Talk:Triangular matrix

This is my first edit and I'm unsure of whether what i've done is right... could someone please confirm for me? The main edit was atomic lower/upper inverses TheFlyingPengwyn 06:46, 30 Mar 2005 (UTC)

Replied on User talk:TheFlyingPengwyn. -- Jitse Niesen 16:01, 30 Mar 2005 (UTC)

Inverses/products of triangular matrices

The article clearly states that products of upper triangular matrices are upper triangular, but it doesn't make the similar (and also true) claim about lower triangular matrices. Further, I only vaguely get the impression that the inverses of upper/lower triangular matrices remain upper/lower triangular. We should probably state these properties more directly, and perhaps clean up the article in general. --Rriegs 04:11, 5 May 2007 (UTC)

Removed false claim

I deleted a line from the article falsely claimed that

"Indeed, we have

$\mathbf {L} ^{-1}={\begin{bmatrix}1&&&&&0\\&\ddots &&&&\\&&1&&&\\&&-l_{i+1,i}&\ddots &&\\&&\vdots &&\ddots &\\0&&-l_{n,i}&&&1\\\end{bmatrix}},$

i.e. the off-diagonal entries are replaced by their opposites."

Except for the first sub-diagonal, the inverse of an atomic lower triangluar is not quite as simple as reversing signs. Consider this counter example:

$\mathbf {L} ={\begin{bmatrix}1&&\\2&1&\\3&4&1\\\end{bmatrix}}$

Notice that ${\begin{bmatrix}1&&\\2&1&\\3&4&1\\\end{bmatrix}}{\begin{bmatrix}1&&\\-2&1&\\-3&-4&1\\\end{bmatrix}}={\begin{bmatrix}1&&\\0&1&\\-8&0&1\\\end{bmatrix}}\neq \mathbf {I}$

Tom Lougheed 01:17, 12 August 2007 (UTC)

going by the artcle's terminology, the matrix in your e.g. is not a "Gauss matrix". article only claims that formula holds when a matrix is Gauss. Mct mht 01:25, 12 August 2007 (UTC)