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Let $ L : \mathbb{R}^n \to \mathbb{R}^m$ be a linear mapping such that $\text{rank}(L) = m$. If $\{v_1, \dots , v_k\}$ spans $\mathbb{R}^n $, then $\{L(v_1), \dots, L(v_k)\}$ spans $\mathbb{R}^m$.

I have to either prove or disprove this statement and I'm not really sure how. Does the $\text{rank}(L) = m$ refer to the standard matrix of $L$? Any hits on how to solve this?

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    $\begingroup$ What's your definition of rank? $\endgroup$ Commented Jun 10, 2015 at 3:36
  • $\begingroup$ the number of columns with leading ones in the matrix $\endgroup$
    – Filip
    Commented Jun 10, 2015 at 4:09
  • $\begingroup$ @Filip that's a very poor and not all-inclusive definition. Look up another definition online that involves linear independence and go from there. $\endgroup$ Commented Jun 10, 2015 at 4:24
  • $\begingroup$ I guess you mean the number of such columns after row reduction $\endgroup$ Commented Jun 10, 2015 at 13:18
  • $\begingroup$ Yes, the rank of a linear mapping is the same as the rank of its standard matrix. $\endgroup$ Commented Jun 10, 2015 at 13:35

2 Answers 2

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You know that if $v_1, \dots, v_k$ span $\mathbb{R}^n$ then we can pick a subset of $n$ of them that are linearly independent, and in particular form a basis for $\mathbb{R}^n$. We know that the rank of $\mathcal{L}$ is $m$. Maybe you have seen that $rk(\mathcal{L}) = dim(im(\mathcal{L}))$? I suggest you look at what could happen to the linearly independent vectors we picked, or perhaps the form of the matrix of $\mathcal{L}$ in this basis.

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if $\{v_1, \dots , v_k\}$ spans ${\mathbb R}^n$ and $L$ linear implies that $B = \{Lv_1, Lv_2, \cdots, Lv_k\}$ spans the image/range of $L.$ now the $rank(L) = \text{ dimesion of } range(L) = m$ shows that $m \le k$ and $B$ spans ${\mathbb R}^m.$

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