TITLE: How to come up with a counter example in linear algebra
QUESTION [5 upvotes]: This came up in a problem I was working on.
Problem:Let $V$ be an $n$ dimensional vector space over a field $F$. Let $T:V\rightarrow V$ be a linear operator and let $W$ be a $T$ invariant subspace of $V$. Prove or give a counter example to the following:

Let $U$ be another subspace of $V$ such that $V=W\oplus U$. Then $U$ is also $T$ invariant.

Solution Well this is false. I came up with a counter example after spending an embarrassing amount of time guessing and checking. Consider the operator 
$$ \left(\begin{array}{ccc}1 & -1 &  \\0 & 1 &\end{array}\right)$$
This does the job because $V=\langle e_1 \rangle \oplus \langle e_2 \rangle$ and the last summand isn't $T$ invertible.
My question is I feel like I've spent too much time on this. I absolutely used no linear algebra knowledge to come up with this. I remotely remember my professor working on $\mathbb{R}^2$ to pull a counter example out of his hat on an unrelated problem. So I just kept plugging away numbers until something worked. Even my grandmother who evidently has never taken linear algebra could have done that. So what I ask is: How would you smart people do this
 problem?. How do you come up with a counter example for this particular problem. Can you explain the Linear Algebra behind your thought process?. What geometric intuition helps here? (and in general for counterexamples of this nature).  
I hope my question makes sense.
Thank you for your comments and answers.

REPLY [4 votes]: It's easier to come up with counter-examples if you understand more about how matrices and linear transformations work, e.g. via Jordan normal form and the special form of complex-valued eigendecompositions of real matrices.  E.g. if you understand the meaning of Jordan normal form, then you know that it is possible to not have a full eigen-decomposition, and instead have a matrix that "shifts" some vector to a combination of itself and some other vector, rather than having a full set of eigenvectors each of which is carried to a multiple of itself.  This is exactly the kind of counter-example you found:  You wrote down a matrix that leaves invariant the $e_1$ subspace but shifts $e_2$ to a combination of $e_1$ and $e_2$ (thus does not leave invariant the $e_2$ subspace).  Realizing this kind of linear transformation is possible, and how to construct it, should be more obvious to you once you know more about the fundamental characterizations of matrices and linear transformations and what they mean.