Thread: Direct sum of a range and T-invariant subspace problem

1. Direct sum of a range and T-invariant subspace problem

Hi!
I've been working for this problem for quite a while, but am seriously stuck now, so I would appreciate any help!
Here's the problem:
Suppose $\displaystyle V=R(T) \oplus W$ and W is T-invariant. Prove that $\displaystyle W \subseteq N(T)$

where R(T) is the range of linear transformation $\displaystyle T:V \rightarrow V$ and N(T) is a null space of T, and W is a subspace of V.
Here's what I got:
I need to show that for all $\displaystyle w \in W, w \in N(T)$ and w is in N(T) if and only if T(w)={0}, so what i need to show is for all $\displaystyle w \in W, T(w) = 0$.
Since V is a direct sum of R(T) and W, then $\displaystyle V=R(T)+W$ and $\displaystyle R(T) \cap W = \{0\}$ . Since V=R(T)+W, i know that for all $\displaystyle v \in V$, there exists $\displaystyle y \in R(T)$ and there exists $\displaystyle w \in W$ such that v=y+w. Since there exists y in R(T), then there exists an $\displaystyle x \in V$ such that T(x)=y. And from W being T-invariant, i know $\displaystyle T(w) \in W$.
Ok, knowing it is good, but i cannot relate that to N(T). What do I do next??

Thank you for any help!!

2. If we take $\displaystyle w \in W$, then from W being T-invariant, we get that $\displaystyle T(w) = w'$ for some $\displaystyle w' \in W$; that is, $\displaystyle T(w) \in W$.

But we also know that $\displaystyle T(w) \in Range(T)$ by definition, therefore we get that $\displaystyle T(w) \in W \cap Range(T)$ but this is possible only if $\displaystyle T(w) = 0 \Rightarrow w \in Null(T) \Rightarrow W \subset Null(T)$

3. Oh, thank you so much! it is so easy now that you explained it! makes perfect sense. Thanks a lot!