In one of the exercises in my textbook we had to prove the following:

Consider $3$ $K$-vectorspaces $U$, $V$ and $W$ and $2$ maps:

  • $f\in\text{Hom}_K(U,V)$
  • $g\in\text{Hom}_K(V,W)$

then : \begin{equation} \dim(\text{im}(f)\cap\ker(g)) = \dim(\text{im}(f))-\dim(\text{im}(gf)) \end{equation}

This proof wasn't addressed during the lectures so I tried to understand it at home but I don't understand most of the steps. The proof is as follows:

We consider the map $g|_{f(U)}:f(U)\rightarrow W$ which is the restriction of $g$ to the image $\text{im}(f)\leqslant V$.

\begin{equation} \begin{split} \dim(\text{im}(f)) & = \dim(\ker(g|_{f(U)})) + \dim(\text{im}(g|_{f(U)})) \\ & = \dim(\{v\in\text{im}(f)\;|\;g(v) = 0\}) + \dim(\{g(v)\;|\; v\in\text{im}(f)\}) \\ & = \dim(\text{im}(f)\cap\ker(g)) + \dim(\{(gf)(w)\;|\;w\in U\}) \\ & = \dim(\text{im}(f)\cap\ker(g)) + \dim(\text{im}(gf)) \end{split} \end{equation}

And then the formula follows from this.

My guess is that in the firs step they use the following:

\begin{equation} \dim(V) = \dim(\ker(f))+\dim(\text{im}(f)) \end{equation}

But I don't quite get why they use $g|_{f(U)}$ as the function in this equation and where it comes from. I also need some help understanding the transition between step 1 and 2 and between 2 and 3. The last step I do understand. Would someone be able to explain this?

  • $\begingroup$ yes, the definition of kernel is: $\ker(f) := \{v\in V\;|\;f(v) = 0\}$ so you could say that after the function the kernel is all the vectors that do not have an image so their image is the nullspace, and the image is: $\text{im}(f) := \{f(v)\;|\; v\in V\}$ if $f:V\rightarrow W$ is a linear map between $2$ $K$-vectorspaces so i guess the image is what isn't in the kernel en vice versa. $\endgroup$ – Viktor Jan 12 at 15:31

Before the first step in your proof, the well known consequence is used:

$$ (*) \space dim(V) = dim(Ker(g)) + dim(Im(g)) $$

Notice that we have used $g$ above, and $V$ is nothing but its domain. Notice also that we introduced a trick by showing $Im(f) \subseteq Dom(g) = V$ therefore, if we study only in the restriction $g\mid _{f(U)=Im(f)}$ we can inject it in $(*)$ to obtain the first line in your argument.


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