Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 32487

Questions where the notion matrix has an important or crucial role (for the latter, note the tag matrix-theory for potential use). Matrices appear in various parts of mathematics, and this tag is typically combined with other tags to make the general subject clear, such as an appropriate top-level tag ra.rings-and-algebras, co.combinatorics, etc. and other tags that might be applicable. There are also several more specialized tags concerning matrices.

0 votes

Sufficient conditions for invertibility of a block tridiagonal matrix

The following theorems generalize those in Tridiagonal matrices: inversion and conditioning to all rational eigenvalues except $1, 1/2, 1/3$, to complex eigenvalues, to block-tridiagonal matrices. … In particular, this covers tridiagonal Toeplitz matrices. …
kaba's user avatar
  • 397
1 vote

Sufficient conditions for invertibility of a block tridiagonal matrix

Conditions for tridiagonal matrices The following conditions are for tridiagonal matrices; i.e. $m_i = 1$ for each $i$. … In particular, this condition covers matrices which may not be strictly diagonally dominated. …
kaba's user avatar
  • 397
5 votes
2 answers
948 views

Sufficient conditions for invertibility of a block tridiagonal matrix

This is in contrast to general block matrices which may require pivoting to complete an LDU-decomposition. … \end{aligned}$$ The minor of a product of matrices $X \in \mathbb{R}^{M \times N}$ and $Y \in \mathbb{R}^{N \times P}$ is: $$\det((XY)_{I, J}) = \sum_{K \subset_{|I|}} \det(A_{I, K}) \det(B_{K, J}),$$ …
kaba's user avatar
  • 397