2x2 Projection matrix of rank 1

[Rik]

I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.

[john]

oops

Answer 1

[Matt J]

a=[1; 2]; n=[3; 4]; x=[5; 6];

r1p(a,n,a)

ans = 2×1

0.1789 0.3578

r1p(a,n,n)

ans = 2×1

1.0e-15 * -0.1986 -0.3972

r1p(a,n,x)

ans = 2×1

-0.1789 -0.3578

function p = r1p(a,n,x)
% computes the action of P, the 2x2 projection matrix of rank 1 having
% a - the sole basis vector for the column space of P
% n - the sole basis vector for the null space of P

    a = normalize(a(:),'n'); 
    b = normalize(null(n(:)'),'n');
    
    p = dot(x,b)*a;

end

[john]

This does not work. I dont understand why

[john]

The anwer for case a=[1; 0.01]; n=[0.01; 1]; x=[1; 0]; results in p = [.01;.0001] whereas the correct answer is p = [.1;.01]

[john]

Thank you so much for continuing to help me out. However, this now works for a=[1; 0.01]; n=[0.01; 1]; x=[1; 0]; (just have to negate it), but does not return the correct answer for a=[1; 2]; n=[3; 4]; x=[5; 6];

[Matt J]

Again, you do not provide what you think is the correct answer, or an explanation of why that answer is correct..

Answer 2

[Moiez Qamar]

%should work for:
a=[1; 0.01]

a = 2×1

1.0000 0.0100

n=[0.01; 1]

n = 2×1

0.0100 1.0000

x=[1; 0]

x = 2×1

1 0

p=r1p(a,n,x)

P = 2×2

1.0001 -0.0100 0.0100 -0.0001

p = 2×1

1.0001 0.0100

p = 2×1

1.0001 0.0100

%and for:
a=[1; 0];  
n=[0; 1];  
x=[1; 0];
p = r1p(a,n,x); 

P = 2×2

1 0 0 0

p = 2×1

1 0

function p = r1p(a,n,x)
    % computes the action of P, the 2x2 projection matrix of rank 1 having
    % a - the sole basis vector for the column space of P
    % n - the sole basis vector for the null space of P
    xi=[0 -1; 1 0]*n;
    chi=xi/(xi'*a);
    P=a*chi'
    p=P*x
end

[Matt J]

P=a*chi'

outer products are not efficient. That's why the exercise asks for you to compute p without computing P.