Determinant of exponential matrix

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Homework Help Overview

The discussion revolves around the determinant of an exponential matrix, specifically the expression Det( ## e^A ## ) and its relationship to the trace of matrix A. Participants explore the implications of similarity transformations and the properties of the trace in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of assuming that ## SAS^{-1} ## is a similarity matrix of A without verification. There are inquiries about the necessity of calculating eigenvalues and eigenvectors to confirm this assumption. Some participants emphasize the importance of the cyclic property of the trace in solving the problem.

Discussion Status

The discussion is active, with participants providing insights into the properties of similarity matrices and the trace. There is a recognition that the cyclicity of the trace is sufficient for the problem at hand, despite some initial uncertainty regarding assumptions made about the similarity matrix.

Contextual Notes

Participants note that the problem does not explicitly state that ## SAS^{-1} ## is a similarity matrix of A, leading to questions about the need for verification. The discussion also highlights the importance of understanding the properties of invertible matrices in this context.

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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Yes.
 
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?
 
##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.
 
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?
 
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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Thanks, I got it.
 

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