Lagmental Vicfred

Orthogonal Projection Solves the Closest-Point Problem by Vicfred

Last updated: Tue 06 September 2022

tags: mathematics, inner products, spectral theorem, quadratic forms

Projection onto a closed subspace is characterized by an orthogonal residual. I will separate the object being defined from the consequence being claimed.

Notation

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.

$$ v=P_Uv+(v-P_Uv) $$

The formulas should not be merged too early. The datum \(\mathsf D\), the conclusion \(\mathsf C\), and the bridge \(\Longrightarrow\) have three different logical jobs.

$$ P_Uv\in U,\qquad v-P_Uv\in U^\perp $$

Stress the formula

The middle display is intentionally dense: it is where signs, bounds, multiplicities, or normalising factors are most likely to be lost.

$$ U=\operatorname{span}\{e_1,\ldots,e_r\}\ \text{orthonormal}\Longrightarrow P_Uv=\sum_{i=1}^{r}\langle v,e_i\rangle e_i $$

Interpretation

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad v=P_Uv+(v-P_Uv),\\[5pt] \mathsf{C}\;&:\quad P_Uv\in U,\qquad v-P_Uv\in U^\perp. \end{aligned} $$

Limit of the argument

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

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] P_Uv\in U,\qquad v-P_Uv\in U^\perp \end{gathered}} $$

The final box is a summary, not a new assumption; the proof still lives in the definitions and the intervening calculation. The source keeps each scope delimiter visible for later inspection.

This article was posted on Tue 24 May 2022. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.