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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?


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I confess to also being confused! It seems to me the right question is to ask for the orthonormal basis of eigenvectors for the given matrix, since the form is symmetric. Thus not by applying Gram-Schmidt to (say) the rows of the matrix, but rather by finding its eigenvalues and respectively normalized eigenvectors. The point would be that the form with respect to that basis would be simply a diagonal matrix. – hardmath Feb 13 '13 at 4:03

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 +,+,-.

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