# If $f:V\to{W}$ and the images of a basis of $V$ span $W$, then the images of every basis are generators in $W$

We have a Linear map $f\colon V\to W$, it is given to us that $V$ is finitely dimensional and the basis is $\{b_1,\dots,b_n\}$. Prove that if $\{f(b_1),\dots,f(b_n)\}$ forms a system of generators for $W$ then every basis $B'$in $V$ is mapped by $f$ to a set of generators for $W$.

My thinking is that since $\{f(b_1),\dots,f(b_n)\}$ is a system of generators of $W$ then the mapping is surjective. It is injective also because the basis and system of generators have the same number of vectors. So it is a bijective mapping. So if we have a dual basis in $V$ $\{b_1',\dots,b_n'\}$ and the mapping is bijective then the basis then $\{f(b_1),\dots,f(b_n)\}$ is a system of generators for $W$

• What do you mean by "dual basis"? (I edited your title since it didn't make sense as it was) – Silvia Ghinassi Dec 1 '15 at 3:57
• en.wikipedia.org/wiki/Dual_space#Finite-dimensional_case – babylon Dec 1 '15 at 3:59
• the finite dimensional case it talks about V* being a dual basis? Thank you for the title – babylon Dec 1 '15 at 4:00
• I know what a dual space is. But the "dual basis" is a basis for $V^*$ (which, yes, it is isomorphic to $V$) which does not appear in your question. – Silvia Ghinassi Dec 1 '15 at 4:01
• The map $f$ is not necessarily injective. The vectors $f(b_i)$ span $W$; nothing guarantees they form a basis of $W$. As an example, consider $f:\mathbf{R}^3\to \mathbf{R}^2$, $(x_1,x_2,x_3)\mapsto (x_1+x_3,x_2-x_3)$. So $(1,0,0)\mapsto (1,0)$, $(0,1,0)\mapsto (0,1)$, and $(0,0,1)\mapsto (1,-1)$. The three vectors span $\mathbf{R}^2$ but are not a basis. – Frédéric Dec 1 '15 at 4:11

1. The vectors $b_i$ and $b_i'$ both form bases of $V$. This means you can write each $b_i$ as a linear combination of the vectors $b_i'$ (and vice versa).
2. The vectors $f(b_i)$ span $W$: any vector $w\in W$ can be written as a linear combination of the vectors $f(b_i)$.
3. Can you write any $w\in W$ as a combination of the vectors $f(b_i')$? Yes. First write $w$ as a combination of the vectors $f(b_i)$. Then write each $b_i$ as combinations of the vectors $b_i'$. By the linearity of $f$ this means that each $f(b_i)$ is written as a combination of the $f(b_i')$ and so ...