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My book states:

if $q: \mathbb{R}^n \to \mathbb{R}$ is a positive definite quadratic form then there exists a basis $B=(v_1,...,v_n)$ such that $q(x)=x_1^2+...+x_n^2=||x||^2$ for every $x=x_1v_1+...+x_nv_n$

But what do I know about the basis $B$? Is it orthonormal? If not, I'm having a hard time justifying the fact that $x_1^2+...+x_n^2=||x||^2$.

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yes, it is orthonormal. –  wildildildlife Jul 4 '11 at 9:42
    
@wild: Care to elaborate? –  daniel.jackson Jul 4 '11 at 9:43
    
@daniel.jackson: For existence of $B$, just use the Gram-Schmidt process, or one of the numerically better processes, to produce an orthonormal basis. –  André Nicolas Jul 4 '11 at 9:51

3 Answers 3

up vote 3 down vote accepted

Yes, this is in fact implied by the statement. Let $B$ denote the associated symmetric bilinear form, i.e. $2B(x,y)=q(x+y)-q(x)-q(y)$. Of course for $x=v_i$ we have $x_j=\delta_{ij}$. Now

$q(v_i)=x_i^2=1^2=1$,

and if $i\neq j$ then

$2B(v_i,v_j)=q(v_i+v_j)-q(v_i)-q(v_j)=(x_i^2+x_j^2)-x_i^2-x_j^2=(1^2+1^2)-1^2-1^2=0$.

Thus the basis is orthonormal.

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The problem here seems to be that you seem to be taking $\lVert x \rVert$ to mean the canonical norm on $\mathbb R^n$, whereas your book uses it to mean the norm defined by $q$. This has nothing to do with the basis $B$; you can just take out the sum of the squares of the coefficients $x_i$ and are left with the equation $q(x)=\lVert x \rVert^2$. The basis $B$ doesn't occur here, and the equation clearly shows that $\lVert x \rVert$ is to be interpreted as the norm defined by $q$ and not the canonical norm. The notion of orthonormality, too, is relative to the norm/quadratic form, so, as wildildildlife has shown, $B$ is indeed orthonormal, but orthonormal with respect to $q$, not in the canonical sense.

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You say the norm defined by q, but I'm not sure I understand that. Does that mean that if $f$ is the associated symmetric bilinear form then $f$ also defines an inner product in $R^n$ such that $f(u,v)=\langle u,v \rangle $ and $q(v)=\langle v,v \rangle = \lVert v \lVert^2$? –  daniel.jackson Jul 5 '11 at 9:09
    
@daniel.jackson: Yes, that's what I meant. –  joriki Jul 6 '11 at 23:37

The basis $B$ of the question can be obtained by the Gram-Schmidt orthonormalization process, or by any one of several other processes that have better numerical stability properties.

With respect to the basis obtained from Gram-Schmidt, the quadratic form collapses to the shape $\sum x_i^2$ (that's what orthonormal basis means), meaning that the norm defined by the quadratic form is exactly $\sum x_i^2$. The fact that this is the norm defined by the quadratic form is not an additional fact, it is built into the assertion that with respect to $B$, the quadratic form assumes a certain shape.

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I don't understand why this is true. By orthogonal diagonalization you can bring $q$ to the form $\sum \lambda_i x_i^2$, how did you conclude that $\lambda_i=1$ in this case? I know that there is a basis where $q(x)=\sum x_i^2$ but why is it necessarily orthogonal? Wikipedia says something similar too. –  daniel.jackson Jul 5 '11 at 10:18
    
@daniel.jackson: We were told that the form is positive definite (of course it it wasn't it couldn't give a norm). So the $\lambda_i$ are positive, and if we choose an orthonormal base, the $\lambda_i$ are $1$. And the base is orthonormal because we used Gram-Schmidt. Or alternately, a very efficient proof is given in the post by wildililife. –  André Nicolas Jul 5 '11 at 13:56

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