Ciro Santilli @cirosantilli 34

When we have a symmetric matrix, a change of basis keeps symmetry iff it is done by an orthogonal matrix, in which case:

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:

Every invertible matrix $M$ can be written as:where:Note therefore that this decomposition is unique up to swapping the order of eigenvectors. We could fix a canonical form by sorting eigenvectors from smallest to largest in the case of a real number.

Intuitively, Note that this is just the change of basis formula, and so:

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.

Related: