- #1

joypav

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It gives the basis dual to the basis of $V$ and proves that this is in fact a basis for $V^*$.

Characterized by $\alpha^i(e_j)=\delta_j^i$

I understand the proof given. But he said a different statement...

If $T: V \rightarrow \Bbb{R}$ is linear, then there is a vector $t$ so that $T(u)=<t,u>$. ($<\cdot, \cdot>$ is the inner product)

He said this is equivalent to showing that the dual basis forms a basis, and that it can be proven using the Gram-Schmidt process. I was wondering what that proof looks like? Even just an outline... I don't need all the details.