Lagmental Vicfred

Sylvester's Law of Inertia Classifies Real Quadratic Forms by Vicfred

A real symmetric form is congruent to a diagonal matrix with only plus one, minus one, and zero. The point is to make the formal expression readable enough to audit line by line.

Set-up

An inner product \(\langle x,y\rangle\) converts algebraic decompositions into orthogonal ones. Self-adjoint maps satisfy \(T=T^\ast\) and have real spectral data.

$$ q(x)=x^{\mathsf T}Ax,\qquad A=A^{\mathsf T} $$

I read the first line as input and the second as output. The symbols \(\forall\) and \(\exists\) are not interchangeable, and neither may be upgraded silently to \(\Longleftrightarrow\).

$$ S^{\mathsf T}AS=\operatorname{diag}(I_p,-I_q,0_r) $$

The calculation

The following line is the smallest calculation that still exercises the mechanism. It keeps nested delimiters and the order of operations explicit.

$$ (p,q,r)\ \text{is invariant under congruence},\qquad p+q+r=n $$

What survives abstraction

The two-row display is also a debugging tool: if the conclusion changes when only notation changes, some hidden choice has entered the argument.

$$ \begin{aligned} \mathsf{D}\;&:\quad q(x)=x^{\mathsf T}Ax,\qquad A=A^{\mathsf T},\\[5pt] \mathsf{C}\;&:\quad S^{\mathsf T}AS=\operatorname{diag}(I_p,-I_q,0_r). \end{aligned} $$

The boundary

Orthogonal diagonalization requires self-adjointness over the real or complex inner-product setting. A general diagonalizable matrix need not have orthogonal eigenvectors.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] S^{\mathsf T}AS=\operatorname{diag}(I_p,-I_q,0_r) \end{gathered}} $$

I would use the boxed line as a reference later, while returning to the full display whenever a hypothesis becomes uncertain. That division keeps compression from becoming ambiguity.

This article was posted on Mon 26 June 2017. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.