The simultaneous reduction to diagonal form of two real symmetric quadratic forms. Consider the two real symmetric quadratic forms $u^{T} A u$ and $u^{T} B u$, where $u^{T}$ stands for the row matrix $(x \quad y \quad z)$, and denote by $\mathrm{u}^{n}$ those column matrices that satisfy
$$
\mathrm{Bu}^{n}=\lambda_{n} \mathrm{Au}^{n}
$$
in which $n$ is a label and the $\lambda_{n}$ are real, non-zero and all different.
(a) By multiplying (E9.1) on the left by $\left(\mathrm{u}^{m}\right)^{\mathrm{T}}$ and the transpose of the corresponding equation for $\mathrm{u}^{m}$ on the right by $\mathrm{u}^{n}$, show that $\left(\mathrm{u}^{m}\right)^{\mathrm{T}} \mathrm{Au}^{n}=0$ for $n \neq m$
(b) By noting that $A u^{n}=\left(\lambda_{n}\right)^{-1} B u^{n}$, deduce that $\left(\mathrm{u}^{m}\right)^{\mathrm{T}} \mathrm{Bu}^{n}=0$ for $m \neq n$. It can be shown that the $u^{n}$ are linearly independent; the next step is to construct a matrix $\mathrm{P}$ whose columns are the vectors $\mathrm{u}^{n}$.
(c) Make a change of variables $u=P v$ such that $u^{\mathrm{T}}$ Au becomes $v^{\mathrm{T}} C v$, and $u^{\mathrm{T}} B u$ becomes $v^{\mathrm{T}} D v$. Show that $C$ and $D$ are diagonal by showing that $c_{i j}=0$ if $i \neq j$ and similarly for $d_{i j}$
Thus $\mathrm{u}=\mathrm{Pv}$ or $\mathrm{v}=\mathrm{P}^{-1} \mathrm{u}$ reduces both quadratics to diagonal form.
To summarise, the method is as follows:
(a) find the $\lambda_{n}$ that allow (E9.1) a non-zero solution, by solving $|\mathrm{B}-\lambda \mathrm{A}|=0$;
(b) for each $\lambda_{n}$ construct $\mathrm{u}^{n}$;
(c) construct the non-singular matrix $\mathrm{P}$ whose columns are the vectors $\mathrm{u}^{n}$;
(d) make the change of variable $\mathrm{u}=\mathrm{Pv}$.