Square matrix to the power of a imaginary unit

In summary: In general, when solving a matrix equation by hand, it is often faster to solve for the eigenvalues and then use the inverse to solve for the elements.Thanks for reply.
  • #1
sunny110
11
0
Hi;

How to raising a square matrix to the power of a complex number?

for example:
[1 2;3 4]^(1+i)

or mathematics software such as Scilab how solve such problems?
-->[1 2;3 4]^(1+%i)
ans =

- 0.1482039 - 0.2030943
- 0.3046414 - 0.4528453


Thanks in advance.

-----
Additational remark:
for element by element power operation (.^) , the output value is equal to the principle value computed as:
z1.^z2 = exp( z2* ( log(abs(z1)) + %i*atan(imag(z1)/real(z1)) ) )


-->[1 2;3 4].^(1+%i)
ans =

1. 1.5384778 + 1.2779226i
1.3644973 + 2.6717311i 0.7338279 + 3.932111i
 
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  • #2
Depending on the matrix, there might not be any sensible way to do this. If the matrix is diagonalizable, then you can apply the exponent to the eigenvalues. In general, you can attempt to define the power using the logarithm of the matrix, ## A^p = \exp ( p \ln A)##. The logarithm can be defined as the matrix ##B## such that ##A = e^B##, or via a power series. In general, the logarithm only exists if ##A## is invertible, and when dealing with complex elements is not necessarily unique.
 
  • #3
Thanks for reply.
fzero said:
In general, you can attempt to define the power using the logarithm of the matrix, ## A^p = \exp ( p \ln A)##. .

yes, for element-wise power operation, in fact the software compute the above expression.
a .^(%i+1) == exp((%i+1)*log(a))

fzero said:
If the matrix is diagonalizable, then you can apply the exponent to the eigenvalues.

Let us return to the main problem i.e: a=[1 2;3 4]; a^(1+i)
eigenvalues and eigenvectors may be get as:
-->[v,d]=spec(a)
d =

- 0.3722813 0
0 5.3722813
v =

- 0.8245648 - 0.4159736
0.5657675 - 0.9093767

This matrix is diagonalizable: inv(v)*a*v == d

So still I don't know what algorithm the software use?


And another question, if one wants to solve this problem by hand, which method should be used?
 
  • #4
sunny110 said:
Let us return to the main problem i.e: a=[1 2;3 4]; a^(1+i)
eigenvalues and eigenvectors may be get as:
-->[v,d]=spec(a)
d =

- 0.3722813 0
0 5.3722813
v =

- 0.8245648 - 0.4159736
0.5657675 - 0.9093767

This matrix is diagonalizable: inv(v)*a*v == d

So still I don't know what algorithm the software use?

I'm not familiar with the software, so I don't know. In terms of computational complexity, multiplication of large matrices takes a similar number of operations as computing the inverse of a matrix. I would guess that using the power series definition of the log (when it converges) is slightly faster than diagonalizing the matrix and then inverting the matrix of eigenvectors. Also, there are invertible matrices that are not diagonalizable, so the software would have to be able to use more than one algorithm anyway.

And another question, if one wants to solve this problem by hand, which method should be used?

For a small matrix I would probably want to diagonalize first whenever possible. For a simple enough matrix, this could lead to an exact result. If only a numerical result is required, then the power series method might be faster, depending on the convergence properties.
 
  • #5


Hello,

Raising a square matrix to the power of a complex number involves using the properties of complex numbers and the properties of matrix operations. The process can be done by hand, but using mathematical software such as Scilab can make it easier and more accurate.

To raise a square matrix to the power of a complex number, we can use the following formula:

A^(a+bi) = A^a * (cos(b*ln(A)) + i*sin(b*ln(A)))

Where A is the square matrix and a+bi is the complex number. This formula uses the properties of complex numbers, specifically De Moivre's formula, and the properties of matrix operations, specifically the power of a matrix.

In Scilab, we can use the "^" operator to raise a matrix to a power, and the "%i" symbol to represent the imaginary unit. For example, [1 2;3 4]^(1+%i) will give us the result -0.1482039 - 0.2030943i -0.3046414 - 0.4528453i.

For element by element power operation, we can use the ".^" operator. In this case, the output value is equal to the principal value computed using the following formula:

z1.^z2 = exp(z2 * (log(abs(z1)) + %i * atan(imag(z1)/real(z1))))

Where z1 and z2 are the elements of the matrix and %i represents the imaginary unit. For example, [1 2;3 4].^(1+%i) will give us the result 1.0000000 1.5384778 + 1.2779226i 1.3644973 + 2.6717311i 0.7338279 + 3.932111i.

I hope this helps. Thank you for your question.
 

1. What is a square matrix?

A square matrix is a type of matrix that has an equal number of rows and columns. It is represented in a grid-like structure and is commonly used in various mathematical and scientific fields.

2. What is the power of a matrix?

The power of a matrix refers to the number of times a matrix is multiplied by itself. For example, if a matrix A is multiplied by itself three times, it can be written as A³.

3. What is an imaginary unit?

An imaginary unit is a complex number that, when squared, gives a negative result. It is represented by the letter "i" and is used in mathematical equations involving complex numbers.

4. How do you raise a square matrix to the power of an imaginary unit?

To raise a square matrix to the power of an imaginary unit, you can use the Euler's formula which states that e^(ix) = cos(x) + i*sin(x). This can be applied to the matrix as (e^(ix))^n = (cos(x) + i*sin(x))^n. Then, the matrix can be multiplied by itself n times to get the final result.

5. What are the applications of raising a square matrix to the power of an imaginary unit?

Raising a square matrix to the power of an imaginary unit is commonly used in various fields such as physics, engineering, and computer science. It is used to solve complex equations, model physical systems, and manipulate data in computer algorithms.

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