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I was reading this post A basis for the dual space of $V$ where given a basis $\mathcal{B} =\{\alpha_{1},\alpha_{2},\dots,\alpha_{n}\}$ for a vector space $V$, and $\mathcal{F} = \{f_{1}, f_{2},\dots, f_{n}\}$ defined by $f_{i}:= \langle v, \alpha_{i}\rangle$. Then $\mathcal{F}$ is a basis for $V^{*}$.

This piqued my curiosity. Instead of finding the set of linear maps $f_{i}(\alpha_{j}) = \delta_{i,j}$, can we just define an inner product and use that to construct a basis instead? If so, what is the impact of the choice of inner product? Given the basis constructed the usual way a dual basis is calculated for a finite dimensional space, can we figure out what the corresponding inner product is that would create it?

As an experiment, I took $\mathcal{B} = \{1, 1-x, x-x^{2}\}$ as a basis of $\mathbb{P}_{2}$ and tried to find a basis for the dual space by taking an arbitrary $p(x)\in \mathbb{P}_{2}$ and defining $$\langle p(x),q(x)\rangle := \int_{-1}^{1} p(x)q(x)dx$$

to be my inner product, but this doesn't seem to produce anything that looks correct because I would have thought that $f_{1}(1-x) = \langle 1-x, 1\rangle$ should equal $0$.

I have a suspicion that I am misunderstanding something in a very big way because the proof by Noah Stein in the above link doesn't seem to depend on a specific inner product.

Community
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JessicaK
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2 Answers2

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You've asked several questions and hopefully this will answer some.

Any inner product will do. Given an inner product space $(V, \langle \cdot, \cdot \rangle)$, if $\{v_i\}$ is a basis for $V$, then $\{\langle v_i, \cdot \rangle\}$ is a basis for $V^*$. I think you see this from the linked proof above, but the point is that $\langle v_i, \cdot \rangle$ picks out the $v_i$-coordinate of a vector. The family $\{\langle v_i, \cdot \rangle\}$ is linearly independent precisely because $\{v_i\}$ is linearly independent.

In fact, having chosen a basis for $V$, every inner product on $V$ can be written $\langle x, y \rangle = x^TAy$ for some (nice) matrix $A$. Given a basis $\{v_i\}$ for $V$, you present two different ways of finding a basis for $V^*$. One is to search for linear maps $\lambda_i$ satisfying $\lambda_i(v_j) = \delta_{ij}$. The other is to define an inner product $\langle \cdot, \cdot \rangle$ and use the basis $\{\langle v_i, \cdot \rangle \}$. You ask if, given the results of the first process, we can find an inner product which will make the second process give identical results. The answer is yes: define $\langle v, w \rangle = \langle \sum a_jv_j, w\rangle = \sum a_j \lambda_j(w)$.

A good, theory-focused linear algebra textbook will address these issues and more. I learned from Hoffman and Kunze but there are lots of others.

Adam Saltz
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Let me check if I am addressing your question correctly.

Suppose you have two vector spaces $V, W$ of the same dimension $n$ over a field $F$.

Fix a basis $v_{1}, \dots, v_{n}$ of $V$.

Now there is a one-to-one correspondence between

  • non-degenerate bilinear forms $\langle \cdot, \cdot \rangle : V \times W \to F$, and
  • bases $w_{1}, \dots, w_{n}$ of $W$,

the correspondence being provided by $\langle v_{i}, w_{j} \rangle = \delta_{ij}$.

Andreas Caranti
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  • I don't think that's quite what I am asking. In particular, the additional space $W$ is confusing me.

    Given a space $V$ and an ordered basis, I would like to generate a basis for the dual space, $V^{}$ by taking the mapping $f_{i}:= \langle v,v_{i}\rangle$ for $v\in V$ as the elements of the basis of $V^{}$. If this is possible, how do I know what inner product to use?

     – JessicaK Jan 19 '14 at 12:01