Find an orthogonal basis for the form on $R^3$ given by the matrix $\begin{pmatrix} 1 & 0 & 1 \\ 0 &2 &1 \\ 1 &1& 1 \end{pmatrix}$ .
I have been trying to find how to find a basis for a bilinear form but I'm still not very sure how – and the fact that the form is given by a matrix is making it more confusing for me.
Do I use Gram Schmidt algorithm on the matrix directly? But what are my $v_1,v_2,v_3$? Or do I compute $v^T \begin{pmatrix} 1 & 0 & 1 \\ 0 &2 &1 \\ 1 &1& 1 \end{pmatrix} w$ and then apply Gram Schmidt algorithm?
Otherwise, is there a more efficient method to find such basis?
Thanks!
Best Answer
There doesn't seem to be any real reason for avoiding the Gram-Schmidt algorithm in this situation. The first two steps terminate easily and the last one is not very complicated.
Start with $u_1=(1,0,0)$ since it is obviously unit length with respect to this form.
Then $(0,1,0)$ turns out to be already perpendicular to this, but it needs to be rescaled because its length is $\sqrt{2}$. Thus $u_2=(0,\frac{1}{\sqrt{2}},0)$.
Finally, the only real calculation is involved when you try to process $(0,0,1)$. Taking away the contributions of $u_1$ and $u_2$, you're left with $(-1,-\frac{1}{2},1)$, whose square norm is $\frac{-1}{2}$, so it would be appropriate to use $u_3=(-\sqrt{2},-\frac{1}{\sqrt{2}},\sqrt{2})$ so that $u_3\cdot u_3=-1$.
That gives an "orthonormal" basis: they are pairwise orthogonal and $u\cdot u\in\{1,-1\}$ for all $u$ in the basis, which is to be expected since the signature of this metric is +,+,-.