# How could I find an orthogonal basis of this bilinear form f?

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?

Edit:
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?

• 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. – amd Jun 1 '16 at 21:15
• First find the eigenvectors, then apply the Graham-Schmidt process. – Doug M Jun 1 '16 at 22:34
• @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. – silverjoe Jun 1 '16 at 22:36
• Thanks as well @Doug M I'll try that way as well and see how it turns out – silverjoe Jun 1 '16 at 22:38
• 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. – amd Jun 2 '16 at 17:13

Diagonalize orthogonally $\;A\;$ , which is possible since $\;A\;$ is symmetric:
$$|xI-A|=\begin{pmatrix}x-1&-1&0\\-1&x-1&-1\\0&-1&x-1\end{pmatrix}=(x-1)^3-2(x-1)=$$
$$(x-1)\left(x^2-2x-1\right)=(x-1)(x-1-\sqrt2)(x-1+\sqrt2)$$