Talk:Constrained generalized inverse

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

Assessment comment

The comment(s) below were originally left at Talk:Constrained generalized inverse/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

If

A\in \mathbb {R} ^{m\times n}

then

x\in \mathbb {R} ^{n}

, not

\mathbb {R} ^{m}

--Prime factor (talk) 06:58, 15 October 2008 (UTC)[reply]

== Bott-Duffin inverse ==

Bott-Duffin inverse is defined in a different context. We are looking for a solution of

$Ax+b=c$ ,

$x\in L$ ,

$b\in L^{\perp }$ ,

where $A\in S^{n\times n}$ is a symmetric matrix of order $n\times n$ , $L$ is a linear subspace of order $n$ , $L^{\perp }$ is an orthogonal compliment to $L$ . Then the solution is given by

$x=P_{L}(AP_{L}+P_{L^{\perp }})^{-1}c$ ,

where $P_{L}$ and $P_{L^{\perp }}$ are orthogonal projectors on $L$ and $L^{\perp }$ .

--Prime factor (talk) 07:16, 15 October 2008 (UTC)[reply]

Last edited at 07:16, 15 October 2008 (UTC). Substituted at 01:55, 5 May 2016 (UTC)

Dimensions

The article does not make sense in its current state: matrix A cannot be m-by-n if both x and b are from ℝ^m. — Preceding unsigned comment added by 209.141.188.140 (talk) 12:52, 20 July 2019 (UTC)[reply]