If A is a tridiagonal Matrix , what does this mean ?

  • Thread starter Thread starter Maths Lover
  • Start date Start date
  • Tags Tags
    Matrix Mean
Click For Summary
A tridiagonal matrix is characterized by having non-zero elements only on the main diagonal and the diagonals directly above and below it. In a 4x4 tridiagonal matrix, this results in three possible non-zero diagonals, while all other entries are zero. The discussion highlights that tridiagonal matrices are efficient for numerical methods, particularly in solving second-order boundary value problems, as they allow for storage of only the three diagonals. In contrast, a dense matrix contains a high proportion of non-zero entries, while a banded matrix has non-zero entries concentrated in diagonal bands. Understanding these matrix types is crucial for efficient computational methods in linear algebra.
Maths Lover
Messages
67
Reaction score
0
if A is a tridiagonal Matrix , what does this mean ?

what does tridiagonal mean in matrix ?
what is the property which A achieve to be tridiagonal ?

what about " dense matrix " and " band matrix " ?

what is the meanning in this words ?
 
Physics news on Phys.org
If A is a diagonal n x n matrix, A has non-zero elements only at indices (k,k), k belongs to {1, n}

so for a 4x4 matrix it looks like this

a 0 0 0
0 b 0 0
0 0 c 0
0 0 0 d

Tridiagonal matrices can have non-zero elements also at indices (k, k+-1)

so a 4x4 tridiagonal matrix would look like this:

a b 0 0
c d e 0
0 f g h
0 0 i j

,a,b,c,d,e,f,g,h,i,j can be anything (also 0)

you can see the pattern, it has 3 possibly nonzero diagonals, one in the middle, one above it and one below it. everything else is 0.

edit: fixed something
 
Last edited:
In practice, there exist methods of solving problems involving tridiagonal matrices so that all succeeding matrices in the solution are also triadiagonal and you only have to store the data on the three diagonals, not all of the 0's.

It is also true that, representing a second derivative numerically, as (f(x+h)- 2f(x)+ f(x-h))/2h, uses only the three values of x, x+h, and x-h. If you solve a second order boundary value problem numerically, you will get a tridiagonal matrix.

A "dense" matrix is the opposite of a "sparse" matrix. A "dense matrix" has a relatively large number of non-zero entries. An n by n matrix has n^2 entries. If it is "tridiagonal" only n+ n-1+ n-1= 3n- 2 of them are non-0, a ratio of (3n- 2)/n^2 so, especially for large n, a sparse matrix.

A "banded matrix" is a matrix in which non-zero entries tend to occur in diagonal "bands". Again, there are methods of working with such matrices that "keep" that property and only the non-zero entries have to be stored.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
13
Views
6K
  • · Replies 33 ·
2
Replies
33
Views
1K
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 20 ·
Replies
20
Views
11K
  • · Replies 8 ·
Replies
8
Views
7K