The information the kernel provides about a linear map In a linear map, if we know the mapping of the basis, we know all the information about the map. On the other hand, if we know the $\ker(f)$, $X/\ker(f)$ is isomorphic to $\mathrm{img}(f)$. I am wondering does the $\ker(f)$ determines the whole structure of the image, i.e. like the map on the basis, what kind of information does the kernel provide for the map?
For example, if $\{e_1,\cdots,e_m\}$ span the kernel space, then we can extend it to the basis of the whole space $e_1,\cdots,e_m,e_{m+1},\cdots,e_n$. Then for any $x\in X$, $f(x)$ is just the projection onto the space spanned by $\{f(e_{m+1}),\cdots,f(e_n)\}$
 A: If you take a subspace $V$ of a finite dimensional $K$-vectorial space $E$ and you take a basis of $E$ $(e_1,...,e_n)$ such that $(e_1,...,e_k)$ is a basis of $V$. I will identify any linear endomorphism of $E$ with matrices $M_n(K)$ using the basis $(e_1,...,e_n)$. Now if $A\in M_n(K)$ we know that we can decompose $A$ with respect to the decomposition :
$$E=V\oplus Vect(e_{k+1},...,e_n) $$
This will give :
$$A=\begin{pmatrix}A_{1,1}&A_{1,2}\\A_{2,1}& A_{2,2}\end{pmatrix} $$
Now basic linear algebra gives you $Ker(A)=V\Leftrightarrow A_{1,1}=0$ and $A_{2,1}=0$ and $A_{2,2}\in GL_{n-k}(K)$. Remark that $A_{2,2}$ is the matrix of the isomorphism between $E/Ker(f)$ and $Im(f)$ induced by $A$. 
Hence once the kernel of $A$ is prescribed to be $V$ we know that $A$ will be written (in the good base) :
$$A=\begin{pmatrix}0&A_{2,1}\\0& A_{2,2}\end{pmatrix} \text{ with } A_{2,2}\in GL_{n-k}(K)\text{ and } A_{2,1}\in M_{k,n-k}(K)$$
In particular, even if you prescribe $A_{2,2}$ there is some choices left (to be precise as many choices as in $M_{k,n-k}(K)$).
