How to perform pseudo inverse with set of elements in a matrix?

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Kalasagarreddi Kottakota
回答： John D'Errico 2021-11-24
A=(1:12).';
B=(1:4).';
C = pinv(reshape(A,length(B),[]))*B;
D = pinv(A(1:4))*B % D should be equal to first element of C
Iam looking for a function that performs multiplication as follows:
Matrix B has 4 elements and matrix A has 12 elements. I want to perform pinv(A)*B, but not to all elements of A at a time. I need pinv(A(1:4))* B and pinv(A(5:6))*B and so on...
I used the formulation mentioned in the script, but getting wrong results? first element of D should be equal to D.
No loops, only vectorization
Could some one help me in this regard?
% answers of D and C
C =
0.8333
0.3333
-0.1667
D =
1.0000

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回答（1 个）

John D'Errico
John D'Errico 2021-11-24
Ran in:
I think you misunderstand the linear algebra here.
A=(1:12).';
B=(1:4).';
A_43 = reshape(A,length(B),[]) % I'll create a variable for A_43, so we can use it several times
A_43 = 4×3
1 5 9 2 6 10 3 7 11 4 8 12
Is that matrix a full rank matrix? No.
rank(A_43)
ans = 2
If you now try to use a tool like pinv to effectively solve a system of equations, then you need to recognize the solution is not unique. So this now:
C = pinv(A_43)*B
C = 3×1
0.8333 0.3333 -0.1667
Yes, you think that it should be able to find the solution [1;0;0]. And it is true that
A_43*[1;0;0]
ans = 4×1
1 2 3 4
does return B. But since that matrix is not full rank, there are infintiely many possible solutions. The solution that pinv does find is one where the result has minimum norm over all possible solutions.
norm([1;0;0])
ans = 1
norm(C)
ans = 0.9129
Hoping to use a version of pinv that applies separately to every column of a matrix separately is NOT the solution. Understanding linear algebra is the solution. Sorry, but it is.
The total set of solutions to the general problem for this rank 2 matrix will be of the form:
syms t
allsols = pinv(A_43)*B + t*null(A_43)
allsols = 
And we can see that when t = 1/sqrt(6), this will yield the solution you think should be the correct one.
simplify(subs(allsols,t,1/sqrt(6)))
ans = 
But there are infintely many possible solutions. Note that \ does find the solution you think to be the correct one, but that is slightly fortuitous. And \ does complain about the singular matrix.
A_43\B
Warning: Rank deficient, rank = 2, tol = 1.875718e-14.
ans = 3×1
1.0000 0 0.0000

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