2

Let $V$ be a vector space over field $K$, $V^*$ is a dual vector space, i.e., the set of all linear transformations from $V$ to $K$.

Let $(x_1, x_2,...,x_n)$ be a basis of $V$. I know the fact that the set of linear transformations $(x^*_1,x^*_2,...,x^*_n)$ forms a basis of $V^*$, it's easy and straightforward.

Conversely, given a basis say $(f_1,f_2,...,f_n)$ of $V^*$, I expect to prove that there exists a basis of $V$ say $(y_1, y_2,...,y_n)$ such that $f_i = y^*_i$ for all $i$.

Please help me.

user26857
  • 52,094
Anh_Rose 1210
  • 634
  • Take $y_i = f_i^\ast$. http://math.stackexchange.com/questions/292353/v-is-isomorphic-to-v-ast-ast-the-double-dual-space-of-v – Derek Allums Aug 17 '16 at 19:10
  • I dont understand your idea: $y_i$ is a element of $V$ while $f^_i$ is a map of $V^$ to $K$. Please be more specific. – Anh_Rose 1210 Aug 17 '16 at 19:34
  • 1
    I intended my comment to be informed by the link I posted with it: take $y_i$ to be the image under the canonical isomorphism of a vector space with it's double-dual, noting that each $f_i$ is some $v^\ast$ for $v\in V$ since $V$ is (non-canonically) isomorphic to $V^\ast$. – Derek Allums Aug 17 '16 at 19:45
  • yes. I got it. Thank you ! – Anh_Rose 1210 Aug 17 '16 at 20:05

1 Answers1

2

This is an application of the fact that the map $\omega_V\colon V\to V^{**}$ defined by $$ \omega_V(x)=\hat{x}, \qquad \hat{x}(f)=f(x),\text{ for }f\in V^* $$ is an isomorphism when $V$ is finite dimensional. (Prove $\omega_V$ is injective.)

Thus you can consider the dual basis of $\{f_1,\dots,f_n\}$ in $V^{**}$ and take the elements in $V$ that map to it.

egreg
  • 238,574