Let $Q(x_1,x_2, \ldots ,x_n)$ be a positive definite real quadratic form in the variables $x_1, \ldots ,x_n$. It is not hard to see that the function $f(x_1, x_2, \ldots ,x_n)=\frac{Q(x_1,x_2, \ldots ,x_n)}{x_n^2}$ attains a minimum value on ${{\mathbb R}^{n-1}} \times {{\mathbb R}^*}$, an that this value (call it $\lambda$) is positive.

Then $R(x_1,x_2, \ldots, x_n)=Q(x_1,x_2, \ldots ,x_n)-{\lambda}x_n^2$ is a nonnegative real quadratic form. Is it true that $R$ can always be written as a sum of $n-1$ squares (of linear forms) ?

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    $\begingroup$ $$A \text{ a real symmetric matrix } \quad\implies\quad A = Q\Lambda Q^{T} \quad\text{ where } \Lambda \text{ diagonal and } Q \text{ orthogonal. }$$ $\endgroup$ – achille hui May 25 '13 at 8:24
  • $\begingroup$ $R$ has 0 as eigenvalue. So, it can be diagonalized and written as $n-1$ squares of linear forms. $\endgroup$ – i707107 May 25 '13 at 8:26
  • $\begingroup$ @i707107 are we sure that $0$ is an eigenvalue of $R$ ? Looks like it, but it’s not clear how to actually prove it. $\endgroup$ – Ewan Delanoy May 25 '13 at 8:29
  • $\begingroup$ That's because the minimum $\lambda$ is obtained for nonzero $x_n$. Thus, for nonzero $x_n$, the quadratic form $R$ attains zero. $\endgroup$ – i707107 May 25 '13 at 8:32

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