Existence of $v$ such that $f_i(v) = \delta_{i j}$

Let $$f_1,...,f_m \in V^*$$ be a linearly independent family ,and F a field. Show that for each $$1\leq j\leq m$$ there is a $$v\in V$$ such that $$f_i (v)= \delta_{i j}$$ for any $$1\leq j\leq m$$

I thought about defining a linear map $$T: V\to F^m$$ such that $$T(v) = (f_1(v),...,f_m(v))$$ and proving that T is surjective using the fact that T is surjective $$\iff Ker(T^t) = 0$$ where $$T^t$$ is the transpose of T. Is it the right way to do it?

Ps: The exercise doesn't say anything about the dimension of V

• This works in infinite dimensions too. Are you assuming $V$ is finite-dimensional? If not, are you comfortable with the notion of codimension? – Theo Bendit Oct 22 '18 at 2:35
• I am not assuming V finite dimensional, I have seen the definition of codimension as dim (V/W) – math.pr Oct 22 '18 at 2:46
• The obvious problem is that you have to be kind of careful about defining the transpose of a linear operator between infinite-dimensional vector spaces. – Joppy Oct 22 '18 at 3:01