2x2 Projection matrix of rank 1
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Rik
2022년 11월 24일
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.
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Matt J
2022년 11월 23일
편집: Matt J
2022년 11월 23일
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
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Matt J
2022년 11월 23일
Again, you do not provide what you think is the correct answer, or an explanation of why that answer is correct..
추가 답변 (1개)
Moiez Qamar
2022년 11월 24일
%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
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Matt J
2022년 11월 24일
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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