Determinant of exponential matrix

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The determinant of the exponential matrix is given by Det(e^A) = e^(trace A), leading to the conclusion that if trace A = 0, then Det(e^A) = 1. The discussion emphasizes that the similarity transformation S A S^{-1} is indeed a similarity matrix of A, which preserves the trace and eigenvalues. It is clarified that checking the eigenvalues and eigenvectors is unnecessary for this problem, as the cyclic property of the trace suffices. The key takeaway is that the properties of similarity matrices are not required for solving this specific determinant problem. Understanding these concepts is essential for correctly applying them in linear algebra contexts.
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Homework Statement



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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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