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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.

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    Could you elaborate what you mean by 'is orthogonal relative both q and q′' please?2012-06-03

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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.

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    @babgen: you don't necessarily need the generalized eigenvalue problem. The fact that two symmetric matrices can be simultaneously diagonalized if one of them is positive definite it is a standard result and cannot (as far as I know) been written down in two lines, so I won't do the trouble and prove it for you. You find it everywhere. After short goggling, I found [this reference](http://staff.science.uva.nl/~brandts/TEACHING/LINALG/PD$F$/hadamard.pdf) where it is proposition 11.2012-06-03