How do you find the basis of the image of a linear transformation $T$, given the transformation of the basis?

Question

Let $T: V \rightarrow W$, Given a basis $(b_1, b_2...)$ of $V$ can one basis of $\operatorname{Im}T$ be formed by taking a subset of $T(b_1), T(b_2)$, etc.? Is there a proof to this or a theorem that it's called?

Answer 1

Yes, a basis for $\operatorname{Im} T$ can be obtained as a subset of $\{Tb_1, Tb_2, \ldots\}$.

Namely, notice that $\operatorname{span}\{Tb_1, Tb_2, \ldots\} = \operatorname{Im} T$.

It is clear that $Tb_i \in \operatorname{Im} T$.

Conversely, take $y \in \operatorname{Im} T$. Hence, there exists $x \in V$ such that $Tx = y$.

Write $x$ in the basis $\{b_1, b_2, \ldots\}$:

$$x = \sum_{i} \alpha_i b_i$$

Therefore:

$$y = Ax = A\left(\sum_{i} \alpha_i b_i\right) = \sum_i \alpha_i Tb_i \in \operatorname{span}\{Tb_1, Tb_2, \ldots\}$$

Note that $\{Tb_1, Tb_2, \ldots\}$ does not have to be a basis, we only know that it can be reduced to a linearly independent subset which is then a basis for $\operatorname{Im} T$.

Answer 2

Definitely not true that the image is a basis in general (think of any matrix from $\mathbb{R}^n\to\mathbb{R}^m$ where $n>m$. There are simply too many basis vectors in the first space to be linearly independent in the second space.

Consider $T$ a linear operator and assume we have that $T(x)=0\iff x=0$. Then, consider,

$$T(\beta_1b_1)+\ldots+T(\beta_nb_n)=0$$

by linearity we have,

$$T(\beta_1b_1+\ldots+\beta_nb_n)=0,\\ \implies \beta_1b_1+\ldots+\beta_nb_n=0$$ and since $\{b_1,\ldots,b_n\}$ is a linearly independent set, we have that $\beta_1,\ldots,\beta_n=0$