LU Decomposition and Linear Algebra

LU Decomposition:

In numerous applications where linear systems appear one needs to solve Ax = b for numerous differentvectors b. For illustration a structure must be tested under several different loads, not just one. As in the example of a truss the loading in such a problem is usually represented by the vector b.

Gaussian eradication with pivoting is the most efficient and accurate way to solve a linear system.Most of the work in this process is spent on the matrix Aitself. If we necessitate to solve several differentsystems with the same A and A is big then we would like to evade repeating the steps of Gaussianelimination on A for every different b. This is able to be accomplished by the LU decomposition whichin effect records the steps of Gaussian elimination.

LU decomposition:

The major idea of the LU decomposition is to record the steps used in Gaussian elimination on A inthe places where the zero is produced. Determine the matrix:

2368_LU decomposition.jpg

The first step of Gaussian eradication is to subtract 2 times the first row from the second row. Inorder to record what we have completed we will put the multiplier 2 into the place it was used to makea zero that is the second row first column. In order to create it clear that it is a record of the step andnot an element of A we will put it in parentheses. This leads to:

457_LU decompostion_2.jpg

There is before now a zero in the lower left corner therefore we don’t need to eliminate anything there. Werecord this fact with a (0). To eradicate the third row second column we require subtracting −2times the second row from the third row. Footage the −2 in the spot it was utilized we have:

530_LU decomposition_3.jpg

Let U be the upper triangular matrix produced as well as let L be the lower triangular matrix with therecords and ones on the diagonal that is:

1507_LU decomposition_4.jpg

Then we have the subsequent mysterious coincidence:

1762_LU decomposition_5.jpg

Therefore we see that A is actually the product of L and U. Here L is lower triangular as well as U isupper triangular. When a matrix be able to be written as a product of simpler matrices we call that adecomposition of A as well as this one we call the LU decomposition.
Using LU to solve equations:

If we as well include pivoting then an LU decomposition for A consists of three matrices P, L and U such that:

PA = LU.

The pivot matrix P is the identity matrix with the alike rows switched as the rows of Aare switched in the pivoting for illustration

7_pivot matrix.jpg

Would be the pivot matrix if the second moreover third rows of A are switched by pivoting. Mat lab will produce an LU decomposition with pivoting for a matrix A with the subsequent command

> [L U P] = lu(A)

Where P is the pivot matrix to utilize this information to solve Ax = b we first pivot both sides by multiplying by the pivot matrix

PAx = Pb ≡ d.

Substituting LU for PA we obtain

LUx = d.

Then we need merely to solve two back substitution problems

Ly = d


Ux = y.

In Mat lab this would work as follows

> A = rand(5,5)
> [L U P] = lu(A)
> b = rand(5,1)
> d = P*b
> y = L\d
> x = U\y
>rnorm = norm(A*x - b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Check the result.

We are able to then solve for any other b without redoing the LU step. Replicate the sequence for a new right hand side c = randn(5,1) you can start at the third line. While this mayn’t seem like a big savings it would be if A were a large matrix from a genuine application.

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