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Let $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ be a quadratic form and

$A = \begin{bmatrix} 7 & -5 & 4 \\ -5 & 7 & 4 \\ 4 & 4 & -2 \end{bmatrix}$ its matrix. Such that $Q=X^TAX$ for $X=\begin{bmatrix} x \\y \\ z \end{bmatrix}$.

Find an $X$ such that $X^TAX=72$.

I found that the eigenvalues of $A$ are $-6,6\text{ and }12$, so the matrix is not positive definite neither negative definite.

How can I solve this kind of problem? Can you give me a hint? Thanks

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    $\begingroup$ Maybe eliminate those crazy crossproduct terms first. I manage to eliminate them with $$P = \begin{bmatrix} -1 &-1 &1 \\ 1 &-1 &2 \\ 0 &2 &1 \end{bmatrix}.$$ $\endgroup$ – IAmNoOne Jun 16 '14 at 9:19
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    $\begingroup$ You could always let $x=\sqrt{72/7}$, $y=z=0$. $\endgroup$ – Gerry Myerson Jun 16 '14 at 9:58
  • $\begingroup$ And for $x=z=0$ you have next solution. $\endgroup$ – Widawensen May 24 '16 at 15:34
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Find an eigenvector $X$ for the eigenvalue $6$, then $Q(X)=6X^TX$, so if $X^TX=a^2\gt 0$, take $Y=\frac {\sqrt {12}}{a}X$ so that $Y^TY=\frac {12}{a^2} X^TX=12$ and $Q(Y)=72$.

You can do something similar for the eigenvalue $12$.

Alternatively, spot an easy answer.

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I found the follow technic for this kind of problem, that is very similar to the Mark Bennet's answer.

If $Q(x,y,z)=7x^2+7y^2-2z^2-10xy+8xz+8yz$ then $A$ is the associated matrix.

Let $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix} \neq0 \text{ and } \lambda \in \mathbb{R}$. If $AX=\lambda X$, then $\lambda$ is an eigenvalue of $A$ and $X$ is an eigenvector of $A$, associated to $\lambda$.

So, $Q(x,y,z)=X^TAX= \lambda X^T X= \lambda (x^2+y^2+z^2)$. One have that $(x^2+y^2+z^2) >0$, so if $\lambda (x^2+y^2+z^2)=72$, then $\lambda>0$.

Between the three eigenvalues of $A$, only $6 \text{ and } 12$ can be applied.

Now by finding the vector space span for the eigenvectors for $6 \text{ and } 12$, one got:

$E_{6}=\langle(1,1,1) \rangle$

$E_{12}=\langle(-1,1,0) \rangle$

About the $X$ that is asked, or $X \in E_{6}$ or $X \in E_{12}$. I found that $X \in E_{12}$:

Let $\begin{bmatrix} -a \\ a \\ 0 \end{bmatrix}$ be a generic vector of $E_{12}$. So , $12 \left((-a)^2+a^2+0^2 \right)=72$. Then $$12 \left(2a^2 \right)=72$$ $$ 24a^2 =72$$ $$a^2=3$$ $$a=\sqrt3$$ Finaly, a $X$ that satisfy what is asked is $X=\begin{bmatrix} -\sqrt3 \\ \sqrt3 \\ 0 \end{bmatrix}$. You can confirm, by computing $Q(x,y,z)$.

$$Q(-\sqrt3,\sqrt3,0)=7 \cdot (-\sqrt3)^2+7\cdot \sqrt3^2 -10 \cdot (-\sqrt3) \cdot \sqrt3$$

$$Q(-\sqrt3,\sqrt3,0)=7 \cdot 3+7\cdot 3+10 \cdot 3$$

$$Q(-\sqrt3,\sqrt3,0)=21+21+30=72$$

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  • $\begingroup$ Is not there for Q(x,y,z) an infinite number of solutions ? Why have you been searching for a particular one ? $\endgroup$ – Widawensen May 24 '16 at 15:34

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