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Consider $W\subseteq V$, a subspace over a field $\mathbb{F}$ and $T:V\rightarrow V$ a linear transformation with the stipulation that $T(W)\subseteq W$. Then we have the induced linear transformation $\overline{T}:V/W \rightarrow V/W$ such that $\overline{T}=T(v)+W$.

I'm supposed to show that this induced transformation is well-defined, and that given $V$ finite and $T$ an isomorphism that $\overline{T}$ is an isomorphism. I'm having a little trouble with this part. Namely, I want to show that the $ker(\overline{T})=W$, and I'm to the point where I realize that this means $T(v)\in W$. How do I know there's not some random $v\in V\setminus W $ such that $ T(v)\in W$.

In particular, is all this true if $V$ is not finite dimensional? I can't immediately think of a counterexample...

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up vote 3 down vote accepted

Hint: Given a linear isomorphism $T : V \rightarrow V'$ of finite-dimensional vector spaces, can you prove that the restricted linear transformation $T|_W : W \rightarrow T(W)$ is an isomorphism?

If $V' = V$ and $T(W) \subseteq W$ as in your problem, can you prove that $T(W) = W$? In that case, you'll have $T^{-1}(W) = W$, which may prove useful in showing that $\ker \overline{T} = \{0\}$.

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Wait, are you asking me to prove something that is given, or am I missing a subtlety to your answer? – AsinglePANCAKE Sep 24 '12 at 2:16
Oops, I edited the second paragraph to indicate what I meant. Both statements are supposed to be relatively straightforward to prove and help you toward proving what you want. – Michael Joyce Sep 24 '12 at 2:38
Ah, okay, so T restricted to W inherits the isomorphism. 1:1ness is easy, and the co-domain is by definition the image...So then in the case that T(W) is a subset of W, we must have that W is isomorphic to this subset, but then they have to be equal by their finiteness. So I'm guessing this breaks down for the infinite dimensional case...Any magical counterexamples you know of offhand? – AsinglePANCAKE Sep 24 '12 at 2:43
Consider an infinite vector space $V$ with basis $e_n$, $n \in \mathbb{Z}$. Define $T: V \rightarrow V$ by $T(e_i) = e_{i+1}$. Consider the subspace $W$ spanned by the $e_n$ with $n \geq 0$. Then $T$ is an isomorphism, $T(W) \subseteq W$, but $T|_W$ is not onto $W$. $T$ is a shift operator and it is a helpful example when trying to understand statements for infinite dimensional vector spaces. – Michael Joyce Sep 24 '12 at 3:20

When you restrict to $W$, you still get a linear transformation -- this much is clear. Since $T$ is injective, its restriction to $W$ is still injective. Therefore, the dimension of the image $T(W)$ is equal to the dimension of $W$. Since $T(W)\subseteq W$, it readily follows that we have an equality. Since $T|_W$ is both injective and surjective, it is an isomorphism.

A few things change when you go to the infinite-dimensional case. First of all, it may be the case that $T(W)\subseteq W$, without equality. For example, this could happen if $W$ is a closed subset of $V$, but $T(W)$ isn't. Also, if topology is involved, a linear isomorphism may not be the kind of isomorphism you are looking for -- you may want bicontinuity as well.

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