Simultaneous Orthogonalization Let $q,q':\mathbb V \longrightarrow \mathbb R$ be two quadratic forms, where $\mathbb V$ is vector space with $\dim \mathbb V \geq3$ and $q(x)+q'(x)>0$ for any $0\neq x\in \mathbb V$. Then there exists a basis for $\mathbb V$ such that is orthogonal relative both $q$ and $q'$.
I have not any idea how to deal with it. Any suggestions ?
Thanks.
 A: Hint:
(1) perform a "basis"-change and introduce two new quadratic forms
$$ p(x) = q(x) + q'(x)\quad \text{and} \quad p'(x) = q(x) -q'(x).$$
(2) most likely you have proven some theorem about simultaneous orthogonalization which can be applied to $p$ and $p'$
(3) think what this means for $q$ and $q'$...
Edit: (some more hints)


*

*Choosing an arbitrary basis $\mathbf e_i$ for $\mathbb{V}$, you can associate a symmetric matrix with each quadratic form via
$$(M_q)_{ij} = \tfrac12[ q(\mathbf{e}_i + \mathbf{e}_j) -q(\mathbf e_i) - q(\mathbf e_j) ]$$
such that $q(\mathbf{x}) = \mathbf{x}^T M_q \mathbf{x}.$

*Matrices $M_q$ and $M_q'$ belonging to different basis choices (but the same quadratic form) are related via congruence,
$$ M_q = T^T M_q' T$$
with $T$ invertible.

*Having two symmetric matrices $M$ and $N$, and $N$ being positive definite they can be simultaneously diagonalized (via congruence relation) by solving the generalized eigenvalue problem $$(M - \lambda N) \mathbf{v}_\lambda =0.$$
See the wiki article.
