= Solution
The <spectral theorem for real symmetric matrices> supplies an <orthonormal basis> of real <eigenvectors>. Put these in the columns of $Q$, so $Q^TQ=I$ and $Q^TAQ=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. In the question's coordinate convention choose $O=Q^T$ and $y=Ox$. Since $x=O^Ty$,
$$
x^TAx=y^TOAO^Ty=\sum_{i=1}^n\lambda_i y_i^2.
$$
Reorder the <eigenvalues> with $m$ positive, $\ell-m$ negative and $n-\ell$ zero. For nonzero eigenvalues set $z_i=\sqrt{|\lambda_i|}y_i$, and retain the remaining coordinates as $z_i=y_i$. This is an invertible real transformation $z=DOx$, where $D$ is the diagonal rescaling; the combined matrix $L=DO$ gives $z=Lx$. It gives
$$
x^TAx=\sum_{i=1}^mz_i^2-\sum_{i=m+1}^{\ell}z_i^2,
$$
with the zero coordinates absent. The general allowed range is \b[$0\leq m\leq\ell\leq n$]. The printed restriction $m\geq1$ is not valid for every real symmetric matrix: a negative-definite or zero form has no positive square. The invariance of these counts is <Sylvester's law of inertia>.
For the specified form, a particularly simple change is
$$
\boxed{z=Lx,\quad L=\begin{pmatrix}1/2&1/2&1/2\\1/2&-1/2&-1/2\\0&1&-1\end{pmatrix},\qquad x_1x_2+x_1x_3=z_1^2-z_2^2.}
$$
Subtracting the first two squared coordinates verifies the identity directly, and $\det L=1$ proves invertibility. Thus $m=1$, $\ell=2$ and there is one zero direction. The final rescaling need not remain orthogonal.
Back to article page
