Determinant of exponential matrix

In summary, the conversation discusses the use of similarity matrices in finding the determinant of ## e^A ##. It is stated that ##SAS^{-1}## is a similarity matrix of A and the cyclic property of the trace can be used to solve the problem without checking all the properties of similarity matrices.
  • #1
Pushoam
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Homework Statement



upload_2017-12-23_1-26-42.png

Homework Equations

The Attempt at a Solution


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Det( ## e^A ## ) = ## e^{(trace A)} ##

## trace(A) = trace( SAS^{-1}) = 0 ## as trace is similiarity invariant.

Det( ## e^A ## ) = 1

The answer is option (a).

Is this correct?

But in the question, it is not given that S AS-1 is a similarity matrix of A. I assumed it without checking.
So, should I check it by calculating eigen values of both S and ## SAS^{-1}## and their eigen vectors?
 

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  • #2
Yes.
 
  • #3
Orodruin said:
Yes.
Pushoam said:
But in the question, it is not given that ##S AS{-1}## is a similarity matrix of A. I assumed it without checking.
So, should I check it by calculating eigen values of both S and ## SAS^{-1} ##and their eigen vectors?
 
  • #4
##S## is clearly a unitary matrix. Anyway, all you really need to know is that ##S## is invertible (otherwise ##S^{-1}## does not exist) and the cyclic property of the trace.
 
  • #5
Orodruin said:
Anyway, all you really need to know is that SSS is invertible (otherwise S−1S−1S^{-1} does not exist) and the cyclic property of the trace.
If there exists a non – singular matrix P, then ## P^{-1} A P ## is known as a similarity matrix of A.

Similarity matrix is the name of ## P^{-1} A P ##.

Why is ## P^{-1} A P ## called a "similiarity matrix" of A? Is there any intuitive reason behind the word " similiarity"?

Now, A and its similiarity matrix have the following properties:

1) They have the same set of eigen values.

2) If x is eigen vector of A, then ## P^{-1} x ## is the eigen vector of corresponding similiarity matrix ## P^{-1} A P ##.

3) Trace( A ) = Trace (## P^{-1} A P ## )
Pushoam said:
But in the question, it is not given that ## SAS^{-1} ##. is a similarity matrix of A. I assumed it without checking.
So, should I check it by calculating eigen values of both S and ## SAS^{-1}## and their eigen vectors?

##SAS^{-1}## is a similarity matrix of A by definition. There is no need to check it.

Is this correct?
 
  • #6
Yes, but the point is that you do not need all those properties to solve this problem. You just need the cyclicity of the trace.
 
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  • #7
Thanks, I got it.
 

1. What is the determinant of an exponential matrix?

The determinant of an exponential matrix is a value that represents the scaling factor of the matrix. It is calculated using the exponential function and can be used to determine if a matrix is invertible.

2. How is the determinant of an exponential matrix calculated?

The determinant of an exponential matrix is calculated by raising the base of the exponential function to the sum of the diagonal elements of the matrix. This can be represented as det(eA) = etr(A) where tr(A) is the trace of the matrix.

3. What is the significance of the determinant of an exponential matrix?

The determinant of an exponential matrix is significant because it can determine if a matrix is invertible. A non-zero determinant indicates that the matrix is invertible, while a zero determinant indicates that the matrix is singular and cannot be inverted.

4. Can the determinant of an exponential matrix be negative?

Yes, the determinant of an exponential matrix can be negative. This can occur when the matrix has a negative determinant and the exponential function is raised to an odd power, resulting in a negative value.

5. How does the determinant of an exponential matrix change with scalar multiplication?

The determinant of an exponential matrix is affected by scalar multiplication. If the scalar is multiplied by a matrix, the determinant is raised to the power of the scalar. For example, if a matrix A is multiplied by a scalar k, the determinant becomes det(ekA) = (det(eA))k.

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