Given a basis $X$ for a finite dimentional vector space $V$, prove that $L(X)$ spans $L(V)$.

Question

Let $L : V \to W$ be a linear map, and suppose $X := \{v_{1},\ldots, v_{n}\}$ is a basis for $V$.

Prove that $Y := \{L(v_{1}),\ldots, L(v_{n})\}$ is a spanning set for $L(V)$.

So I am a struggling on this question. I was told that the answer is it cannot be proven.

So the reasoning I gave is that "the basis $X$ is not the entire preimage of the transformation, so it can not reliably be transformed to span the image of $L$".

But I am not confident in my understanding of the questions and concepts involved.

So any help would be appreciated.

Answer 1

Actually, the statement is true: if $w\in\operatorname{im}(L)$, then $w=L(v)$, for some $v\in V$. And $v$ can be written as $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n$, for some scalars $\alpha_1,\alpha_2,\ldots,\alpha_n$. But then$$w=\alpha_1L(v_1)+\alpha_2L(v_2)+\cdots+\alpha_nL(v_n)\in\operatorname{span}\bigl(\{L(v_1),L(v_2),\cdots,L(v_n)\}\bigr).$$

Answer 2

Let $w\in\operatorname{img}(L)$, then there exists a $v\in V$ with $L(v)=w$. Since $X=\{v_1,\ldots,v_n\}$ is a basis of $V$, you find scalars $\lambda_1,\ldots,\lambda_n\in\mathbb{R}$, so that: \begin{equation} v=\sum_{i=1}^n\lambda_iv_i. \end{equation} Using linearity of $L$, we have: \begin{equation} w=L(v) =L\left(\sum_{i=1}^n\lambda_iv_i\right) =\sum_{i=1}^n\lambda_iL(v_i). \end{equation} Therefore $Y=\{L(v_1),\ldots,L(v_n)\}$ is a spanning set of $\operatorname{img}(f)$. Notice that it does not have to be a basis. Take $L\colon\mathbb{R}^2\rightarrow\mathbb{R},\begin{pmatrix} x \\ y \end{pmatrix}\mapsto x$, then the basis vectors $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ of $\mathbb{R}^2$ are both mapped to $1$.