Proving that if $L(\vec{v}_1),\ldots,L(\vec{v}_k)$ spans $W$ that $\dim(V) \geq \dim(W)$

Question

Let $\vec{v}_1,\ldots,\vec{v}_k$ be vectors in a vector space $V$ and $L\colon V \rightarrow W$ be a linear mapping. How would you prove that if $L(\vec{v}_1),\ldots,L(\vec{v}_k)$ spans $W$ that $\dim(V) \geq \dim(W).$

Answer 1

Based on your comments, I'll give hints on how to show $L$ is surjective (i.e. its range is $W$).

Suppose that $w \in W$. We need to show that $w$ is in the range of $L$. Because $\{L(v_1), L(v_2), \ldots, L(v_k)\}$ spans $W$, we have

$w = c_1L(v_1) + c_2L(v_2) + \cdots + c_kL(v_k)$ for some scalars $c_1,c_2,\ldots,c_k$.

Now use linearity to show that $w$ is in the image of $L$.

Answer 2

If $\dim V\lt\dim W$, then there aren't enough vectors to span $W$...

For $$\operatorname{rank}L+\operatorname{null} L=\dim V.$$

But $\operatorname{rank} L\ge\dim W$.

Answer 3

The map $L$ is surjective, so the rank-nullity theorem says $$ \dim V=\dim\ker L+\dim\operatorname{range}L=\dim\ker f+\dim W $$ Why is $L$ surjective? Because the span of $\{L(v_1),\dots,L(v_k)\}$ is contained in the image, for any set of vectors $\{v_1,\dots,v_k\}$. But, by assumption, the span of that particular set of vectors is $W$.