Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|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

1 Answer 1

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.