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I would like to show that if $V$ is finite dimensional and $T$ is an isomorphism, $\overline{T}$ is an isomorphism, and if $V$ is infinite dimensional, $\overline{T}$ may not be an isomorphism. I think I've shown that $\overline{T}$ is an isomorphism, but have not used the finite dimensional assumption, so I've been sloppy somewhere - I cannot identify where.

Setup: $W\subset V$ are vector spaces over field $F$, $T:V\to V$ a linear transformation so that $T(W)\subset W$. Let $\overline{T}:V/W\to V/W$ be given by $\overline{T}(v+W)=T(v)+W$, where $V/W=\{v+W:v\in V\}$.

I have verified that $\overline{T}$ is a well-defined linear transformation on $V/W$, so I omit this part.

My argument for showing $\overline{T}$ is an isomorphism, which fails to use the finite dimension of $V$:

Pick $v\in V$; then there exists $u\in V$ so that $T(u)=v$ since $T$ is onto. So $\overline{T}(u+W)=T(u)+W=v+W$. $v+w$ is an arbitrary element of $V/W$ and $u+W\in V/W$, so $\overline{T}$ is onto.

Note that the zero element of $V/W$ is $W$. $\overline{T}(u+W)=T(u)+W$ iff $u=0_V$, the zero element of $V$, since $T$ is 1-1. So $\ker \overline{T}=W$, and $\overline{T}$ is 1-1.

I would greatly appreciate any guidance! Thanks in advance.

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  • $\begingroup$ Why do you think that $\;\overline T(u+W):=Tu+W=W\iff u\in W\;$ , which is what you need in order to get $\;\overline T\;$ is injective? This isn't necessarily true... or else prove it . $\endgroup$ – DonAntonio Oct 1 '16 at 22:06
  • $\begingroup$ Sorry I forgot to add this condition of the problem statement. Thanks for bringing this up. I'll add it to the original question. We are given that $T(W)\subset W$. $\endgroup$ – manofbear Oct 1 '16 at 22:07
  • $\begingroup$ Of course we're given that, otherwise $\;\overline T\;$ couldn't be well-defined. The problem still remains, though.... and here precisely is where the finite dimensionality kicks in! $\endgroup$ – DonAntonio Oct 1 '16 at 22:08
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    $\begingroup$ Of course you did (and you lost me with your other comment): when you (try to) say that $\;\overline T(u+W)=Tu+W=W\implies u=0_V\;$ you're doing that...and even this is wrong. What is true in finite dimension is that $\;Tu+W=W\iff u\in W\;$ with your other conditions. In infinite dimension even this last thing is false. $\endgroup$ – DonAntonio Oct 1 '16 at 22:13
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    $\begingroup$ This is an excellent question, and the answers have been really nice. $\endgroup$ – Andres Mejia Oct 2 '16 at 9:39
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I'll try to add my view on this.

First, $\;T:V\to T\;$ is an isomorphism, and $\;W\le V\;$ being $\;T\,-$ invariant means $\;TW\subset W\;$ ,and because of both sides have the same finite dimension, this means $\;TW=W\;$ ...and this is the crucial point: if for some $\;v\in V\;$ we have that $\;Tv=w\in W\;$ , then since $\;TW=W\;$ and also $\;T|_W: W\to W\;$ is an ismomorphism (why?), then there exists

$$w\in W\;\;s.t.\;\;Tw'=w=Tv\implies v=w'\in W\;,\;\;\text{by injectivity}$$

and thus, as commented:

$$\overline T(u+W):=Tu+W=W\iff Tu\in W\implies u\in W\implies u+W=W$$

and $\;\overline T\;$is injective.

Counter example in the infinite dimensional case: For example, take the real linear space, with the usual termwise operations

$$V=\left\{\;\{a_n\}_{n\in\Bbb N}\subset\Bbb R\;\right\}$$

and define $\;T:V\to V\;$ by

$$\;T\{a_1,a_2,a_3,...\}:=\{a_1,a_1-a_2,a_2-a_3,a_3-a_4,...,a_n-a_{n+1},...\}\;$$

I'll leave it to you to check this is an isomorphism, and

$$W:=\left\{\;\{a_n\}\in V\;/\;\lim_{n\to0}a_n=0\right\}\;\;\text{is a $\;T\,-$ invariant subspace}$$

Observe now that

$$T\{1,1,1,....,...\}=\{\;1,0,0,0,...\;\}\xrightarrow[n\to\infty]{}0$$

and thus $\;T\{1,1,...\}\in W\;$ eventhough $\;\{1,1,...\}\notin W\;$ . Try to finish now from here your counter example.

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  • $\begingroup$ I was able to work through the details to get the counterexample. Thanks a bunch for the detailed response! $\endgroup$ – manofbear Oct 2 '16 at 18:41
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You have the wrong condition for injectivity : $\overline{T}(u+W)=T(u)+W=W$ iff $T(u)\in W$ (and not $T(u)=0$). So you get that $\overline{T}$ is injective iff $T(u)\in W$ implies $u\in W$. This is true if $W$ is finite-dimensional ; indeed, if $T(u)\in W$, then $T(W+\langle u\rangle)\subset W$, and since $T$ is injective $$\dim(W+\langle u\rangle) \leq \dim W,$$ which implies that $u\in W$.

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  • $\begingroup$ I am not the OP, but I just found this question somewhat interesting, but is the last fact hinged on the subjectivity of $T$? So that $T(W) \subset W$ in the invariance? $\endgroup$ – Hawk Oct 1 '16 at 22:20
  • $\begingroup$ I didn't use the surjectivity of $T$ in my answer (but of course we need it to prove that $\overline {T}$ is surjective!) $\endgroup$ – Arnaud D. Oct 1 '16 at 22:33

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