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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 http://mathworld.wolfram.com/MatrixSignature.html.

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    @user02138: Thanks!2012-10-17
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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 $q(x)=x^TAx$.2011-12-01