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Let $V$ be a finite dimensional vector space and let $T: V\to V$ be a linear transformation. Assume that $T$ is a projection - i.e.,

$T^2(v) = T(v)$ for every $v \in V$.

Assuming that $v \in \mathrm{range}(T)$, how do we show that $T(v) = v$?

I thought that it would be straight-forward to say that

$v \in \mathrm{range}(T)$ which equals $\{w \in V : w = T(v), v \in V\}$

so $v$ obviously $= T(v)$ since it must fit the condition $w = T(v)$. However, I was told that we must show this:

$T(v) = T^2(w) = T(w) = v$ where $v = T(w)$, $w \in V$.

But I don't understand why. Help appreciated! (I'm sure the explanation is something quick I'm missing)

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up vote 2 down vote accepted

If $v \in \mathrm{range}(T)$ then there exists some $w \in V$ such that $T(w)=v$. Applying $T$ to both sides we have that $T^2(w)=T(v)$, but $T^2(w)=T(w)=v$.

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Oh I see now. It's strange how the solution feels.. oddly 'circular' since you start with T(w) = v and then use it again at the end. But thanks! (I guess I'm still getting used to these problems) – dmonopoly Dec 15 '12 at 8:17

The problem in your argument is you have $w = T(v)$ for some $v\in V$ not the specific $v$ you started with.

The confusion is coming because you're using the same symbol $v$ to represent two things:

  1. your vector you want to show has $T(v) = v$
  2. a random vector in $V$

if you change the notation on the set I think it will be more clear why your argument doesn't work.

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Ah, that's a good point - this helps explain where I went wrong. So I need to be careful how I declare my variables and not to conflate the names. – dmonopoly Dec 15 '12 at 8:13

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