General method for finding the basis of the image of a linear transformation

Question

Let $f: V \rightarrow W$ be a linear transformation, and let $\mathfrak{B}_V=\{b_i\}$ be a basis of $V$. If the images of the elements of the basis $\mathfrak{B}_V=\{b_i\}$ are $\{f(b_i)\}$, then how would the base of the image of $f$ be found?

My attempt at a solution

I have tried to work out a general method this way: as $f$ is a linear transformation, then, $\forall x \in V$, $f(x)=f(\lambda_1b_1+\lambda_2b_2+...+\lambda_mb_m)=\lambda_1f(b_1)+\lambda_2f(b_2)+...+\lambda_mf(b_m)$. Therefore, the image of any element in $V$ can be expressed as a linear combination of the vectors $\{f(b_i)\}$. My doubt is this: will this set be linearly independent (i.e., a basis) or this is not necessarily the case (i.e., a generating set)?

Answer 1

This is not the case. For a simple example, take $V, W$ finite dimensional with positive dimensions and $f = 0$.

You can check that $f(\mathfrak{B}_V)$ is a basis of the image of $f$ if and only if $f$ is injective.

Edit: I just noticed that I didn't actually answer your question. To find a basis, just do row reduction on the matrix with the vectors of $f(\mathfrak{B}_V)$ as columns and remove linearly dependent vectors.

Edit 2: Changed $\mathfrak{B}_V$ to $f(\mathfrak{B}_V)$

Answer 2

Let the basis of $\mathrm{ker}(f)\subseteq V$ be $\{v_1,v_2\cdots,v_k\}$, where $k=\dim(\mathrm{ker}(f))$. Extend this basis of $\mathrm{ker}(f)$ into a basis of the entire space $V$, let it be $\{v_1,\cdots,v_k,v_{k+1},\cdots,v_n\}$, where $n=\mathrm{dim}(V)$. Then $\{f(v_{k+1}),\cdots,f(v_n)\}$ form a basis of $\mathrm{Im}(f)\subseteq W$.

This is because $\lambda_1 f(v_{k+1})+\cdots+\lambda_{n-k}f(v_n)=0\Rightarrow f(\lambda_1v_{k+1}+\cdots+\lambda_{n-k}v_n)=0$. As $v_{k+1},\cdots,v_n\not\in\mathrm{ker}(f)$, then this can hold only if $\lambda_1v_{k+1}+\cdots+\lambda_{n-k}v_n-0$, which implies $\lambda_i=0$ for all $i=1,\cdots,n-k$.

Answer 3

In general, to find a basis for a set of vectors, put them as the columns of a matrix, and then row-reduce. It's easy to read off a basis for the column space of the reduced matrix. Now the corresponding columns of the original matrix form the basis you are looking for.

If you go all the way to the RREF, then a basis will be formed by the pivot columns (the columns which contain leading $1$'s).

Now to answer the question in the title, note that you can take the basis $\beta=\{e_1,\dots,e_n\}$ to be the standard basis, in case $V$ is $n$-dimensional. You get a matrix whose columns are the $f(e_i)$. Now recall that the image of $f$ will be the column space. Follow the procedure outlined above, and you will have your basis.