Matrix inverse with svd

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Elie Hatem
Elie Hatem el 7 de Jun. de 2021
Comentada: Arnab Paul el 27 de Oct. de 2023
Hello,
I am trying to invert a matrix using svd and \ and comparing the result with inv().
For this matrix:
A = [ 2 4 6; 8 5 7 ; 0 3 1]
I did the svd decomposition as follows:
[U,S,V] = svd(A);
And, I believe the inverse of A can be expressed as :
So, I tried it by writing the code below:
disp('inverse with svd:')
disp(V*S\U')
And, I got this answer:
inverse with svd:
0.0236 0.0440 -0.0528
0.1765 0.1361 0.1922
0.4277 -0.3787 -0.1247
Then, I tried to invert A using inv() as follows:
disp('inverse with inv:')
disp(inv(A))
However, I got a different answer as seen below:
inverse with inv:
-0.2000 0.1750 -0.0250
-0.1000 0.0250 0.4250
0.3000 -0.0750 -0.2750
Can someone tell me what I am doing wrong?
My main goal is that I want to use svd to invert large matrices that may be close to becoming singular.
Thank you

Respuesta aceptada

Christine Tobler
Christine Tobler el 7 de Jun. de 2021
Use one of
A = [ 2 4 6; 8 5 7 ; 0 3 1]
A = 3×3
2 4 6 8 5 7 0 3 1
[U,S,V] = svd(A);
V*inv(S)*U'
ans = 3×3
-0.2000 0.1750 -0.0250 -0.1000 0.0250 0.4250 0.3000 -0.0750 -0.2750
V*(S\U')
ans = 3×3
-0.2000 0.1750 -0.0250 -0.1000 0.0250 0.4250 0.3000 -0.0750 -0.2750
% The formula you're using above puts the parentheses in a different place
% than you think:
V*S \ U'
ans = 3×3
0.0236 0.0440 -0.0528 0.1765 0.1361 0.1922 0.4277 -0.3787 -0.1247
(V*S) \ U'
ans = 3×3
0.0236 0.0440 -0.0528 0.1765 0.1361 0.1922 0.4277 -0.3787 -0.1247
  2 comentarios
Elie Hatem
Elie Hatem el 7 de Jun. de 2021
Thank you so much!
Arnab Paul
Arnab Paul el 27 de Oct. de 2023
What happens to non-square matrix?
is it same as pinv(A)?

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