Proof of Decomposition of a Vector Space, Need Answer I'm in a rather peculiar situation right now. I'd come to MSE with the exact same question which this person asked right here:
Proof of Existence of Decomposition of a Vector Space
Problem is, the single person who answered the question only offered a hint, and I can't figure out the proof. There are no other answers or comments, and I don't have enough reputation to comment asking for help, so I've decided to repost the question here and ask for help. I'd greatly appreciate any input from you guys!
 A: The answerer is suggesting the use of strong induction. The predicate that should be proven is
\begin{align*}
P(n) : \;&\text{Given any } n\text{-dimensional invariant subspace } U \text{ of } V, U \text{ can be expressed}\\
&\text{as a direct sum of indecomposable invariant subspaces with respect to } T.
\end{align*}
Note that $P(\dim V)$ is the result you want proven.
The base case is to note that $1$-dimensional invariant subspaces $U$ of $V$ are automatically indecomposable, and so we can write $U$ as the sum containing only itself.
For strong induction, we assume, for some $n$, $P(k)$ is true for all $1 \le k < n$. Suppose $U$ is $n$-dimensional and invariant under $T$. If $U$ is indecomposable, then we are done, much like in the base case: $U$ is equal to the sum containing only itself.
Otherwise $U = U_1 \oplus U_2$, where $U_1, U_2$ are invariant with respect to $T$ and non-trivial. Thus, both $U_1$ and $U_2$ have dimensions at least $1$ and strictly less than $n$. Therefore, $P(\dim U_1)$ and $P(\dim U_2)$ are both true, so $U_1$ and $U_2$ are both the direct sum of indecomposable invariant subspaces, hence the same is true of $U_1 \oplus U_2 = U$.
This completes the proof of $P(n)$, and hence $P(\dim V)$ by strong induction.
(Compare and contrast: the fundamental theorem of arithmetic!)
