4
$\begingroup$

Does a quadratic form always come from symmetric bilinear form ? We know when $q(x)=b(x,x)$ where $q$ is a quadratic form and $b$ is a symmetric bilinear form. But when we just take a bilinear form and $b(x,y)$ and write $x$ instead of $y$,does it give us a quadratic form ?

3 Answers 3

1

Yes, every quadratic form (over a finite-dimensional $\mathbb{R}$-space) can be expressed in terms of a symmetric bilinear form, because if your quadratic form $Q(x)$ (for $x \in \mathbb{R}^n$) is written as $$ Q(x) = \sum_{i\le j} c_{ij} x_i x_j $$ then $Q(x) = x^T Ax$, where $A$ is a symmetric matrix given by $A = (a_{ij})$ with $$a_{ij} = a_{ji} = \frac{c_{ij}}{2}, i < j, \quad a_{ii} = c_{ii}$$

  • 0
    yes but my question is a little general.i mean if we have any quadratic form does it have to be come from symmetric bilinear form ?2012-11-18
  • 0
    I just constructed a symmetric matrix $A$ such that $Q(x) = x^T Ax$, so then the map $B(x,y) = x^T A y$ is a symmetric bilinear form.2012-11-18
  • 1
    but when we think as a function,if we have a quadratic form then do we have to have a bilinear form ?2012-11-18
  • 0
    The construction I gave shows that for every quadratic form (on a finite-dim space) $Q(x)$, we can write $Q(x) = B(x,x)$ for a symmetric form.2012-11-18
  • 0
    ok but what about non-symmetrics ? :)2012-11-18
0

Sure. In finite dimension over the real numbers you do not even need the bilinear form symmetric, because any matrix is the sum of a symmetric and a skew-symmetric matrix, and these are easy to find.

Meanwhile, as long as characteristic of the underlying field is not $2,$ you can go back with $$ c(x,y) = \frac{1}{2} \left( q(x+y) - q(x) - q(y) \right). $$ Since this recipe gives a symmetric bilinear form, those are preferred.

Even in infinite dimension, I suppose you can symmetrize with $$ f(x,y) = \frac{1}{2} \left( h(x,y) + h(y,x) \right). $$ There won't be any matrices.

  • 0
    I see but there is something missing i guess.because if we have just bilinear form (not symmetric) and write x instead of y then we get quadratic forms.am i wrong ?2012-11-18
  • 0
    @TurkuKirli, if you start with non-symmetric you still get a quadratic form. It may be the constant $0$ form, for example. If your original form is not symmetric you cannot recover it from the quadratic form.2012-11-18
  • 0
    But i still can not get it :/ i mean when we have bilinear form yes we may get a quadratic form.but what about the inverse ? when we have a quadratic form ( yes we can get a symmetric bilinear form) but can not we get any bilinear form ?2012-11-18
  • 0
    @TurkuKirli, I sent you an email with some preliminary observations. You should probably delete the message with your email address now.2012-11-18
-1

If we have b symmetric bilinear form we can get q quadratic form
$q\colon V \to \mathbb{R}$
q(v)=b(v,v)

conversely if q is a quadratic form
$q\colon V \to \mathbb{R}$
we can define
$\frac 12$(q(v+w)-q(v)-q(w)):=b(v,w)

the vital answer is you just get a bilinear form not always a symmetric bilinear form. because the definition $\frac 12$(q(v+w)-q(v)-q(w)) leads us to $\frac 12$(b(v,w)+b(w,v))

  • 1
    For your converse part, you need to mention that the characteristic of the field needs to be different from 2, other wise your 1/2 will not exist.2013-01-05
  • 1
    Sorry I didn't get the point.Don't i do my works on the field $\mathbb{R}$? So do I have to say anything else?2013-01-05
  • 0
    Well, the question does not mention that the underlying field is $\mathbb{R}$, so, as Will Jagy mentions in his answer, we have make sure that the characteristic is not 2.2013-01-05
  • 3
    @SerkanHafızYaray Why do you say "not always a symmetric bilinear form"? The bilinear form defined by b(v,w)=(q(v+w)-q(v)-q(w))/2 is *always symmetric*, by the commutativity of addition.2013-03-06
  • 0
    @rschwieb I am not able to check, that the $b:V\times V\rightarrow \mathbb R$ associated to $q$ (a quadratic form) is actually bilinear, (As you commented, symmetric is obvious).2016-10-02
  • 0
    @Babai Sunds pretty serious if you're **not able to check**. I can't help if something is physically preventing you from checking. If you were just stuck at some particular point, I'm sure I would be able to help (if you told me where you were stuck.)2016-10-02
  • 0
    @rschwieb can you tell me what is the definition of a quadratic form.2016-10-03