# Proof of Existence of Decomposition of a Vector Space

I want to prove the following,

Let $$T$$ be a linear transformation on a finite dimensional vector space $$V$$ with dim$$V \neq 0$$. Then for some integer $$m \geq 1$$, there exist $$T$$-invariant subspaces $$U_1,...,U_m$$ which are indecomposable with respect to $$T$$ , such that $$V = U_1 \oplus \ ... \oplus \ U_m$$.

$$\oplus$$ denotes a direct sum.

Some background:

Let $$U$$ be a subspace of a finite dimensional vector space $$V$$. Let $$T : V \rightarrow V$$ be a linear transformation.$$\ U$$ is said to be $$T$$-invariant if $$T(U) \subseteq U$$, where $$T(U)$$ denotes the image of the subspace $$U$$ under $$T$$.

Let $$U$$ be a subspace of a finite dimensional vector space $$V$$. Let $$T : V \rightarrow V$$ be a linear transformation. A subspace $$U$$ of $$V$$ is said to be decomposable with respect to $$T$$ if there exist proper subspaces $$U_1,U_2$$ of $$U$$, invariant under $$T$$ such that $$U = U_1 \oplus U_2$$.

I have been given a hint, to consider the proof that any integer can be factorised into primes, and to imitate this. I am referring to the proof here: https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

The reason why I want to prove this lemma is that it is useful for proving that any linear mapping can be represented as a Jordan matrix. It is a chain of one of many lemmas to arrive at the result. It's also a pretty nice result that draws parallels with the factorisation of an integer into prime numbers.

My attempt:

Let $$V$$ be a non-zero, finite dimensional vector space. Let $$T : V \rightarrow V$$ be a linear transformation. We can write $$V = V \oplus \{\textbf{0}\}$$ where $$\textbf{0}$$ denotes the zero element of $$V$$. It is clear that $$\{\textbf{0}\}$$ is indecomposable. If $$V$$ is indecomposable, then we are done, so we suppose $$V$$ is decomposable, so we can write $$V = U_1 \oplus U_2$$ where $$U_1$$ and $$U_2$$ are proper subspaces of $$V$$, and $$U_1$$ and $$U_2$$ are $$T$$-invariant. I could keep going with this line of reasoning, saying if $$U_1$$ and $$U_2$$ are indecomposable, we are done. So suppose at least one of $$U_1$$ and $$U_2$$ are decomposable.

I know that this process can't go on indefinitely, since $$V$$ is finite dimensional, we have $$\sum_{i = 1}^{m} \text{dim}(U_i) = \text{dim}(V)$$ This suggests to me that I may need to use induction, but I am not sure how to implement it here.

If $V=U_1\oplus U_2$ is a nontrivial decomposition then both $U_1$ and $U_2$ have smaller $\dim$ than $V$.