$ is not linearly independent for some $v \in \mathbb{R}^{n}$, choose that $v$.
According to Cayley-Hamilton Theorem, we have the characteristic polynomial of A, $p(\lambda)$ such that $p(A)=0 \implies p(A)v=0 \forall v \in \mathbb{R}^{n}$. $p(A)$ can be factored into irreducible polynomials of degree $1$ or $2$ by the Fundamental Theorem of Algebra.
Thus, there is some linear or quadratic polynomial $q(\lambda)$ such that $q(\lambda)$ is a factor of $p(\lambda)$ and $q(A)v = 0$. If $q(A)$ is linear, we are done.
If $q(A)$ is quadratic, there are some $s,r \in \mathbb{R}$ such that
\begin{equation}
\begin{split}
  & A^{2}v + sAv + rv = 0\\
  &\quad\implies  A^{2}v = -sAv-rv \\
  &\quad\implies  T^{2}v = -sTv-rv \label{eq:1}
\end{split}
\end{equation}
Now, for some $v' \in \operatorname{span}\{v,Tv\}$, $v' = av + bTv$ for some $a,b \in \mathbb{R}$.
So,
\begin{equation}
\begin{split}
  T(v') & = T(av + bTv) \\
  & = aTv + bT^{2}v \\
  & = aTv + b(-sTv-rv) \\
  & = (a-bs)Tv - rbv \in \operatorname{span}\{v,Tv\} \\
\end{split}
\end{equation}
Thus $\operatorname{span}\{v, Tv\}$ is the required 2-dimensional invariant subspace.
