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

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


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 $Q(x,y,z)$.

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

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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.