Where $f : \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ corresponding to the quadratic form $q : \mathbb{R}^3 \rightarrow \mathbb{R}$,

$q(x,y,z) = x^2 + 2xy + y^2 + 2yz + z^2$

I found that if $u = (x,y,z)$ and $q(u,u) = u^tAu = q(x,y,z) = x^2 + 2xy + y^2 + 2yz + z^2$,

$A= \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} $

I've tried using the Gram-Schmidt Process however I am getting the problem of dividing by zero. What would be the best way for me to find the orthogonal basis in this case?

Let $B = \{b_1, b_2,b_3\}$ where $b_1 = (1, 0, 0)$, $b_2 = (0, 1, 0)$, $b_3 = (0, 0, 1)$. Now let $C = \{c_1, c_2,c_3\}$ be an the orthonormal basis generated iteratively using the Gram-Schmidt method.

Let $f(u,v) = u^tAv$ and $c_1 = b_1$
$c_2 = b_2 - \frac{f(b_2, c_1)}{f(c_1, c_1)}c_1$ $= (0,1,0) - \frac{1}{1}(1,0,0)$
$c_2 = (-1, 1, 0)$
$c_3 = b_3 - \frac{f(b_3, c_1)}{f(c_1, c_1)}c_1 - \frac{f(b_3, c_2)}{f(c_2, c_2)}c_2$

At this point $f(c_2, c_2) = 0$, would I maybe need to retrace and pick another $c_2$ or is this one ok?

  • $\begingroup$ You should provide more information if you’d like us to show you where you’re going wrong. To what basis are you applying the G-S process? How did you get it? Where are you getting division by zero, which shouldn’t happen unless you’ve made a mistake along the line. $\endgroup$ – amd Jun 1 '16 at 21:15
  • $\begingroup$ First find the eigenvectors, then apply the Graham-Schmidt process. $\endgroup$ – Doug M Jun 1 '16 at 22:34
  • $\begingroup$ @amd Oh sorry about that I've just added some details to the question, the main problem is computing the third vector in the orthogonal basis. $\endgroup$ – silverjoe Jun 1 '16 at 22:36
  • $\begingroup$ Thanks as well @Doug M I'll try that way as well and see how it turns out $\endgroup$ – silverjoe Jun 1 '16 at 22:38
  • $\begingroup$ The procedure that you tried will produce a set of orthogonal vectors relative to the scalar product $f(u,v)=u^TAv$, but that’s not what you’re being asked to do. You should be looking for an orthogonal basis of $\mathbb R^3$ in which the cross terms of $q$ vanish. A way to do that is to find the eigenvectors of $A$ as others have suggested. In addition, the reason that the $G-S$ process as you’re using it breaks down is that $f$ is not positive-definite. $\endgroup$ – amd Jun 2 '16 at 17:13

Diagonalize orthogonally $\;A\;$ , which is possible since $\;A\;$ is symmetric:



Now just calculate eigenvectors of the above eigenvalues, which will be orthogonal (by symmetry, again). If you want the basis orthonormal just divide each of those eigenvectors by its length, and there you have an orthonormal basis.


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