1
$\begingroup$

I'm learning about quadrics. My textbook says that a quadric is a quadratic form $X^TAX = 0$ with $A$ en symmetric matrix and $X$ the homogeneous coordinates in a orthonormal basis. But if we're working in an orthonormal basis, isn't the scalar product always just $X^TX$ (so $A=I_n$)?

$\endgroup$
  • 2
    $\begingroup$ The text never said that a quadratic form must be a scalar product. It is the converse: the scalar product is a particular quadratic form (with the unit matrix). $\endgroup$ – Yves Daoust Jun 17 '16 at 21:12
  • $\begingroup$ Thank you! So if $A$ is also positive definite, it's a scalar product and it has to be $I_n$? $\endgroup$ – Ruben Van Belle Jun 17 '16 at 21:16
  • $\begingroup$ In that case, it will be the regular scalar product, possibly expressed in a different basis, yes. $\endgroup$ – Arthur Jun 17 '16 at 21:18
  • $\begingroup$ @RubenVanBelle: no, I don't see how you deduce that. And a matrix can't be a scalar product. $\endgroup$ – Yves Daoust Jun 17 '16 at 21:20
  • $\begingroup$ You seem to be confusing the (canonical) scalar product and a scalar (or inner) product, or more generally a symmetric bilinear form, corresponding to a symmetric matrix. $\endgroup$ – Bernard Jun 17 '16 at 21:53
3
$\begingroup$

You're mixing up a number of things.

  1. The quadric (often also called “conic section” or just “conic”) is the set of points $X$ which satisfy $X^T\cdot A\cdot X=0$. It's a convention (and required for some operations) to choose $A$ as a symmetric matrix.

  2. The $X$ in this formula is meant to represent homogeneous coordinates. Which usually means that you pick one vector as a representative, but it is implied that multiples of that vector represent the same point. (If you do things more formally, you must phrase everything in equivalence classes all of the way, which makes notation more complicated for little gain.)

  3. For some reason, your text book adds a requirement that the basis used for these homogeneous coordinates should be orthonormal. I must say that I consider this a strange requirement here, since in my opinion it only makes sense if you think of the projective plane as being embedded into some 3-dimensional containing space. While this may be a useful idea when starting with projective geometry, my experience is that by the time you get to quadrics, thinking about a containing space is more confusing than helpful most of the time. And since a projective transformation (which is an invertible linear transformation of the containing space) maps quadrics to quadrics, I see no benefit in adding the orthogonality requirement. Particularly since normalization to length $1$ fits in poorly with the arbitrary choice of representatives you have with homogeneous coordinates.

    If it helps you, forget about the “orthonormal” part of your text and continue. You may even skip all the following points of this answer.

  4. If you want to have a closer look at orthonormality, you find that a basis is orthonormal if $\langle X,Y\rangle=X^T\cdot I_n\cdot Y=\sum_{k=1}^nx_ky_k$. In words, if your vectors $X$ and $Y$ are expressed with respect to the same orthonormal basis, then computing the scalar product is the same as computing the bilinear form encoded by the identity matrix. But all of this has very little to do with your original setup. Yes, the standard scalar product is a symmetric bilinear form, just like that matrix $A$ is supposed to be symmetric. If you set $Y=X$ you obtain a corresponding quadratic form, just like your quadric is described by a quadratic form. But here that form describes the metric properties of your containing vector space. It does not describe the conic mentioned originally.

  5. You are of course free to pick the identity matrix and look at the quadric it describes. For $X=[x:y:z]$ you get the equation $x^2+y^2+z^2=0$. Obviously this has no real solutions. So while it makes sense to think of this as a (non-degenerate) complex quadric, there are no points on this conic which you could actually draw in your (real) drawing plane.

  6. Some comments explored the fact that any positive-definite symmetric bilinear form may be called a scalar product. While this is true, I feel that it is taking you further away from the actual issue at hand. It means that you'd most likely have not just one basis, but two of them. The identity matrix in one of them corresponds to some arbitrary symmetric positive definite matrix in the other. But you don't intend to do any coordinate conversions, so this view doesn't help. Again, you can plug the matrix of any such scalar product into the formula for a quadric, and come up with a different quadric which is just as complex as the one I just described for the standard scalar product.

    So when you want something to imagine, I suggest you imagine a conic where the matrix $A$ is indefinite. That way you get real solutions and can actually imagine the resulting conic section as a figure in your projective plane. Keep in mind that definite and semi-definite matrices satisfy your formal requirements as well, and may prove relevant corner cases in some situations, but for intuition I'd skip those for now.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.