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Let the following be a ternary quadratic form:

$$A = \begin{pmatrix}0 & a & b \\ a & 0 & c \\ b & c & 0 \end{pmatrix}$$

with $a,b,c\in\mathbb{Q}$. If at least one term in the diagonal was nonzero, one could transform it into a canonical form by "completing squares".

How would one proceed in this case?

In essence, I need to find two matrices $R$ and $D$ where $D$ is diagonal so that $R^TDR = A$. How would one find such matrices?

Also, could one ensure that the entries on $R$ and $D$ are rational?

Darth Geek
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  • As $A$ is symmetric you can find a set of orthogonal eigenvectors and write them as columns of $R^t$, then $AR^t = R^tD$. – flawr Jul 12 '16 at 21:38
  • The part saying $R$ symmetric is incorrect. The usual condition is determinant one, in which case the before and after matrices are called congruent, the quadratic forms called equivalent. Hmmmm...well, those are over the ring of rationals. If everything in sight is an integer, there may be no way to accomplish this with $ \det R = 1$ and all entries of $R$ integers. – Will Jagy Jul 12 '16 at 23:17

1 Answers1

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There is no agreed upon canonical form for indefinite ternary integer/rational quadratic forms. In the order you requested,

$$ \left( \begin{array}{rrr} \frac{1}{2} & -1 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{b+c}{2a} & \frac{b-c}{a} & 1 \end{array} \right) \left( \begin{array}{rrr} 2 a & 0 & 0 \\ 0 & -\frac{a}{2} & 0 \\ 0 & 0 & - \frac{2bc}{a} \end{array} \right) \left( \begin{array}{rrr} \frac{1}{2} & \frac{1}{2} & \frac{b+c}{2a} \\ -1 & 1 & \frac{b-c}{a} \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{rrr} 0 & a & b \\ a & 0 & c \\ b & c & 0 \end{array} \right) $$

See reference for linear algebra books that teach reverse Hermite method for symmetric matrices where you are using the inverse of the matrix found in their algorithm.

? q = matadjoint(p)
%17 = 
[1/2 1/2 (b + c)/(2*a)]

[-1 1 (b - c)/a]

[0 0 1]

? d
%18 = 
[2*a 0 0]

[0 -1/2*a 0]

[0 0 -2*c*b/a]

?  mattranspose(q) * d * q
%19 = 
[0 a b]

[a 0 c]

[b c 0]

?
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Will Jagy
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