Finding a basis for the image set of a linear transformation Given a linear transformation $t : V \rightarrow W$, my book suggests the following way to find a basis for $\text{Im } t$, given that we know the basis $\{ \mathbf{e_1}, \ldots, \mathbf{e_n} \}$ for the domain $V$:


*

*Find $S=\{ t(\mathbf{e_1}), \ldots, t(\mathbf{e_n}) \}$.

*If there is a vector $\mathbf{v}$ in $S=\{ t(\mathbf{e_1}), \ldots, t(\mathbf{e_n}) \}$ that is a linear combination of the other vectors in $S$, discard $\mathbf{v}$.

*Continue step 2 until we reach a linearly independent set $S$. This is a basis for $\text{Im } t$.



*

*Why does this method actually work?

*Is there a shortcut in step 2? Usually I try to solve the equation $\alpha_1 t(\mathbf{e_1}) + \cdots + \alpha_n t(\mathbf{e_n}) = \mathbf{0}$. Sometimes I can simplify the previous equation to a system of equations that has more unknowns than equations, thus having infinitely many solutions - if that's the case, can I then discard any $\mathbf{v}$ instead of having to figure out which $\mathbf{v}$ is a multiple of the other vectors in $S$ (which I find difficult and time consuming.)

 A: To begin with, notice that $S = \{T(e_{1}),T(e_{2}),\ldots,T(e_{n})\}$ spans $\text{Im}(T)$.
Here I assume that $\dim V = n$ and $\dim W = m$.
Indeed, if $w\in\text{Im}(T)$, then there exists $v \in V$ such that $w = T(v)$. This means that:
\begin{align*}
w = T(v) & = T(a_{1}e_{1} + a_{2}e_{2} + \ldots + a_{n}e_{n}) = a_{1}T(e_{1}) + a_{2}T(e_{2}) + \ldots + a_{n}T(e_{n})\in\text{span}(S)
\end{align*}
The second step is necessary to eliminate redundant directions.
Once you have eliminated them, what remains is a set of LI vectors which spans $\text{Im}(T)$.
Hence it is a base.
Hopefully this helps!
A: If you put your basis candidates $t(e_i)$ into the columns of a matrix you can use a standard Gaussian elimination to find the linear relations between the columns, since the transformation
$$Ax=0\iff A^{'}x=0$$
keeps the linear relations between the columns $A_1,\dots,A_n$ and  $A^{'}_1,\dots,A^{'}_n$ intact, like this
$$x_1A_1+\dots+x_nA_n=0\iff x_1A^{'}_1+\dots+x_nA^{'}_n=0$$
Let us try a random example with $t(e_1)=(1, 3, 2)$, $t(e_2)=(4, 1, 0)$ and $t(e_3)=(-5, 7, 6)$. The matrix is
$$A=\left(
\begin{array}{ccc}
 1 & 4 & -5 \\
 3 & 1 & 7 \\
 2 & 0 & 6 \\
\end{array}
\right)
$$
And the row reduced matrix $A^{'}$ is
$$A^{'}=\left(
\begin{array}{ccc}
 1 & 0 & 3 \\
 0 & 1 & -2 \\
 0 & 0 & 0 \\
\end{array}
\right)
$$
The first thing we notice is that $\mathrm{rank}(A^{'})=\mathrm{rank}(A)=2$ which means the dimension of the column space is $2$. So we are looking for a basis of two linearly independent vectors.
Looking more closely we see that the first column $\left(
\begin{array}{c}
 1  \\
 0  \\
 0  \\
\end{array}
\right)$ and the second column $\left(
\begin{array}{c}
 0  \\
 1  \\
 0  \\
\end{array}\right)$ are linearly independent. There is no way to add them together and get $\mathbf{0}$ unless we take a zero amount of each vector.
This means that the first two columns of the original matrix $A$ must also be linearly independent (because the linear relationships between the columns of $A$ and $A^{'}$ are conserved). In other words $[t(e_1), t(e_2)]$ is a basis. More generally, each column with a pivot element present is linearly independent.
Since the matrix was transformed to reduced row echelon form it is also easy to express the last remaining column as a linear combination of the first two columns. From the third column $\left(
\begin{array}{c}
 3  \\
 -2  \\
 0  \\
\end{array}
\right)$ we see that we need to take 3 of the first column and -2 of the second column to get the third column. This must also be true for the original matrix $A$. Hence $t(e_3)=3t(e_1)-2t(e_2)$ (the linear relations are intact).
This approach is definitely easier when you have access to a tool (like a pocket calculator) that does the Gaussian elimination for you. It might not be a shortcut if you have to do the elimination by hand.
