Lagmental Vicfred

A Quotient Space Removes a Subspace Direction by Vicfred

tags: mathematics, vector spaces, linear maps, dual spaces

Vectors differing by an element of U become equal in V modulo U. The formulas are more useful when each symbol has a job rather than merely decorating the theorem.

Definitions first

A linear map \(T:V\to W\) is organized by its kernel \(\ker T\) and image \(\operatorname{im}T\). Quotients and duals express the same information without choosing bases.

$$ V/U=\{v+U:v\in V\} $$

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 \(\Longleftrightarrow\).

$$ \dim(V/U)=\dim V-\dim U $$

A small case in full

The computation below is not a second theorem. It is a checksum for the definitions and a place to inspect the difficult LaTeX at full size.

$$ 0\longrightarrow U\xrightarrow{\ i\ }V\xrightarrow{\ \pi\ }V/U\longrightarrow0 $$

The reusable statement

The invariant statement is the one that does not depend on a convenient choice of coordinates, representatives, basis, or enumeration.

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

A nearby false statement

Dimension formulas below assume finite-dimensional spaces. Infinite-dimensional vector spaces require cardinal arithmetic and may not identify naturally with their double duals.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] \dim(V/U)=\dim V-\dim U \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Thu 16 December 2010. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.