Suppose V is a vector space, W is a subspace of V. Suppose a linear map T from V to V is injective. Suppose the image of W under T, T(W), is contained in W. My question is: is there an isomorphism between V/T(V) and W/T(W)?

If V/T(W) is of finite dimension, by a calculation for the dimensions of these two spaces, I can conclude that they are isomorphic. I would like to know if there is more elegant argument to show these two spaces are isomorphic, perhaps there is a canonical one between them in general?


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    $\begingroup$ Suppose $V = \ell^1 \oplus \ell^2$, $W = \ell^2$, and $T$ shifts the $\ell^1$ component but fixes $W$. Then $V/T(V) \cong \mathbb{C}$, but $W/T(W) \cong \{0\}$. $\endgroup$ – Daniel Fischer Mar 25 '14 at 16:46
  • $\begingroup$ You need to suppose $T$ bijective to make your finite dimensional case work out with this context. $\endgroup$ – Patrick Da Silva Mar 25 '14 at 16:55
  • $\begingroup$ @PatrickDaSilva: For linear maps I think injective implies bijective; if f is 1-1 ,kerT is trivial, and then by Rank-nullity, Dim(RankT)=n=DimV. $\endgroup$ – user99680 Mar 25 '14 at 18:01
  • $\begingroup$ @user99680 : My point was that in the finite dimensional case, injective and bijective are the same. So if he wants to work with a general vector space of non-necessarily finite dimension, then the correct assumption (as the counter examples have shown) is not injective, but rather bijective. In this case both spaces are zero though, so the isomorphism is quite canonical. I admit my first comment didn't realize the only example gave zero... -_- $\endgroup$ – Patrick Da Silva Mar 26 '14 at 1:42
  • $\begingroup$ @Patrick Da Silva: Ah, I see, I isunderstood/misunderestimated your point, sorry. $\endgroup$ – user99680 Mar 26 '14 at 2:00

No. For a counterexample, let $V = \bigoplus_{i \in \mathbb{Z}}\mathbb{R}$ be the direct sum of countably infinite copies of $\mathbb{R}$, Denote by $e_i$ the element with a $1$ in the $i^{\textrm{th}}$ place and zeros elsewhere, and let $W$ be the subspace generated by $e_0$. Let $T$ be defined by $T(e_i) = e_{2i}$. Then $T$ is injective on $V$ and is the identity on $W$, so $W/T(W)=0$, but $V/T(V)$ is isomorphic to $V$, since the image does not contain $e_i$ for any odd $i$.


Since $$V/T(V) = V/(W+T(V))\oplus (W+T(V))/T(V) = V/(W+T(V))\oplus {\color{blue} {W/(T(V)\cap W)}}$$ and $$W/T(W)={\color{blue}{W/(T(V)\cap W)}} \oplus (T(V)\cap W)/T(W).$$

Thus $$V/T(V)\cong W/T(W)$$ if (${\color{red}{\text{ BUT NOT ONLY IF }}}$) $$V/(W+T(V))\cong (T(V)\cap W)/T(W).$$

However, $T(V)/T(W)\cong V/W=V/(W+T(V))\oplus (W+T(V))/W= V/(W+T(V))\oplus T(V)/(W\cap T(V))$

and $(T(V)\cap W)/T(W)\oplus T(V)/(T(V)\cap W)=T(V)/T(W)$.

So if $V/W=T(V)/T(W)$ is of finite dimension, we do have $V/T(V)\cong W/T(W)$.

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    $\begingroup$ Oh man you should really format this better. This is painful to read. $\endgroup$ – Cameron Williams Mar 26 '14 at 2:05

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