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

  1. #1
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    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!!
    Last edited by vanishingpoint; Mar 30th 2010 at 03:49 PM.
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  2. #2
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    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 $\displaystyle T(w) = 0 \Rightarrow w \in Null(T) \Rightarrow W \subset Null(T)$
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  3. #3
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    Oh, thank you so much! it is so easy now that you explained it! makes perfect sense. Thanks a lot!
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