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 ?

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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 $$a_{ij} = a_{ji} = \frac{c_{ij}}{2}, i < j, \quad a_{ii} = c_{ii}$$

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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 ? – Turku Kirli Nov 18 '12 at 21:50
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. – Christopher A. Wong Nov 18 '12 at 22:08
but when we think as a function,if we have a quadratic form then do we have to have a bilinear form ? – Turku Kirli Nov 18 '12 at 22:17
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. – Christopher A. Wong Nov 18 '12 at 22:22
ok but what about non-symmetrics ? :) – Turku Kirli Nov 18 '12 at 22:25

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.

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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 ? – Turku Kirli Nov 18 '12 at 21:46
@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. – Will Jagy Nov 18 '12 at 21:54
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 ? – Turku Kirli Nov 18 '12 at 21:59
@TurkuKirli, I sent you an email with some preliminary observations. You should probably delete the message with your email address now. – Will Jagy Nov 18 '12 at 23:33
okey thank you :) – Turku Kirli Nov 19 '12 at 5:32

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))

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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. – Old John Jan 5 '13 at 14:33
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? – Serkan Yaray Jan 5 '13 at 14:41
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. – Old John Jan 5 '13 at 14:46
@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. – rschwieb Mar 6 '13 at 19:42