Write a MATLAB function to estimate the condition number of

Write a MATLAB function to estimate the condition number of the square matrix A in the two-norm using the symmetric power iteration. The function should be of the form [cond_est,u1,v1,u2,v2]=cond2(L,U,A) where L and U are from [L, U]=lu(A). Your function cannot call cond or inv or use y=A\\v or z= A\'\\u

Solution

You are to write a MATLAB function to implement to estimate the condition
number of the square matrix A in the two-norm using the symmetric power
iteration.
Your function should have the form [cond est,u1,v1,u2,v2]=cond2(L,U,A). where L and U are the out- put from the from the command [L,U]=lu(A) which does Gaussian elimination with partial pivoting.

The value cond est is your estimate for the
value
2 (A) = kA1 k2 kAk2 .
Thus, you will have to estimate both kA1 k2 and kAk2 using the power
method. The vectors u1,v1, u2 and v2 satisfy
ku1k2 = kv1k2 = ku2k2 = kv2k2 ,
kAv1k2 kAk2 ,
kAT u1k2 kAk2 ,
kA1 v2k2 kA1 k2 ,
kAT u2k2 kAT k2 .
These four vectors should emerge out of the power method for approximating
kAk2 and kA1 k2 . To find kAk2 you are implicitly performing the power
method on AT A, to find kA1 k2 you are doing the same for (AT A)1 =
A1 AT .
To estimate kAk2 , you should perform k power iterations to produce (k) ,
u1 and v1 such that
A v1 = (k) u1
and either
| (k) (k1) | (1e 3) (k) 1 or
kA0 u1 (k) v1k2 (1e 3) (k)
or
k = 10.
To estimate kA1 k2 , you should perform ` power iterations to produce (`) ,
u2 and v2 such that
A1 v2 = (`) u2
and either
| (`) (`1) | (1e 3) (`)
or
kAT u2 (`) v1k2 (1e 3) (`)
or
` = 10.
Your estimate of the condition number is thus, cond est = (`) (k) . Our
course, you can get away with storing only the two most recent values of (k)
and (`) .
In order to implement this method, you will need four matrix operations
with A.
1. A v for a vector v. this is just implmented as is in MATLAB.
2. AT u for a vector u.this is just A0 u in MATLAB.
3. A1 v for a vector v
4. AT u for a vector u.
To do the last two operations above, use the results from lu, namely a
factorization of A into A = LU where L is a permutation of a lower triangular
matrix and U is an upper triangular matrix. Remember that y = A1 v may
be computed from
w=L\\v
y=U\\w
and z = AT u may be computed from 2 w=U’\\u
z=L’\\w

function [ v d] = power_method( A )

% for finding the largest eigen value by power method

disp ( \' Enter the matrix whose eigen value is to be found\')

% Calling matrix A

A = input ( \' Enter matrix A :   \ \')

% check for matrix A

% it should be a square matrix

[na , ma ] = size (A);

if na ~= ma

    disp(\'ERROR:Matrix A should be a square matrix\')

    return

end

% initial guess for X..?

% default guess is [ 1 1 .... 1]\'

disp(\'Suppose X is an eigen vector corresponding to largest eigen value of matrix A\')

r = input ( \'Any guess for initial value of X? (y/n):   \',\'s\');

switch r

    case \'y\'

        % asking for initial guess

    X0 = input(\'Please enter initial guess for X :\ \')

        % check for initial guess

    [nx, mx] = size(X0);

    if nx ~= na || mx ~= 1

        disp( \'ERROR: please check your input\')

        return

    end

    otherwise

    X0 = ones(na,1);

end

%allowed error in final answer

t = input ( \'Enter the error allowed in final answer:  \');

tol = t*ones(na,1);

% initialing k and X

k= 1;

X( : , 1 ) = X0;

%initial error assumption

err= 1000000000*rand(na,1);

% loop starts

while sum(abs(err) >= tol) ~= 0

    X( : ,k+ 1 ) = A*X( : ,k); %POWER METHOD formula

    % normalizing the obtained vector

    [ v i ] = max(abs(A*X( : ,k+ 1 )));

    E = X( : ,k+ 1 );

    e = E( i,1);

    X(:,k+1) = X(:,k+1)/e;

    err = X( :,k+1) - X( :, k);% finding error

    k = k + 1;  

end

%display of final result

fprintf (\' The largest eigen value obtained after %d itarations is  %7.7f  \ \', k, e)

disp(\'and the corresponding eigen vector is \')

X( : ,k)

use the above code to produce power iteration.