# Span of a linear mapping

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?

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

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.
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 ${\mathbb R}^m.$