# Homework Help: Determinant of exponential matrix

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1. Dec 22, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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

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

Det( $e^A$ ) = 1

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?

2. Dec 22, 2017

### Orodruin

Staff Emeritus
Yes.

3. Dec 22, 2017

### Pushoam

4. Dec 22, 2017

### Orodruin

Staff Emeritus
$S$ 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. Dec 22, 2017

### Pushoam

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

Is this correct?

6. Dec 22, 2017

### Orodruin

Staff Emeritus
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.

7. Dec 22, 2017

### Pushoam

Thanks, I got it.