Determine the cokernel of a linear transformation between $\mathbb Q$ vector spaces Let $f:E\longmapsto V$ be a linear map between finite dimensional $\mathbb Q$-vector spaces with bases $\{e_1,\cdots,e_n\}$ and $\{v_1,\cdots,v_m\}$ 
Define $coker(f)$ to be the quotient vector space $V/Im(f)$. This is a $\mathbb Q$-vector space of dimension $m-rank(f)$. Knowing that a basis for $Im(f)$ is already determined and it is composed of the $rank(f)$ vectors among the $n$ vectors $\{f(e_1), \cdots,f(e_n)\}$ that are linearly independant,  I want to give an explicit basis for $coker(f)$, possibly in terms of the basis of $V$ and the basis of $Im(f)$. thank you for your help!
 A: This is an application of the exchange lemma: if $X$ is a linearly independent set in $V$ and $B$ is a basis of $V$ (actually it's sufficient that $B$ is a spanning set is), then we can substitute $|X|$ elements of $B$ with the elements of $X$ and the resulting set is still a basis.
So, if $X=\{y_1,\dots,y_k\}$ is a basis of $\mathrm{Im}(f)$ (determined as you outline), you can find $m-k$ elements among $\{v_1,\dots,v_m\}$ such that
$\{y_1,\dots,y_k,v_{i_{1}},\dots,v_{i_{m-k}}\}$ is a basis of $V$. Set $w_j=v_{i_j}$ $(i=1,\dots,m-k)$ to keep notation simple.
You need only to show that $\{\pi(w_1),\dots,\pi(w_{m-k})\}$ is linearly independent, where $\pi\colon V\to V/\mathrm{Im}(f)$ is the canonical projection, as it is clearly a spanning set. Saying that
$$\alpha_1\pi(w_1)+\dots+\alpha_{m-k}\pi(w_{m-k})=0$$
means that
$$\alpha_1w_1+\dots+\alpha_{m-k}w_{m-k}\in\mathrm{Im}(f)$$
so
$$\alpha_1w_1+\dots+\alpha_{m-k}w_{m-k}=\beta_1y_1+\dots+\beta_ky_k$$
which is only possible if all the coefficients are zero.
Without a knowledge of $f$ it's impossible to say more than this, so this is as much “explicit” as you can get.

Now, assume we know the map $f$ and that we are able to compute the coordinates with respect to a basis of $V$ (it can be the given one or any other, for instance the canonical basis if $V=\mathbb{Q}^m$). So, if $x_1,\dots,x_n$ are the coordinate vectors of $f(e_1),\dots,f(e_n)$ and $y_1,\dots,y_m$ are the coordinate vectors of $v_1,\dots,v_m$, you can build the matrix
$$[x_1\ x_2\ \dots\ x_n\ |\ y_1\ y_2\ \dots\ y_m]$$
and proceed to find its reduced echelon form. The bar is just to separate the two sets of vectors.
At the left of the bar you'll find $k=\mathrm{rank}(f)$ dominant columns, at the right $m-k$ dominant columns. The $k$ columns corresponding to the dominant columns at the left are a basis for $\mathrm{Im}(f)$, the canonical projections of the $m-k$ at the right give a basis for $\mathrm{coker}(f)$. This is just like using the exchange lemma.
A: Here's one way to find a basis.  Extend the basis of Im$(f)$ to a basis of $V$ (one can always extend a basis of a subspace to a basis of the entire space).  Then the images of the extra basis elements (i.e. the basis elements in your basis of $V$ that are not in the basis of Im$(f)$) in coker$(f)$ give a basis for coker$(f)$.
EDIT: If you want your basis of the cokernel to be in terms of your fixed basis of $V$, then you can use the elements $\{v_1,\dots,v_n\}$ to extend your basis of Im$(f)$ to a basis of $V$.  I.e. add an appropriate subset of $\{v_1,\dots,v_n\}$ to your basis of Im$(f)$ to get a basis of $V$ (add a $v_i$ that is not in Im$(f)$, then add a $v_j$ that is not in Im$(f)$ + Span $\{v_i\}$, etc. until you have a basis of $V$).
