Ciro Santilli @cirosantilli 34

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 $x\cdot x=1$. But after a rotation of 90 degrees, it becomes $x_2=(0,1)$, and now $x_2\cdot x_2=2$! Therefore, we have to search for an isomorphism between the two sets of matrices.

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:

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.

The main interest of this theorem is in classifying the indefinite orthogonal groups, which in turn is fundamental because the Lorentz group is an indefinite orthogonal groups, see: all indefinite orthogonal groups of matrices of equal metric signature are isomorphic.

It also tells us that a change of basis does not the alter the metric signature of a bilinear form, see matrix congruence can be seen as the change of basis of a bilinear form.

The theorem states that the number of 0, 1 and -1 in the metric signature is the same for two symmetric matrices that are congruent matrices.

For example, consider:

The eigenvalues of $A$ are $1$ and $4$, and the associated eigenvectors are:symPy code:and from the eigendecomposition of a real symmetric matrix we know that:

Now, instead of $P$, we could use $PE$, where $E$ is an arbitrary diagonal matrix of type:With this, would reach a new matrix $B$:Therefore, with this congruence, we are able to multiply the eigenvalues of $A$ by any positive number $e_1^2$ and $e_2^2$. Since we are multiplying by two arbitrary positive numbers, we cannot change the signs of the original eigenvalues, and so the metric signature is maintained, but respecting that any value can be reached.

Note that the matrix congruence relation looks a bit like the eigendecomposition of a matrix:but note that $D$ does not have to contain eigenvalues, unlike the eigendecomposition of a matrix. This is because here $S$ is not fixed to having eigenvectors in its columns.

But because the matrix is symmetric however, we could always choose $S$ to actually diagonalize as mentioned at eigendecomposition of a real symmetric matrix. Therefore, the metric signature can be seen directly from eigenvalues.

Also, because $D$ is a diagonal matrix, and thus symmetric, it must be that:

What this does represent, is a general change of basis that maintains the matrix a symmetric matrix.

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