Ciro Santilli $$Sponsor Ciro$$ 中国独裁统治 China Dictatorship 新疆改造中心、六四事件、法轮功、郝海东、709大抓捕、2015巴拿马文件 邓家贵、低端人口、西藏骚乱
🔗

# Sylvester's law of inertia

| nosplit | ↑ parent "Eigendecomposition of a matrix" | words: 527 | descendant words: 626 | descendants: 5
🔗
🔗
🔗
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: $$A=[22​​2​3​] (125)$$
🔗
The eigenvalues of are and , and the associated eigenvectors are: $$v1​=[−2​,1]Tv4​=[2​/2,1]T (126)$$ symPy code:
A = Matrix([[2, sqrt(2)], [sqrt(2), 3]])
A.eigenvects()

and from the eigendecomposition of a real symmetric matrix we know that: $$A=PDPT=[−2​1​2​/21​][10​04​][−2​2​/2​11​] (127)$$
🔗
Now, instead of , we could use , where is an arbitrary diagonal matrix of type: $$[e1​0​0e2​​] (128)$$ With this, would reach a new matrix : $$B=(PE)D(PE)T=P(EDET)PT=P(EED)PT (129)$$ Therefore, with this congruence, we are able to multiply the eigenvalues of by any positive number and . 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: $$D=SMST (130)$$ but note that does not have to contain eigenvalues, unlike the eigendecomposition of a matrix. This is because here is not fixed to having eigenvectors in its columns.
🔗
But because the matrix is symmetric however, we could always choose to actually diagonalize as mentioned at eigendecomposition of a real symmetric matrix. Therefore, the metric signature can be seen directly from eigenvalues.
🔗
Also, because is a diagonal matrix, and thus symmetric, it must be that: $$ST=S−1 (131)$$
🔗
What this does represent, is a general change of basis that maintains the matrix a symmetric matrix.
🔗

🔗