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Is the rank and signature of a quadratic form $x^TAx$ basically the rank and signature of $A$? Thanks.

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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

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How do you define a rank for $x^TAx$ ? – user13838 Dec 1 '11 at 2:51
@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. – Grumpy Parsnip Dec 1 '11 at 2:54
Yes, 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 $A$. – user13838 Dec 1 '11 at 2:55
@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. – Grumpy Parsnip Dec 1 '11 at 3:07
I 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. – user13838 Dec 1 '11 at 3:14

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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Are 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 $q(x)=x^TAx$. – Grumpy Parsnip Dec 1 '11 at 1:53

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