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We've recently learned about metric signatures following the proof of Sylvester's law of inertia but we didn't quite say which properties does the signature of a given matrix $A\in \mathcal{M}_n\left(\mathbb{R}\right) $ have. Let's call it $(n_{_+} ,n_{_-} ,n_{_0} )$. What I'm asking is what are the main things I can conclude based on a signature?

For example given $B:\mathbb{R}^6\times\mathbb{R}^6\to \mathbb{R}$ and a singature $(3 ,3 ,0 )$ what does it tell me?

Is there a subspace $U\subseteq \mathbb{R}^6$ on which $B\Bigl|_{U} \equiv0$ ?

What's the maximal (dimension-wise) subspace of $\mathbb{R}^6$ on which $ B$ is positive/negative definite ?

When reading Wikipedia it seems like this is the kind of information one can conclude from a signature, but unfortunately I could not find a list of properties nor proofs to those properties.

Edit: It seems that the $n_{_0}$ is not like the others in the sense that for example considering $\left[\mathbf{B^{\phantom{}}}\right] =\left(\begin{matrix}1 & 0 & 0\\ 0 & -1 & 0 \\ 0 &0 &1\end{matrix}\right)$, the signature is $(2,1,0)$ but if we take $ U = \mbox{span} \left \lbrace e_1 + e_2 \right \rbrace$ in that basis we get that:

$v,u\in U \Rightarrow v=\alpha\left(e_1 + e_2 \right) \land v=\beta\left(e_1 + e_2 \right),\ \forall v,u\ \ \ B(v,u) = B(\alpha(e_1 + e_2 ),\beta(e_1 + e_2 ))= \alpha \beta B(e_1 + e_2 ,e_1 + e_2 ) = \alpha \beta \left(B(e_1,e_1) + \overbrace{B(e_1,e_2)}^0 + \overbrace{B(e_2,e_1)}^0 + B(e_2,e_2)\right) = \alpha \beta (1-1) = 0 \Rightarrow B\left(U\right)\equiv 0$

Proposition $\min(n_{_+},n_{_-}) + n_{_0} $ is the maximal dimension of $U\subseteq \mathbb{R}^n$ on which $B\Bigl|_{U} \equiv0$ ?

Is there a simple proof or counter example to this property ?

share|cite|improve this question
What do you understand by $B\big|_U$? $\forall u\in U\, Bu=0$ (in this case your example is invalid), or $\forall u\in U\, (Bu,u)=0$? (in this case it's easy to study your hypothesis by the technique I used for $n_+$) – TZakrevskiy Jun 11 '13 at 9:00
@TZakrevskiy well obviously not the first one, I don't think I ever mentioned $\forall u\in U Bu=0$ :) and not the second one either (at least not directly...), what I mean is : is there some sub-space of $\mathbb{R}^n $ on which $B$ if confined to this subspace will be the zero Bi-linear form – Scis Jun 11 '13 at 14:18
I edited my post to reflect this definition. Your hypothesis is right. – TZakrevskiy Jun 11 '13 at 19:51
up vote 1 down vote accepted

Let's take a matrix $A=A^\ast$ with signature $(n_+,n_-,n_0)$. It means we have an eigenvalue $0$ of multiplicity $n_0$; this allows to conclude on the existence of your subspace $U$ ($U\ne 0\iff n_0>0$). In the same spirit, if $n_+>0$, then we have positive eigenvalues, and on the subspace generated by corresponding eigenvectors we have our matrix $A$ as positive definite; apparently, the maximum dimension is equal to $n_+$, because that's exactly the number of eigenvectors for positive eigenvalues. Same goes for the case $n_->0$.

Edit A more formal approach. 1) $A=A^\ast$, hence our matrix is diagonalisable, has an orthonormal basis of eigenvectors. It's signature is $(n_+,n_-,n_0)$. The eigenvectors $\vec e_k^+$ correspond to positive eigenvalues, similarly for $\vec e_k^-$ and $\vec e_k^0$.

2) Let $E_+$ be a subspace on which $A$ is positive definite and its dimension is $N$. We have $N$ independent vectors $\vec u_j\in E_+$, each of them has non-zero coordinates with respect to $\vec e_k^+$ (otherwise they wouldn't belong here). We call $\vec u^+_j$ the orthogonal projection of $\vec u_j$ on $\text{span}\{\vec e_k^+\}$. Suppose that the family $\vec u^+_j$ is linearly dependent, then there exists a linear combination of these vectors which is zero. The same linear combination of vectors $\vec u_j$ would have its components only in $\text{span}\{\vec e_k^-\}+\text{span}\{\vec e_k^0\}$, which is impossible in $E_+$. This implies that the family $\vec u^+_j$ is independent. We can have a family of at most $n_+$ linearly independent vectors in $\text{span}\{\vec e_k^+\}$, thus $N\le n_+$.

We can conclude by taking $E_+=\text{span}\{\vec e_k^+\}$ that $dim\,E_+$ can be equal to $n_+$.

3) Same reasoning applies for $E_-$.

4) Let's define $E_0$ as a maximum subspace where $A$ is zero a bilinear map. Clearly, all $\vec e_k^0$ are in it. This allows to write $E_0=\text{span}\{\vec e^0_k\}\oplus E$ where $E$ is a subspace generated only by vectors that do not contain $\vec e^0_k$ in their development. Suppose we have a basis $\vec u_j$ in $E$, and its orthogonal projections $\vec u^+_j$, $\vec u^-_j$. Clearly, if the family $\vec u^+_j$ is not linearly independent, then there exist a linear combination of $\vec u^+_j$ which is zero. The same linear combination $v$ of $\vec u_j$ lies in $\text{span}\{\vec e^-_k\}$, which can't happen in $E_0$ since $(Av,v)<0$. Hence, the family $\vec u^+_j$ is linearly independent; similarly, the family $\vec u^-_j$ is independent, too. Given that $\,dim\, \text{span}\{\vec e^\pm_k\} =n_\pm$, we conclude that $\,dim\, \text{span}\{\vec u_j\}\le \min(n_-,n_+)$ and $\,dim\, E_0\le \min(n_-,n_+)+n_0$. It is quite easy to build such $E_0$ with its dimension precisely equal to $\min(n_-,n_+)+n_0$, which concludes the proof.

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Thanks but I was looking for a more rigorous proof for those properties, because I did understand what do $ n_{+} ,n_{-} ,n_{_0} $ mean from their names really (they're quite indicative I'll give 'em that :) ). I mean that I do find this intuitive but I'm having hard time thinking about a good way to formalize it. – Scis Jun 10 '13 at 20:21
I'll edit my post to make it more "formal". – TZakrevskiy Jun 10 '13 at 20:23
Thanks, I agree on the $\pm$ cases but it seems that you're wrong about the 0 case (see the counter example in the edit). – Scis Jun 11 '13 at 7:25

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