Determinant of exponential matrix

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
Pushoam
Messages
961
Reaction score
53

Homework Statement



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

Homework Equations

The Attempt at a Solution


[/B]


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?
 

Attachments

  • upload_2017-12-23_1-26-42.png
    upload_2017-12-23_1-26-42.png
    10.2 KB · Views: 2,504
on Phys.org
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?
 
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?