2x2 Projection matrix of rank 1
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2 Comments
Rik
on 24 Nov 2022
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.
Accepted Answer
Matt J
on 23 Nov 2022
Edited: Matt J
on 23 Nov 2022
a=[1; 2]; n=[3; 4]; x=[5; 6];
r1p(a,n,a)
r1p(a,n,n)
r1p(a,n,x)
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
4 Comments
Matt J
on 23 Nov 2022
Again, you do not provide what you think is the correct answer, or an explanation of why that answer is correct..
More Answers (1)
Moiez Qamar
on 24 Nov 2022
%should work for:
a=[1; 0.01]
n=[0.01; 1]
x=[1; 0]
p=r1p(a,n,x)
%and for:
a=[1; 0];
n=[0; 1];
x=[1; 0];
p = r1p(a,n,x);
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
1 Comment
Matt J
on 24 Nov 2022
P=a*chi'
outer products are not efficient. That's why the exercise asks for you to compute p without computing P.
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