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logm and expm function

7 vues (au cours des 30 derniers jours)
mingcheng nie
mingcheng nie le 13 Sep 2022
Modifié(e) : Torsten le 13 Sep 2022
Hi there,
%If we consider a diagnal matrix defined as:
delta=diag(exp(1i*2*pi*(0:10)/11));
%and we define another matrix C as:
C=diag(1i*2*pi*(0:10)/11);
%so, if we compute this:
D=logm(delta);
%D should be equal to C, right? but the result is not, part of elements in D is equal to C but not all of them. I can't fix this issue.
Many thanks to you!
Charlie
  2 commentaires
Dyuman Joshi
Dyuman Joshi le 13 Sep 2022
Modifié(e) : Dyuman Joshi le 13 Sep 2022
Edit - Check Walter Roberson's answer for more details on the question.
In your matrix delta, all terms except for the diagonal terms are 0. And thus using logm() would not be ideal.
delta=diag(exp(1i*2*pi*(0:10)))
delta =
1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 - 0.0000i
Though you can use expm() on C and compare it with delta
C=diag(1i*2*pi*(0:10));
D=expm(C);
isequal(delta,D)
ans = logical
1
mingcheng nie
mingcheng nie le 13 Sep 2022
thank you for your answer, I make sense now!

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Réponse acceptée

Walter Roberson
Walter Roberson le 13 Sep 2022
The problem you are running into is that you are assuming that log(exp(X)) == X for all X. However, that is only true for real numbers. For complex numbers, with the sin() going on, values are mapped to the primary branch. exp(2i*pi*N) for integer N is going to involve 1i*sin(2*pi*N) which is going to have a complex part of 0 for all integer N.
format long g
Pi = sym(pi);
%If we consider a diagnal matrix defined as:
delta = diag(exp(1i*2*Pi*(0:10)))
delta = 
%and we define another matrix C as:
C = diag(1i*2*Pi*(0:10))
C = 
%so, if we compute this:
D=logm(delta)
D = 
C - D
ans = 
  7 commentaires
mingcheng nie
mingcheng nie le 13 Sep 2022
what if we set
delta=diag(exp(1i*2*pi*(0:10)/11));
then
C=diag(1i*2*pi*(0:10)/11);
as delta vary from 0 to 10/11, we might not get the N is integer, so I think the expm(C) == delta right?
Torsten
Torsten le 13 Sep 2022
Modifié(e) : Torsten le 13 Sep 2022
expm(diag(v)) = diag(exp(v))
for every vector v.
Here, you have no restrictions.
v = [log(675) log(i) log(-8)];
expm(diag(v))
ans =
1.0e+02 * 6.7500 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0100i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -0.0800 + 0.0000i
diag(exp(v))
ans =
1.0e+02 * 6.7500 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0100i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -0.0800 + 0.0000i

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