Suppose $V$ is a vector space over $\Bbb F, T:V\to V$ is a linear operator and $U$ is a T-invariant subspace of V. is there necessarily another T-invariant subspace$-\,W\,\, S.T. \,U\oplus W = V?$
I suspect this is in fact not true but cannot come up with a counter example.
Work out the details in the following example:
$$T:\Bbb R^2\to\Bbb R^2\;,\;\;T\binom xy:=\begin{pmatrix}1&1\\0&1\end{pmatrix}\binom xy$$
You can check the only eigenvalue of $\;T\;$ is $\;\lambda=1\;$, and thus there is a $\;T\,-$ invariant space, $\;U:=$ Span$\,\left\{\,\binom x0\,\right\}\;$, which is the eigenspace of the above eigenvalue.
Well, if there were another $\;T\,-$ invariant space $\;W\;$ s.t. $\;\Bbb R^2=U\oplus W\;$ , then $\;T\;$ would have another eigenvalue different from $\;1\;$ ....but this is not so.
