Simultaneous Orthogonalization

Let $q,q':\mathbb V \longrightarrow \mathbb R$ be two quadratic form 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? –  copper.hat Jun 3 '12 at 7:16

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 $q(\mathbf{x}) = \mathbf{x}^T M_q \mathbf{x}.$

• Matrizes $M_q$ 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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I have not seen any theorem about simultaneous orthogonalization. I was wondering if you could help me with it –  Babak Miraftab Jun 3 '12 at 7:12
@babgen: are you following a course or reading a book? How did you come up with this question? –  Fabian Jun 3 '12 at 7:18
@babgen: The standard result about simultaneous diagonalizabilty of two quadratic forms should be covered in your book/lecture note. Alternatively, you can read the wiki article. By the way, wiki is a bit strange. First they assume $M$ to be positive definite and then they say they don't need is: fact is that only $N$ has to be positive definite. –  Fabian Jun 3 '12 at 7:24
I am reading Classical groups and Geometric algebra by Larry grove, but this book doesn't have theorem about this subject –  Babak Miraftab Jun 3 '12 at 7:25
@babgen: I added some more information. I hope you manage to put the remaining pieces together. –  Fabian Jun 3 '12 at 7:34