# Prove $\{f(v_{m+1}),\dots ,f(v_{n})\}$ is linearly independent

I was wondering if someone could give me a quick proof or counterexample to the following statement.

Let $f:V \rightarrow W$ be a linear map between finite dimensional vector spaces $V$ and $W$, both over a field $\mathbb{F}$. Let $\{v_{1},\dots ,v_{m}\}$ be a basis for $\mathrm{ker}(f)$. Extend to a basis $\{v_{1},\dots ,v_{m},v_{m+1},\dots ,v_{n}\}$ for V. Is the set $\{f(v_{m+1}),\dots ,f(v_{n})\}$ linearly independent?

I want to believe it's true but I'm not sure, so any help would be greatly appreciated.

• You can actually prove that $B:= \{f(v_{m+1}),\dots ,f(v_{n})\}$ is a basis of $\mathrm{image}(f)$. You would need to show that also $\mathrm{span}(B)=\mathrm{image}(f)$. Take $y \in \mathrm{image}(f)$. Then $y = f(x)$ for some $x \in V$. Now write $x$ in terms of the basis vectors and use linearity of $f$ to get $y = c_{k+1} v_{k+1} + \ldots + c_{n} v_{n}$. Therefore $y \in \mathrm{span}(B)$ and the result follows. – philmcole Mar 24 '18 at 16:38

Suppose $\alpha_{m+1}f(v_{m+1})+\dots+\alpha_nf(v_n)=0$. This can be rewritten as $$f(\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n)=0$$ that is, $$\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n\in\ker(f)$$ In particular, $$\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n= \beta_{1}v_1+\dots+\beta_mv_m$$ for some scalars $\beta_1,\dots,\beta_m$. Therefore $$-\beta_{1}v_1-\dots-\beta_mv_m +\alpha_{m+1}v_{m+1}+\dots+\alpha_nv_n=0$$ Can you finish?