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Given a linear mapping $T:V\rightarrow W$ between the two vector spaces, $\operatorname{Im}(T)$ and $V/\operatorname{ker}(T)$ are isomorphic. I wonder if $\operatorname{Ker}(T)$ and $W/\operatorname{Im}(T)$ are also isomorphic?

If yes, then we also have $\dim{(W)}=\dim{(\operatorname{Ker}(T))} + \dim{(\operatorname{Im}(T))}$?

Thanks!

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This is not true. Consider the map $T: \mathbf{R} \rightarrow \mathbf{R}^3$ given by $T(x)=(x,0,0).$ –  ShawnD Aug 18 '11 at 22:34
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This cannot be true since one can replace W by a larger vector space and keep the same linear mapping. Then Ker(T) stays the same but W/Im(T) changes. –  Did Aug 18 '11 at 22:35
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$W/\text{im}(T)$ is known as the cokernel. The linear map fits in an exact sequence $0 \to \ker{(T)} \to V \to W \to \text{cokernel}(T) \to 0$ –  Juan S Aug 19 '11 at 1:31
    
Take $0\to K$ or $K\to0$, where $K$ is the base field. In general, you have an exact sequence $$0\to\mathrm{Ker}\to V\to W\to\mathrm{Coker}\to0,$$ and the alternate sum of the dimensions is $0$. –  Pierre-Yves Gaillard Aug 19 '11 at 2:44

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up vote 3 down vote accepted

Another way to see that this cannot be true is to use the correct rank-nullity theorem, i.e., $\dim{(V)} = \dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})}$. If we assume $\dim{(W)}=\dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})}$, then we must have $\dim{(W)} = \dim{(V)}$!

Now, if we assume that $V$ and $W$ are vector spaces of the same dimension, and $T:V\to W$ is a linear map, then $V/\operatorname{Ker}{(T)}$ and $\operatorname{Im}{(T)}$ are canonically isomorphic via $[v] \to T(v)$, where $[v]$ is the class of $v\in V$ in $V/\operatorname{Ker}{(T)}$.

Since $$\dim{(W)} = \dim{(\operatorname{Im}{(T)})} + \dim{(W/\operatorname{Im}{(T)})}$$ and by rank-nullity $$\dim{(W)}=\dim{(V)} = \dim{(\operatorname{Ker}{(T)})} + \dim{(\operatorname{Im}{(T)})},$$ we may subtract both equations and obtain $\dim{(W/\operatorname{Im}{(T)})}=\dim{(\operatorname{Ker}{(T)})}$. Hence, $W/\operatorname{Im}{(T)}$ and $\operatorname{Ker}{(T)}$ are vector spaces of the same dimension, therefore isomorphic... but the isomorphism is not canonical! They are isomorphic simply because their dimensions coincide.

Although the term "canonical" does not have a very precise meaning in general, in this entry by canonical I mean "natural", or "coordinate-free". For instance, if $T:V\to W$ is any linear map of $\mathbb{F}$-vector spaces, then the isomorphism $\phi:V/\operatorname{Ker}{(T)}\to \operatorname{Im}{(T)}$ is canonical because I can describe it without having to choose bases for the vector spaces involved. The map is simply $\phi([v])= T(v)$, for any $v\in V$. However, as we have shown above, if $V$ and $W$ have the same dimension, we also have an isomorphism $\psi: W/\operatorname{Im}{(T)} \to \operatorname{Ker}{(T)}$, but we cannot describe this isomorphism unless I define a basis $\{w_1,\ldots,w_n\}$ of $W/\operatorname{Im}{(T)}$ and a basis $\{v_1,\ldots, v_n\}$ of $\operatorname{Ker}{(T)}$, and then, once in coordinates, we choose an arbitrary isomorphism $\mathbb{F}^n \to \mathbb{F}^n$.

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Thanks! What do "canonical" and"non-canonical"mean in this case and in its most generaility (I have seen them many times)? –  Tim Aug 18 '11 at 23:59
    
Good question! Perhaps the Wikipedia entry for canonical may help here: en.wikipedia.org/wiki/Canonical#Mathematics -- I will also add something to my entry... –  Álvaro Lozano-Robledo Aug 19 '11 at 0:07

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