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I’ve been trying to prove the following:

Let $k$ be an algebraically closed field with $\text{char}(k)\neq2$, and let $Q$ be a non-singular quadratic form on $k^n$. Show that for some choice of basis of $k^n$ we can write $Q=r_1X_1^2+\cdots+r_nX_n^2$.

Here is my approach so far:

We can construct an $n\times n$ matrix $A=(a_{ij})$ such that $Q(v)=v^TAv$ by setting $a_{ii}$ as the coefficient of $X_i^2$ in $Q$, and $a_{ij}=a_{ji}$ as $\frac{1}{2}$ the coefficient of $X_iX_j$. We also have an associated bilinear form $$b(u,v)=\frac{1}{2}[Q(u+v)-Q(u)-Q(v)]=v^TAu$$

Then with respect to the standard basis $e_1,\ldots,e_n$, we have $a_{ij}=b(e_i,e_j)$.

Now, by induction on $n$ we can find a basis $f_1,\ldots f_n$ of $k^n$ which is orthogonal with respect to $b$.

(The $n=0$ case is trivial, then if $b(v,v)=0$ for all $v\in k^n$ we have $Q(v)=0$ for all $v$ and so any basis is orthogonal. Otherwise we have some $b(f_1,f_1)\neq0$, consider $kf_1\oplus(kf_1)^\perp$. It can be shown that $n=\dim{kf_1}+\dim{(kf_1)^\perp}$ since $Q$ is non-singular, so $k^n=kf_1\oplus(kf_1)^\perp$ and we apply the induction hypothesis to $(kf_1)^\perp$.)

Let $S$ be the change of basis matrix for this basis. I would like to conclude that $$(Sv)^TA(Sv)=v^TDv$$ for some diagonal matrix $D$. But I believe all we have shown is that $S^{-1}AS=D$.

If we could prove that $S^T=S^{-1}$ then we could make this final deduction, but I’m struggling to show that it is. Any help would be much appreciated.

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I think you're overcomplicating matters. You have $\ f_1, f_2, \dots, f_n\ $ orthogonal with respect to $\ b\ $, so if you write an arbitrary member $\ x\ $ of $\ k^n\ $ as $\ x=\sum_{i=1}^n X_i\,f_i\ $, then you have $\ Q\left(x\right) = b\left(x,x\right)\ $. What happens if you now substitute $\ x=\sum_{i=1}^n X_i\,f_i\ $ into $\ b\left(x,x\right)\ $ and use the bilinearity of $\ b\ $ to expand this expression out in terms of the $\ X_i\ $?

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  • $\begingroup$ Ah so then $Q(x)=\sum_{i=1}^{n}x_i^2b(f_i,f_i)$, and so using our new coordinates we have $Q=\sum_{i=1}^{n}b(f_i,f_i)\bar{X_i}^2$. Many thanks $\endgroup$ – Dave Mar 10 at 10:51

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