How do I write out symbolically what this function does in the case n= 5 for the iteration with j = 3, Using ordinary element wise vector and matrix notation

function [L, U] = lufacto(A)

% LUFACT LU factorization

% Input:

% A square matrix

% Output:

% (L, U) unit lower triangular and upper triangular such that LU = A

n = length(A);

L = eye(n); % ones on diagonal

% Gaussian elimination

for j = 1:n-1

L(j+1:n, j) = A(j+1:n, j) / A(j, j);

A(j+1:n, j:n) = A(j+1:n, j:n) - L(j+1:n, j) * A(j, j:n);

end

U = triu(A);

end

[Dyuman Joshi]

n = length(A);

Using length() for non-vector, non-square arrays can be confusing (for the lack of a better word).

Instead use the more robust option - size, where the output for a 2D array is

[number_of_rows number_of_columns]

Answer 1

[Pratyush]

Hi Muhideen,

I understand you want to represent the calculation steps for n=5, in the iteration j=3 using Matrix notation.

Symbolically, the calculations for the given iteration can be represented as:

• L(4, 3) = A(4, 3) / A(3, 3)
• L(5, 3) = A(5, 3) / A(3, 3)
• A(4, 3:5) = A(4, 3:5) - L(4, 3) * A(3, 3:5)
• A(5, 3:5) = A(5, 3:5) - L(5, 3) * A(3, 3:5)

Note: The above notation assumes 1-based indexing.

[Muhideen Ogunlowo]

I tried doing this but for the line

• L(5, 3) = A(5, 3) / A(3, 3),

It says "Index in position 1 exceeds array bounds, Index must not exceed 4"

[Dyuman Joshi]

What is the size of A?

[Muhideen Ogunlowo]

A is a 3 X 3 matrix

[Dyuman Joshi]

You are trying to access the 4th and the 5th row of a 3x3 matrix, which is not possible, hence the error.