I was solving exercises and I found the following problem: Suppose you have $V$ an $\mathbb{R}$-vector space (with finite dimension) and $f_1,f_2,f_3$ belonging to the dual space of $V$ $(V^*)$. Let $T$ be the linear transformation
$$T: V \rightarrow \mathbb{R}^3$$ given by
$$ T(v ) = (f_1(v),f_2(v),f_3(v)) $$ ($v$ belonging to $V$) Prove that $T$ is surjective if and only if ${f_1,f_2,f_3}$ are linearly independent as vectors in $ V^* $.
My first attempt was to find a relation between being surjective (ephimorphism) and injective (monomorphism),
since I remember that they are equivalent if a condition related to the dimension is met, but I am not sure if this is the correct way.
On the other hand, I was looking to write $f_1,f_2,f_3$ as a linear combination and try to show that it is equal to $0$ only if the coefficients are $0$, but I can't see how the surjectivity of $T$ can help me prove it.
$$ af_1(v) + bf_2(v) + cf_3(v) = 0 $$
In general, I can see that if $f_1,f_2,f_3$ were linearly dependent, they couldn't generate all of $\mathbb{R}^3$ so $T$ wouldn't be surjective, but I can't find the right way to prove it.
I'd appreciate some hints.
Best Answer
If $ a.f_1(v) + b.f_2(v) + c.f_3(v) = 0 $ with at least one of $a,b,c$ not zero then $T$ cannot have $a,b,c$ in its range. This is bacause there would then be a $v$ such that $f_1(v)=1,f_2(v)=b$ and $f_3(v)=c$ which gives $a^{2}+b^{2}+c^{2}=0$ which implies $a=b=c=0$.
Conversely, suppose $f_i$'s are independent. If $T$ is not surjective, then its range would have dimension less than $3$. This implies that there is a non-zero vector $(a,b,c)$ orthogonal to the range. This gives $ a.f_1(v) + b.f_2(v) + c.f_3(v) = 0 $ for all $v$ contradicting linear independence.