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My understanding of canonical form is very limited, and so may require some help.

Suppose a quadratic of the form:

$$ x_1*x_2+x_1*x_3=Q.$$

How would one go about putting that into canonical form, i.e. $$z_1^2+z_2^2+\cdots.$$

Many thanks.

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Usually one uses rotations to remove cross-terms... –  J. M. Jan 6 '12 at 11:25
    
Ok, sorry to be naive, but how would you go about doing that? Many thanks. –  Richard Jan 6 '12 at 14:01
    
If you've studied conic sections, you should have gone through the bit on rotating coordinates‌​; one can always pick an angle such that there is no cross term in the new coordinate system. –  J. M. Jan 6 '12 at 14:05
    
Do you mean by finding the eigenvalues of the characteristic matrix for the equation? If so, then this would not leave it in the required form unfortunately. –  Richard Jan 6 '12 at 14:12

1 Answer 1

I change the notation to $a,b,c$ just for the typing convenience. I hope you don't mind.

You can do this the fun way: Complete the square as $$ \begin{align} 2ab + 2bc &= 2Q \\ 2b(a+c)&=2Q\\ 2b(a+c)+(a+c)^2+b^2-(a+c)^2-b^2 &=2Q\\ (a+b+c)^2-(a+c)^2-b^2 &=2Q\\ \frac{1}{2}\left((a+b+c)^2-(a+c)^2-b^2\right) &=Q \end{align} $$ or you use the matrix notation $$ \begin{align} 2ab + 2bc &= 2Q \\ \begin{pmatrix}a\\b\\c\end{pmatrix}^T\begin{pmatrix} 0 &1 &0\\1 &0 &1\\0 &1 &0 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}&=2Q\\ \begin{pmatrix}a\\b\\c\end{pmatrix}^T \begin{pmatrix} 1 &1 &0\\1 &0 &1\\1 &1 &0 \end{pmatrix} \begin{pmatrix} 1 &0 &0\\0 &-1 &0\\0 &0 &-1 \end{pmatrix} \begin{pmatrix} 1 &1 &1\\1 &0 &1\\0 &1 &0 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix} &=2Q\\ \begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix}^T \begin{pmatrix} 1 &0 &0\\0 &-1 &0\\0 &0 &-1 \end{pmatrix}\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix} &=2Q\\ \begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix}^T \begin{pmatrix} 0.5 &0 &0\\0 &-0.5 &0\\0 &0 &-0.5 \end{pmatrix}\begin{pmatrix}a+b+c\\a+c\\b\end{pmatrix} &=Q \end{align} $$ as this shows explicitly what the coordinate rotations (that J.M. mentions in the comments) are. You might want to grease those congruence transformation gears, it takes some affinity to write quickly the matrix product. But it is not that difficult anyway.

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