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

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SUMMARY

A tridiagonal matrix is defined as a square matrix where non-zero elements are present only on the main diagonal and the diagonals immediately above and below it. For example, a 4x4 tridiagonal matrix can be represented as having non-zero entries at indices (k, k), (k, k-1), and (k, k+1). This structure allows for efficient storage and computation, particularly in numerical methods for solving second-order boundary value problems. 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.

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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 ?
 
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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.
 

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