Inner products

June 7th, 2006 ~ Posted in: Mathematics

I’m auditing a matrix theory class this summer. The first problem, assigned yesterday, was to find non-trivial examples of inner products in \R^2 and \R^3.

Recall a (real-valued) inner product on a vector space is a function \langle \cdot, \cdot \rangle: V \times V \rightarrow \R with the following properties:

  1. \langle x, y \rangle = \langle y, x \rangle (symmetry)
  2. \langle x, x \rangle \geq 0 and \langle x, x \rangle = 0 \Leftrightarrow x = 0 (positive definiteness)
  3. \langle x + y, z \rangle = \langle x, z\rangle + \langle y, z \rangle (linearity)
  4. \langle \alpha x, y \rangle = \alpha \langle x, y \rangle (linearity).

The canonical inner product in \R^n is \langle x, y \rangle = \sum_{i=1}^n x_i y_i ; you can show that in the case n=2, this satisfies the relationship \cos \theta_{xy} = \frac{\langle x, y \rangle}{\|x\| \|y\|} where \|x\| denotes the usual length of a vector. Similarly, for any inner product, if we define the norm of a vector appropriately, \|x\| = \sqrt{\langle x, x \rangle}, then we have \left | \frac{\langle x, y \rangle}{\|x\| \|y\|} \right| \leq 1 , so the inner product supplies a concept of the angle between two vectors in an arbitrary vector space (you can take the angle to be arccos of that quotient).

The first result that came to me is that inner products on finite dimensional vector spaces are the bilinear maps determined by positive definite matrices, in the sense that if we take \mathcal{B} = \{ e_1, \ldots, e_n\}[tex] to be a basis of [tex]V, a real vector space, then by linearity, we have

\langle x, y \rangle = \sum_{i=1}^n \sum_{j=1}^n x_i y_j \langle e_i, e_j \rangle = x^t Ey,

where in the last equality, I’m identifying x and y with their representations as column vectors relative to \mathcal{B}, and E_{ij} = \langle e_i, e_j \rangle . The fact that E is positive definite follows from property 2 of inner products. It is easy to see that a positive definite matrix A determines an inner product, hence the two objects (inner products and positive definite matrices) are in a sense equivalent.

I don’t know much about positive definite matrices in general, so I couldn’t find any more concrete conditions on matrices that would allow you to construct an inner product, but I was able to find a sufficient condition in the case of \R^2:
if A = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}, then it is sufficient that a_{11}, a_{22} > 0 and a_{11},a_{22} > |a_{12} + a_{21}| for A to be positive definite and therefore determine an inner product.

But that is a pretty strong condition, in my opinion, and I don’t see any obvious way to extend it to n>2. Shihka had the idea of just modifying the arguments to the standard inner product by applying an invertible linear transformation to get a new inner product:

\langle x, y \rangle_\prime = \langle Ax, Ay \rangle = x^tA^tAy

As you can see, this neat idea has the advantage of showing rather simply that A^tA is positive definite (is this easier than the standard proof?), and gives a quick way to generate nontrivial inner products in any finite dimensional vector space.

Now, the question is, is every positive definite matrix of the form A^tA for some invertible matrix A? That is, are there exactly as many inner products as invertible linear transformations, or more?

Mathematics ~

This entry was posted on Wednesday, June 7th, 2006 at 11:10 am and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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