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

1. Nov 21, 2012

### Maths Lover

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

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

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

what is the meanning in this words ?

2. Nov 21, 2012

### aija

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: Nov 21, 2012
3. Nov 22, 2012

### HallsofIvy

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.