logm and expm function

Question

mingcheng nie 2022-9-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1803995-logm-and-expm-function

编辑： Torsten 2022-9-13

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

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Dyuman Joshi 2022-9-13

编辑：Dyuman Joshi 2022-9-13

在 MATLAB Online 中打开

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 = 
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
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
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
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
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
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
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
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
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
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
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 2022-9-13

thank you for your answer, I make sense now!

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Answer 1

Walter Roberson 2022-9-13

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1803995-logm-and-expm-function#answer_1052280

在 MATLAB Online 中打开

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 =

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mingcheng nie 2022-9-13

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 2022-9-13

编辑：Torsten 2022-9-13

在 MATLAB Online 中打开

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