Do $T$-invariant subspaces necessarily have a $T$-invariant complement?

Solution 1:

No. For example consider $\begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}$ as a linear map from $\mathbb{R}^2$ to itself, and the $T$-invariant subspace generated by $\begin{bmatrix}0 \\ 1 \end{bmatrix}$. If there is a complement, it must have some element of the form $\begin{bmatrix} a \\ b \end{bmatrix}$ with $a \neq 0$, but then apply $T$, you see that $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ also lies in that subspace, which means the subspace is not a complement.

The point is that if a $T$-invariant subspace always has complement, this automatically implies that $T$ is always diagonalizable provided that the eigenvalues lie in the field you are working with - and you can easily find something non-diagonalizable.