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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 $\dim(V) \geq \dim(W).$

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  • $\begingroup$ What is $k$? Is $k=\dim(V)$? Are $v_1,\ldots,v_k$ linearly independent? $\endgroup$ Commented Jun 14, 2018 at 16:28
  • $\begingroup$ no, you're supposed to use the rank nullity theorem, and in the solutions it is claimed that Range(L)=W which I have no idea how that impliciation is made $\endgroup$ Commented Jun 14, 2018 at 16:29
  • $\begingroup$ Essentially this says that if $L$ is surjective then the dimension of the domain is at least as big as the dimension of the image. $\endgroup$ Commented Jun 14, 2018 at 16:30

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

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  • $\begingroup$ If you write \operatorname{rank}L instead of \text{rank}L then you automatically get proper spacing in things like $A\operatorname{rank}B$ and $A\operatorname{rank}(B),$ and I include both of those examples to show the context-dependent nature of the spacing: you see less space to the right of $\operatorname{rank}$ in the second example than in the first. $\qquad$ $\endgroup$ Commented Jun 14, 2018 at 17:02
  • $\begingroup$ $\operatorname{rank}L=\dim W$. $\endgroup$
    – egreg
    Commented Jun 14, 2018 at 17:34
  • $\begingroup$ @egreg it seems that in case the vectors $v_i$ are not independent we might not have equality... $\endgroup$
    – user403337
    Commented Jun 14, 2018 at 17:42
  • $\begingroup$ @ChrisCuster So what? Can the rank of $L$ be greater than $\dim W$? $\endgroup$
    – egreg
    Commented Jun 14, 2018 at 17:43
  • $\begingroup$ @MichaelHardy thanks for the tip $\endgroup$
    – user403337
    Commented Jun 16, 2018 at 22:37
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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 $L$.

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  • $\begingroup$ can you expand on this please? $\endgroup$ Commented Jun 14, 2018 at 16:37
  • $\begingroup$ What does it mean for $L$ to be linear? Use this with the expression in the spoiler to try and write $w = L(v)$ for some $v$. $\endgroup$ Commented Jun 14, 2018 at 16:38
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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 $W$.

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