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

  • $\begingroup$ This works in infinite dimensions too. Are you assuming $V$ is finite-dimensional? If not, are you comfortable with the notion of codimension? $\endgroup$ – Theo Bendit Oct 22 '18 at 2:35
  • $\begingroup$ I am not assuming V finite dimensional, I have seen the definition of codimension as dim (V/W) $\endgroup$ – math.pr Oct 22 '18 at 2:46
  • $\begingroup$ 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. $\endgroup$ – Joppy Oct 22 '18 at 3:01

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