Find a change of variable that will reduce the quadratic form $x_1^2-x_3^2-4x_1x_2+4x_2x_3$ to a sum of squares, and express the quadratic form in terms of the new variable.
Asked
Active
Viewed 7,485 times
4
-
1You can complete the squares to do this. – Seth Mar 26 '14 at 01:46
-
what would be the next step after completing the square? thank you for your help – user0692 Mar 26 '14 at 02:06
-
1Any thoughts about my answer? – Gerry Myerson Mar 29 '14 at 23:12
-
Yes that's what I got to in the end. So Q(Py)=3y1^3-3y2^2? – user0692 Mar 30 '14 at 00:33
-
1A quadratic form is not going to have a $y_1^3$ in it. – Gerry Myerson Mar 31 '14 at 13:23
-
yes that was a typo sorry, meant to put 3y1^2-3y2^2 – user0692 Apr 01 '14 at 20:28
2 Answers
3
Call the quadratic form, $Q(x)$. Write down the symmetric matrix $A$ such that $Q(x)=x^tAx$; that would be $$A=\pmatrix{1&-2&0\cr-2&0&2\cr0&2&-1\cr}$$ Since $A$ is symmetric, there is an orthogonal matrix $P$ such that $P^tAP=D$ is diagonal. Define new variables $y=(y_1,y_2,y_3)$ by $x=Py$. Then $$Q(x)=Q(Py)=(Py)^tAPy=y^tP^tAPy=y^tDy=\lambda_1y_1^2+\lambda_2y_2^2+\lambda_3y_3^2$$
Do you know how to find $P$?
Gerry Myerson
- 179,216
-
This is more of a number theory question. These $\lambda_1, \lambda_2, \lambda_3 $ are the eigenvalues of $A$. If we know the matrix $A$ then they can easily be computed. But suppose I have a quadratic form $f(x)= \sum_{i=1}^{n-1}a_i x_i ^2$ s.t. it represents $0$ non trivially, then how will I get this $P$? My $A$ will be a diagonal matrix with $a_i$ s entries. – Dark_Knight Mar 14 '17 at 03:56
-
-
1If the $a_i$ aren't given, then what is? If $A$ is already diagonal, then you can take $P$ to be the identity matrix. I'm probably not understanding what it is you want to ask. But whatever it is you want to ask, maybe you should ask it as a new question. – Gerry Myerson Mar 14 '17 at 04:19
-
http://math.stackexchange.com/questions/2186265/find-a-new-quadratic-form-through-change-of-variables I have posted my question here. – Dark_Knight Mar 14 '17 at 13:10
2
$$ x^2-z^2-4xy-4yz = x^2 +4y^2 -4xy - \left(4y^2+z^2+4yz\right) $$ and now you can trivially complete the square. Now substitute to get the sum of squares.
Note it's not expected to be the same sum of squares as @Gerry Myerson's answer.
-
could you go into more detail, I don't understand how you obtained this. – user0692 Mar 26 '14 at 02:15
-
Yes, rename $\left(x_1,x_2,x_3\right) \to (x,y,z)$ for convenience, then add and subtract $y^2$ and then factor out the $-$ in front of the $y,z$ terms. – gt6989b Mar 26 '14 at 02:17
-
-
after you are done with arithmetic you end up with $(...)^2 + (*)^2$ so change variables to have $a = ...$ and $b = *$ and you get $a^2 + b^2$ – gt6989b Mar 26 '14 at 02:23