See: eigenvalues and eigenvectors.

Set of eigenvalues of a linear operator.

Unlike the simple case of a matrix, in infinite dimensional vector spaces, the spectrum may be continuous.

The quintessential example of that is the spectrum of the position operator in quantum mechanics, in which any real number is a possible eigenvalue, since the particle may be found in any position. The associated eigenvectors are the corresponding Dirac delta functions.

The general result from eigendecomposition of a matrix:becomes:where $O$ is an orthogonal matrix, and therefore has $O^{-1}=O^T$.

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:

Two symmetric matrices $A$ and $B$ are defined to be congruent if there exists an $S$ in $GL(n)$ such that:

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.

This is the possibly infinite dimensional version of a Hermitian matrix, since linear operators are the possibly infinite dimensional version of matrices.

There's a catch though: now we don't have explicit matrix indices here however in general, the generalized definition is shown at: en.wikipedia.org/w/index.php?title=Hermitian_adjoint&oldid=1032475701#Definition_for_bounded_operators_between_Hilbert_spaces

A good definition is that the sparse matrix has non-zero entries proportional the number of rows. Therefore this is Big O notation less than something that has $N^2$ non zero entries. Of course, this only makes sense when generalizing to larger and larger matrices, otherwise we could take the constant of proportionality very high for one specific matrix.

Of course, this only makes sense when generalizing to larger and larger matrices, otherwise we could take the constant of proportionality very high for one specific matrix.

Forms a normal subgroup of the general linear group.

Forms a normal subgroup of the general linear group.