Quadratic form as generalized distance?

Question

In the book A Linear Systems Primer (by Antsaklis and others), they first mention squared distance of a point x from the origin:

$$x^{T}x = ||{x}||^2$$ which represents the square of the Euclidean distance of the state from the equilibrium $x=0$.

So far, so good this is basic linear algebra. Then they go on to say:

In the following discussion, we will employ as a "generalized distance function" the quadratic form given by $${x}^TPx , P={P}^T $$ where $P$ is a real $n\times n$ matrix.

I am familiar with this definition of a quadratic form from linear algebra : we interpose a symmetric matrix to weight the variables in different ways.

But I am not familiar with this as a distance function. Is that mainly to say that it satisfies the requirements of a metric space? Is there a geometric (intuitive) discussion of the sense in which the quadratic form generalizes the more basic notion of a distance?

Answer 1

This determines a norm $\|\_\|_P$ iff $P$ is positive definite, then it naturally defines a metric by $d(x,y):=\|y-x\|_P$.

Else the above $d$ function would not be defined or would fail to be a metric, as e.g. there could be $x\ne 0$ with $x^TPx\le0$.

All in all, what it basically says is that the quadratic form $x\mapsto x^TPx$ can be viewed as a generalisation of $x\mapsto \|x\|^2$.

For more details and geometric insights, see for example Pseudo Euclidean spaces.

Answer 2

I think some examples from physics might help provide the geometric (intuitive) sense you seek, in which quadratic forms generalize distance, though I doubt whether it’s useful to think of quadratic forms as providing a “more basic notion of distance” in quite the way that I think you're expecting. I see them as providing measures of distance like the usual (Pythagorean/Euclidean) notion of distance, but applied in slightly more general spaces.

A simple example of a generalisation of distance encoded by a quadratic form is given by the Minkowski metric for space-time; for the squared length of a vector from the origin:

$$d^2 = x^2 + y^2 + z^2 –(ct)^2$$

… which I’m sure you can see can be expressed as $X^TPX$ where $P$ is a $4\times 4$ matrix with all off-diagonal elements zero and the diagonal $\{1,1,1,-1\}$. See also http://mathworld.wolfram.com/MinkowskiMetric.html where the matrix is illustrated.

And so for the slightly less-clear case of a distance between the endpoints of vectors v1 and v2, I think you will now see this also as a quadratic form (using an overly-simplistic notation):

$$d^2(v1-v2) = (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2 –(ct1-ct2)^2$$

In General Relativity, however, the matrix $P$ is called the Metric Tensor, and can have all non-zero components. As these components can change from point to point, this metric provides for a quadratic form that can encode distances in curved space-times, which is a generalisation of the "flat" Minkowski space-time.

So, as Berci pointed out via https://en.wikipedia.org/wiki/Pseudo-Euclidean_space , there’s a clear geometric interpretation, but it’s “simply” a generalisation of the familiar types of distances that apply in Pythagorean/Euclidean/pseudo-Euclidean spaces, rather than a generalisation of the more “disjointed” type that would be needed to include, say, Manhattan, Chessboard or Mahalanobis distances.

(Clearly, I’m not a mathematician, but I think the physics examples help provide the geometric insight you seek, and if I’m wrong in this, would appreciate correction from those more expert. Put into MathJax with assistance acknowledged below.)