Canonical form of a Matrix

Question

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.

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.