For my linear algebra course at university I was given this task:
Let $f :\Bbb R^n \to \Bbb R^m$ be a linear map. Let further $A := (v_1, \ldots , v_k)$ be a family of linearly independent vectors in $\Bbb R^n$ and let $B := (f(v_1), \ldots, f(v_k))$ be their images. Also denote by $M \subset \Bbb R_n$ an arbitrary subset of $\Bbb R^n$ such that $\operatorname{Span}(M) = \Bbb R^n$. Show that
a) $f$ injective $\implies$ The family B is linearly independent,
b) $f$ surjective $\implies$ $\operatorname{Span}(f(M)) =\Bbb R^m$.
My problem is that I can't imagine a useful approach for this task and I would be very thankful for some hints how to start my proofs.
Thank you guys!
a) If there are $\alpha_1,\ldots,\alpha_k \in \mathbb{R}$ such that $$\sum\limits_{i=0}^k\alpha_if(v_i) = 0,$$ then, since $f$ is linear, $$f\left(\sum\limits_{i=0}^k\alpha_iv_i\right) = 0,$$ and since $f$ is injective, $$\sum\limits_{i=0}^k\alpha_iv_i = 0,$$ but the $v_i$ all lie in $A$, so are linearly independent, so no such $\alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $x\in\mathbb{R}^m\setminus \mathop{\mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y \in \mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $\mathbb{R}^n$, there are some $m_1,\ldots,m_k\in M$ and some $\alpha_1,\ldots,\alpha_k\in\mathbb{R}$ such that $$y = \sum\limits_{i=0}^k\alpha_im_i.$$ But then \begin{align*}x = f(y) &= f\left(\sum\limits_{i=0}^k\alpha_im_i\right) \\&=\sum\limits_{i=0}^k\alpha_if(m_i)&\mbox{since }f\mbox{ is linear} \\&\in\mathop{\mathrm{Span}}\{f(m_1),\ldots,f(m_k)\} \\&\subseteq\mathop{\mathrm{Span}}(f(M))&\mbox{since }m_i \in M\mbox{ for all }i.\end{align*} This contradicts our definition of $x$, so no such $x$ exists, so $\mathrm{Span}(f(M)) = \mathbb{R}^m$.