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".

$O(m,n)=O∈M(m+n)∣∀x,yx_{T}Ay=(Ox)_{T}A(Oy)$

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.

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Following the definition of the indefinite orthogonal group, we want to show that only the metric signature matters.

First we can observe that the exact matrices are different. For example, taking the standard matrix of $O(2)$:
and:
both have the same metric signature. However, we notice that a rotation of 90 degrees, which preserves the first form, does not preserve the second one! E.g. consider the vector $x=(1,0)$, then $x⋅x=1$. But after a rotation of 90 degrees, it becomes $x_{2}=(0,1)$, and now $x_{2}⋅x_{2}=2$! Therefore, we have to search for an isomorphism between the two sets of matrices.

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For example, consider the orthogonal group, which can be defined as shown at the orthogonal group is the group of all matrices that preserve the dot product can be defined as:

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