I want to use the method "completing squares" for this term:
$x^2-2xy +y^2+z^2*a+2xz-2yz$
The result should be $(x-y+z)^2 +(a-1)*z^3$
Is there a "recipe" behind how to do this? Hope someone could help
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$\begingroup$There is a method that is entirely algorithmic; it is Hermite's method done somewhat backwards, so that we diagonalize in the form $P^T M P = D,$ but $P$ is not orthogonal. Many books refer to this as "congruence," this should not be confused with similarity, which uses $R^{-1}M R.$
Your problem is just one step, $$ \left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{array} \right) \left( \begin{array}{rrr} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & A \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & A - 1 \end{array} \right) $$
This is of the form $P^T M P = D,$ as you can see. You actually want $Q^T DQ = M,$ which means we need $Q = P^{-1}.$ Since $\det P = 1$ and $P$ is upper triangular, this is not difficult, it is just the adjoint matrix, $$ Q = \left( \begin{array}{rrr} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) $$ Reading rows from $Q,$ this means that your quadratic form is $$ 1 \cdot (x-y+z)^2 + (A-1) \cdot (z)^2 $$ which is what you wanted.
See reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$\endgroup$ $\begingroup$There is no 'recipe', there's just practice which builds intuition.
Here the formula $x^2 - 2xy + y^2$ should immediately be noticible. You can write this as $(x-y)^2$. Furthermore, the $2xz - 2yz$ term can be factored to $2z(x-y)$. Thus your expression becomes
$$ (x-y)^2 + z^2 a + 2z(x-y)$$
Since you want to complete the square, you might notice that there is a $z^2$ term (with a constant $a$ in front of it but whatever for now), an $(x-y)^2$ term and a $2z(x-y)$ term. So by using a trick and writing $z^2 a = z^2 + z^2 (a-1)$ with the motivation of getting a $z^2$ term alone, we can write the expression as
$$ (x-y)^ 2 + z^2 + 2z(x-y) + (a-1)z^2$$
and completing the square yields
$$ (x-y + z)^2 + (a-1)z^2$$
which is the desired result.
$\endgroup$ 0 $\begingroup$Aras' answer is really good. But you might be wondering how one develops intuition and what happens if one doesn't immediately see $x^2 - 2xy + y^2$ is a square. (Part of developing intuition is that this does become obvious.)
This'll get messy. But
So you have:
$\color {blue}{x^2}-2xy +y^2+z^2*a+2xz-2yz$
And you have an $x^2$ so we'll "complete" x first so take all the x terms and put them nearby.
$\color {blue}{x^2}\color {purple}{-2xy +2xz}+y^2+z^2*a-2yz$
We know that completing the square, the expression will need to start something like this: $x^2 \pm 2*x*something + ....$ so we factor out the $2x$
$\color {blue}{x^2 + 2x}\color {purple}{(z - y)}+y^2+z^2*a-2yz$
Now we know to "complete the square" we must do $x^2 \pm 2x*something \color{green}{+ something^2} \color {red}{- something^2} +...$
So
$\color {blue}{x^2 + 2x(z - y)} \color{green}{ + (z -y)^2} \color{red}{- (z -y)^2}+y^2+z^2*a-2yz=$
$\color {blue}{(x + (y -z))^2} \color{red}{- (z^2 - 2zy + y^2) }+y^2+z^2*a-2yz$
So let's clean up a little.
$\color {blue}{(x + y -z)^2} \color{red}{- z^2 + 2zy - y^2 }+y^2+z^2*a-2yz$
$\color {blue}{(x + y -z)^2} -z^2 +z^2*a$
So now we need to deal with just the $ ... - z^2 + z^2*a$. Well we factor out common terms.
$\color {blue}{(x + y -z)^2} +z^2(-1 +a)$
$\color {blue}{(x + y -z)^2} +z^2(a-1)$
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