Ciro Santilli @cirosantilli 34

Given a matrix $A$ with metric signature containing $m$ positive and $n$ negative entries, the indefinite orthogonal group is the set of all matrices that preserve the associated bilinear form, i.e.:Note that if $A=I$, we just have the standard dot product, and that subcase corresponds to the following definition of the orthogonal group: Section "The orthogonal group is the group of all matrices that preserve the dot product".

As shown at all indefinite orthogonal groups of matrices of equal metric signature are isomorphic, due to the Sylvester's law of inertia, only the metric signature of $A$ matters. E.g., if we take two different matrices with the same metric signature such as:and:both produce isomorphic spaces. So it is customary to just always pick the matrix with only +1 and -1 as entries.

If $C$ is the change of basis matrix, then the matrix representation of a bilinear form $M$ that looked like:then the matrix in the new basis is:Sylvester's law of inertia then tells us that the number of positive, negative and 0 eigenvalues of both of those matrices is the same.

Proof: the value of a given bilinear form cannot change due to a change of basis, since the bilinear form is just a function, and does not depend on the choice of basis. The only thing that change is the matrix representation of the form. Therefore, we must have:and in the new basis:and so since:

Related:

From effect of a change of basis on the matrix of a bilinear form, remember that a change of basis $C$ modifies the matrix representation of a bilinear form as:

So, by taking $S=C^T$, we understand that two matrices being congruent means that they can both correspond to the same bilinear form in different bases.

Like the matrix representation of a bilinear form, it is a matrix, but now the matrix has to be a symmetric matrix.

We can then immediately see that the matrix is symmetric, then so is the form. We have:But because $B(x,y)$ is a scalar, we have:and: