Is the rank and signature of a quadratic form $x^TAx$ basically the rank and signature of $A$? Thanks.
Rank and signature of a quadratic form
2 Answers
I'm not sure what percusse is talking about, but yes, that's how these invariants of quadratic forms are defined. In more detail, changing variables in the quadratic form corresponds to changing the matrix $A$ by congruence: $A\mapsto P^TAP$. Any quadratic form over $\mathbb R$ can be diagonalized by an orthogonal matrix to $q(x_1,\ldots,x_n)=x_1^2+\cdots+x_k^2-x_{k+1}^2-\cdots -x_n^2$. The signature of such a quadratic form is defined to be $(k,n-k)$, which is also the signature of the diagonal matrix with k $+1$'s and $n-k$ $-1$'s on the diagonal. It is a well-defined invariant by Sylvester's Law of Inertia. See http://mathworld.wolfram.com/MatrixSignature.html.
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1How do you define a rank for $x^TAx$ ? – 2011-12-01
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1@percusse: the rank of a quadratic form is the largest dimension of a subspace on which it is nonzero. Equivalently, after diagonalizing, it is the number of nonzero diagonal terms. – 2011-12-01
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1Yes, how do you diagonalize $x^TAx$, that's what I am asking. Because that's precisely the definition of rank $A$ using the inertia of $A$. – 2011-12-01
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1@percusse: The change of variables $x\mapsto Px$ corresponds to the transformation $(Px)^TAPx=x^T(P^TAP)x$, which is exactly a similarity transformation of the matrix $A$. So diagonalizing a quadratic form is equivalent to diagonalizing a matrix up to similarity, which can be done for a real symmetric matrix. – 2011-12-01
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1I really don't see it. First that's congruence (not similarity) what you perform and second, you are operating on the matrix and in the end evaluate the quadratic form. Then, using the properties of the matrix, you attach properties to the quadratic form. But for vector $x$, the term $x^TAx$, by definition, has rank 1. There is no way around it. It is a scalar and the matrix has rank $m\geq 1$. You can stretch this definition and say, well the matrix associated with this quadratic form has rank $m$, hence I will call it a rank-$m$ quadratic form but that you have to define beforehand. – 2011-12-01
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1@percusse: "You can stretch this definition and say, well the matrix associated with this quadratic form has rank m, hence I will call it a rank-m quadratic form but that you have to define beforehand." You got it. That is the standard definition of the rank of a quadratic form, just like the OP asked. – 2011-12-01
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1Is this definition equivalent to the definition saying that the signature of a matrix is. The number of positive eigenvalues minus the number of negative eigenvalues? – 2011-12-01
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1@percusse: thanks for asking those questions. Those were points that confused me too. – 2011-12-01
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1@Greg No problem at all. I have never seen such usage before but apparently it is the standard of some other field. Feels awfully awkward though. – 2011-12-01
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1The map $A \mapsto P^{\top} A P$ is a _congruence map_, not a similarity map. A similarity involves the inverse of $P$ not its transpose. – 2012-10-17
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1@user02138: Thanks! – 2012-10-17
No, (assuming lower case variables mean vectors or matrices of smaller sizes) $x^TAx$ is typically a scalar whereas $A$ is obviously a matrix. If $A$ is not a scalar itself then there is no relation.
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1Are you sure? The OP is not asking about the rank and signature of the scalar $x^TAx$, but rather what the definition is of these invariants for the quadratic form $q(x)=x^TAx$. – 2011-12-01