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The question is:

Let $V$ be the real vector space of functions from $\mathbb{R}$ to $\mathbb{R}$. And $W \subset V$ so that the subspace is spanned by the functions $f_1, f_2, f_3$ with $f_1 = \cos(x),\: f_2 = \sin(x)\: f_3 = \sin(2x)$. For $k=1,2,3$ evaluate $\phi_k \in W^*$ given by $\phi_k(f) = f((k-1)\frac{\pi}{4})$.
1) Calculate the $3\times3$ matrix $( \phi_i(f_j))_{i,j}$.

2) Derive from this that $f_1, f_2, f_3$ are a basis for $W$ and that $\phi_1,\phi_2,\phi_3$ are a basis for the dual vectorspace $W^*$ of $W$

So part 1 is just filling in the matrix, but I'm not quite sure how to start part 2. By the definition of dual vector spaces $v_i^*(v_j)=\delta_{ij}$. So am I supposed to figure out a basis of $W$ and show that the forementioned equality holds up? And given the question isn't the matrix calculated in part 1 a basis for $W^*$ as long as it is linearly independent?

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  • $\begingroup$ They don't form a dual basis (since $\phi_2(f_1) \neq 0$), but you can just check that the matrix obtained in (1) is invertible. $\endgroup$ – Prahlad Vaidyanathan Oct 28 '13 at 17:18
  • $\begingroup$ Hmm good point. It is suprising though, since the question insinuates that $\phi$ does in fact form a basis for $W^*$. How does showing that matrix in 1 is invertible show that $f_1,f_2,f_3$ is a basis for $W$? $\endgroup$ – user103846 Oct 28 '13 at 20:49
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Since the number of vectors matches the dimension, the issue of being a basis reduces to their linear independence. Two observations:

  • if $f_j$ are linearly dependent, then the columns of $(\phi_i(f_j))$ are linearly dependent.
  • if $\phi_i$ are linearly dependent, then the rows of $(\phi_i(f_j))$ are linearly dependent.

Since the matrix $(\phi_i(f_j))$ is invertible, neither of the above happens.

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