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I'm on my journey to understanding linear algebra. I've reached the point where I have to understand the dual basis and bilinear forms. This is something I didn't find in the book that I learn from (David Poole's A Modern Introduction to Linear Algebra). My teacher at the university did some examples so I want to understand them.

I know what a basis is. I know also what the geometrical interpretation is. I also know what is a quadratic form (is it related in any way to bilinear forms?)

Could you please summarize what is a bilinear form and a dual base? Also, if there is any geometrical interpretation, please list it. Thank you a lot!

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Have you read the Wikipedia articles on these topics? I like talking about bilinear forms, but it seems like you want definitions and there are lots of places to find those. – Dylan Moreland Dec 5 '11 at 20:57
@DylanMoreland are the standard basis vectors always used to create a dual basis? – Andrew Dec 5 '11 at 21:01
I would say no. Many vector spaces do not come with a standard basis. – Dylan Moreland Dec 5 '11 at 21:04
up vote 3 down vote accepted

(I'm going to assume that we're talking about real finite-dimensional vector spaces here, although what I'm about to say mostly generalizes.)

Every vector space $V$ has a dual $V^*=Hom(V,\mathbb{R})=\{\mbox{ linear }x:V\to\mathbb{R}\}$.

Exercise: $V^*$ is a vector space.

In fact, $V^*$ is a finite-dimensional vector space. See this by choosing a basis $\{b_i\}$ of $V$, and defining $\beta_i$ to be the linear function such that $\beta_i(b_j)=\delta_{ij}$.

Exercise: The $\beta_i$ form a basis of $V^*$.

These form the "dual basis" of $\{b_i\}$, and of course implies that $V^*$ has the same dimension as $V$.

Exercise: A choice of basis induces an isomorphism $V\leftrightarrow V^*$.

But a priori any isomorphism depends on a choice of basis! $V$ is isomorphic to $V^*$, but without extra structure, one cannot say that $V$ is equal to $V^*$.

A bilinear form $\langle\cdot,\cdot\rangle$ is a bilinear map* $V\times V\to\mathbb{R}$. The dot product on $\mathbb{R}^n$ is a bilinear form. A choice of symmetric, nondegenerate* bilinear form determines an isomorphism $V\leftrightarrow V^*$ via $v\mapsto \langle -,v\rangle$.

*N.B.: Your professor's definition of "bilinear form" may vary from mine here. The term is often taken to mean symmetric and positive-definite right out of the box.

Exercise: The map $v\mapsto\langle-,v\rangle$ induced by a symmetric, nondegenerate bilinear form is an isomorphism.

You'll want to make your peace someday with tensor algebras, which are the natural homes of symmetric bilinear forms. Introducing you to them is beyond the scope of this post.

For an abstract, perhaps unreadable, introduction, you might consult chapter 2 of Warner, Introduction to Differentiable Manifolds and Lie Groups. I would myself welcome any suggestions of advanced linear algebra textbooks.

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To answer just one of your questions, let me work over a field $k$ of characteristic $\neq 2$, such as $\mathbf R$. Then one can pass back and forth between quadratic forms and symmetric bilinear forms as follows. A quadratic form $f\colon k^n \to k$ on $k^n$ (thought of as column vectors) corresponds to a (unique) symmetric $n \times n$ matrix $A$ such that $$ f(x) = {}^txAx. \qquad \text{[${}^tx$ is the transpose of the column vector $x$]} $$ But we can also define a map $g\colon k^n \times k^n \to k$ by $$ g(x, y) = {}^txAy, $$ and this is a symmetric bilinear form (take the transpose of the right-hand side). Can you see how to go in the other direction?

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You might have to read a bit more about bilinear forms before this makes sense, but this is the connection -- you were right to suspect that there is one! – Dylan Moreland Dec 5 '11 at 22:00

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