Congruence for Symmetric and non-Symmetric Matrices for Quadratic Form

[CGandC]

I learned that for a bilinear form/square form the following theorem holds:
matrices ## A , B ## are congruent if and only if ## A,B ## represent the same bilinear/quadratic form.

Now, suppose I have the following quadratic form ## q(x,y) = x^2 + 3xy + y^2 ##. Then, the matrix representing this quadratic form can be ## B = \pmatrix{1 & 3 \\ 0 & 1 } ## and also ## A = \pmatrix{1 & \frac{3}{2} \\ \frac{3}{2} & 1 } ##.

I tried showing that ## A, B ## are congruent, meaning that there exists an invertible matrix ## M ## such that ## B = M^T \cdot A \cdot M ##, but such matrix ## M ## does not exist, here's the calculation for example:
## M = \pmatrix{ a & b \\ c & d } ##

And according to Wolfram:

Yet, ## A,B ## represent the same quadratic form, but they are not congruent, so this is a contradiction to the theorem above.

Question:
How is this possible? These matrices should be congruent since they do represent the same quadratic form.
( And yes, I know that for every matrix ## A ##, ## \frac{1}{2} ( A + A^T ) ## is symmetric, but this doesn't answer why the above congruence fails to hold )

 

[martinbn]

##B## does not represent the quadratic form.

 

[CGandC]

##B## does not represent the quadratic form.

Yes it does because: ## \pmatrix{ x & y } \pmatrix{1 & 3 \\ 0 & 1 } \pmatrix{ x \\ y } = \pmatrix{ x & y }\pmatrix{x + 3y \\ y } = x(x+3y) +y^2 = x^2 + 3xy + y^2 ##

 

[Office_Shredder]

I think you have some terminology confusion.

Given a bilinear form on a vector space, if you pick a basis it has a unique symmetric representation.

Two matrices are congruent if they are representatives of the same bilinear form with different choices of basis.

Note the representatives are always symmetric.

 

[CGandC]

I think you have some terminology confusion.

Given a bilinear form on a vector space, if you pick a basis it has a unique symmetric representation.

Two matrices are congruent if they are representatives of the same bilinear form with different choices of basis.

Note the representatives are always symmetric.

You mean a quadratic form?, because bilinear form isn't necessarily symmetric

 

[Office_Shredder]

You mean a quadratic form?, because bilinear form isn't necessarily symmetric

Sorry, you are correct, quadratic form

 

[CGandC]

Ok but if
## \pmatrix{ x & y } B\pmatrix{ x \\ y } = \pmatrix{ x & y } \pmatrix{1 & 3 \\ 0 & 1 } \pmatrix{ x \\ y } = \pmatrix{ x & y }\pmatrix{x + 3y \\ y } = x(x+3y) +y^2 = x^2 + 3xy + y^2 ##
And
## \pmatrix{ x & y } A\pmatrix{ x \\ y } = \pmatrix{ x & y } \pmatrix{1 & \frac{3}{2} \\ \frac{3}{2} & 1 } \pmatrix{ x \\ y } = \pmatrix{ x & y }\pmatrix{x + \frac{3}{2}y \\ \frac{3}{2}x+y } = x(x+\frac{3}{2}y) +y(\frac{3}{2}x+y) ##
## = x^2 + \frac{3}{2}xy + \frac{3}{2}yx + y^2 = x^2 +3xy+y^2 ##

Then according to what you said ## A ## is a representative of the quadratic form since it is symmetric.
But what is ## B ## then? is it also defined as a representative of the quadratic form?

 

[CGandC]

Ok, I've asked on mathexchange and I was told that when dealing with symmetric bilinear-forms/quadratic-forms then when using the theorems related to these, one should assume he's looking at the symmetric representation of these forms, otherwise most of the theorems for symmetric bilinear-forms/quadratic-forms will fail to hold and it'll be nonsense.

For example, in the above example if one assumes ## B = M^T A M ## to be symmetric then ## B^T = M^T A^T M = M^T A M = B ##, but this is a contradiction since ## B^T \neq B ##, so ##A,B ## are not congruent.
So everything's clear now, thanks for the help!