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This is a seemingly simple question but I'm having some problems with it:

Suppose $q:V \to R$ is a positive semidefinite quadratic form.

How can I show that $$L_0=\{v\in V | q(v)=0\}$$ is a subspace of dimension $n-\rho$ where $\rho$ is $q$ s rank?

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2 Answers

up vote 1 down vote accepted

Let $\varphi$ be the symmetric bilinear form such that $q(v)=\varphi(v,v)$. If $q(v_1)=q(v_2)=0$, then $q(v_1+v_2) = 2 \varphi(v_1,v_2)$ and $q(v_1-v_2) = -2 \varphi(v_1,v_2)$ and both have to be nonnegative, so $\varphi(v_1,v_2)=0$ and so $v_1 + v_2 \in L_0$. It is obvious that $0 \in L_0$ and that $L_0$ is stable under scalar multiplication, so $L_0$ is indeed a vector space.

The assumption about the rank should be easy now.

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By the spectral theorem, you can reduce the matrix of $q$ to a diagonal matrix; since $q$ is positive, all its nonzero eigenvalues are positive. The rank of $q$ is equal to the number of positive eigenvalues (counted with multiplicity). The $v\in V$ such that $q(v)=0$ are then in the subspace spanned by the basis whose entries in the matrix of $q$ are zero; there are $n-\rho$ such vectors.

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Thanks. But I think this: "The $v \in V$ such that $q(v)=0$ are then in the subspace" requires further explanation, such as the one provided by Plop. –  daniel.jackson May 28 '11 at 18:10
    
I don't this so. Since you have reduced $q$ to a diagonal matrix, in the basis, you have $q(v) = \sum_{k=1}^n \lambda_i v_i^2$. All the terms in the sum are nonnegative, so if $q(v) = 0$, then all the terms are zero; for nonzero $\lambda_i$, $v_i=0$, thus $q(v) = 0 \Leftrightarrow v \in ker(Q)$ (where $Q$ is the matrix of $q$ in the eigenbasis). –  Najib Idrissi May 28 '11 at 19:17
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