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I’m studying linear algebra, and i saw the following statement: “As we associated a norm to an inner product, we can associate a quadratic form to a bilinear form, and the quadratic form behaves like the square of a norm of a vector.”

Context: this statement is made after talking about bilinear forms and before defining the associated quadratic form.

There are two proposition that I think “support” this statement:

  • Let $\varphi$ be a bilinear form and $\phi$ be the associated quadratic form. Let $u$, $v$ be vectors and $t \in \mathbb{R}$. Then we can define the polarization identities, in a similar way we do for the norm and for the inner product: \begin{gather*} 4\varphi (u, v) = \phi (u + v) - \phi (u - v) \,,\\ 2\varphi (u, v) = \phi (u + v) - \phi (u) - \phi(v) \,. \end{gather*}

  • Let $\phi$ be an associated quadratic form. Then $\phi(0) = 0$ and $\phi(tv)=t^2\phi(v)$.

Still, I don’t understand the use of saying that a quadratic form is similar to a norm: I’d expect that this would be useful if, as the norm associated to a inner product lets us measure lengths and angles, the quadratic form is a kind of generalization of measure of lengths and angles (since bilinear forms are a generalization of inner products). But I can’t find anything that supports this, and I can’t really have any intuition of what a quadratic form is, except from the fact that you can graph it as with any other function (for example, real quadratic forms represent quadrics).

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    $\begingroup$ Perhaps this comment might be of help. If you have a matrix $A= BB^\top$, then the quadratic form $x^\top A x = x^\top B B^\top x = y^\top y = \|y\|_2^2$, where $y= B^\top x$. So what you are doing is projecting the vector $x$ into the column-space of $B$ and taking the norm in that new space. $\endgroup$
    – yes
    Commented Dec 30, 2022 at 14:26
  • $\begingroup$ @VanBaffo ok thanks, so the idea is that a norm is a special case of a quadratic form which occurs when the Gram matrix is made by the product of a matrix and its transposition $BB^T$? Because if it’s that, it seems to me that it’s still the case that, in general, quadratic form doesn’t have a geometrically significant interpretation related to the norm: that happens just in the “special case” $\endgroup$
    – selenio34
    Commented Dec 30, 2022 at 14:36
  • $\begingroup$ Well it just makes the parallel that as an inner product gives us a norm, a bilinear product gives a quadratic form (in characteristic different from 2). If you want a simple connection to physics, the spacetime metric is Lorentzian $ds^2=-dt^2+d\vec x^2 $ so this is not a norm in the usual sense, but still has an associated bi-linear form (which is not an inner product) and everything works pretty much more as if the quadratic form were positive definite. $\endgroup$ Commented Dec 30, 2022 at 14:37
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    $\begingroup$ @JacopoSenoner You can show that the contour lines of the function $x^\top A x$ define ellipses (as long as $A$ is positive semidefinite), so in that sense the matrix $A$ transform the vectors in the space, so it transforms the space. If $A=I$, i.e. identity matrix, the level curves are circular, there is a notion of "isotropic" meaning that to compute the length of a vector, you can simply "take a ruler" and measure the vector straight. If there is a matrix, that is no more the case. $\endgroup$
    – yes
    Commented Dec 30, 2022 at 14:45
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    $\begingroup$ @VanBaffo Btw, if you’re interested in posting your comments as an answer, I’ll accept them. $\endgroup$
    – selenio34
    Commented Dec 30, 2022 at 15:09

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If you have a matrix $A=BB^\top$, then the quadratic form $$x^\top Ax=x^\top BB^\top x=y^\top y=\|y\|_2^2 $$ measures the length of the vector $y=B^\top x$ and squares it. So what you are doing is projecting the vector $x$ into the column-space of $B$ and taking the norm in that new space.

Geometrical Interpretation

You can show that the contour lines of the function $x^\top Ax$ define ellipses (as long as $A$ is positive semidefinite), so in that sense the matrix $A$ transform the vectors in the space (it transforms the space). If $A=I$, i.e. identity matrix, the level curves are circular, there is a notion of "isotropic" meaning that to compute the length of a vector, you can simply "take a ruler" and measure the vector straight. If there is a matrix, that is no more the case.

To be an ellipse the matrix of the quadratic form has to be positive semidefinite. Otherwise it can still have a geometric illustration, for instance the hyperbola. This might be of great help: https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections#Standard_form_of_a_central_conic

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