# Raising e to a matrix?

1. Apr 22, 2017

### Vitani11

1. The problem statement, all variables and given/known data
Find all components of the matrix eiaB. a is a constant and B is a 3x3 matrix whose first row is 0,0,-i second row is 0,0,0 and third row is i,0,0. The taylor expansion of eiaB gives 1+iaB-a2B2/2! - ....

2. Relevant equations
The taylor expansion of eiaB gives 1+iaB-a2B2/2! - ....

3. The attempt at a solution
I don't know what to do from here. If this is a diagonal matrix I would be able to multiply each element in B by ia and then raise e to the power of whatever the result is for each element in the matrix, but this doesn't qualify as a diagonal matrix. I have looked online and can't find any resources that speak about raising e to a non-diagonal matrix.

2. Apr 22, 2017

### Vitani11

Right now I have that the element in place of -i is eB and the element in place of i is e-B, the 0's stayed 0

3. Apr 22, 2017

### Staff: Mentor

Continue what you're doing here with the Maclaurin expansion. The powers of B are cyclical, oscillating between two values.

4. Apr 23, 2017

### Vitani11

I see that it is also cos(aB)+isin(aB), but how does this help? What do you mean by oscillating between two values?

5. Apr 23, 2017

### StoneTemplePython

With respect to all square matrices, the way to think about it is: act as if they were diagonal (and failing that settle for just upper triangular).

Do you know how to diagonalize a matrix? It can definitely be done in this case.

6. Apr 24, 2017

### vela

Staff Emeritus
The Cayley-Hamilton theorem says that B satisfies its characteristic polynomial. In this case, that means $B^3-B = 0$. It follows then that all odd powers of B are equal to B and all even powers are equal to $B^2$. If you calculate the first few powers of B, you can verify this is the case.

Last edited: Apr 24, 2017
7. Apr 25, 2017

### Vitani11

Okay so I can diagonalise this thing and find its matrix of eigenvalues and then I can just raise the elements in this diagonal matrix to the power e like I would for a matrix which was originally diagonal since it is equivalent to B?

8. Apr 25, 2017

### Dick

You aren't quite done then. You still have to take the resulting diagonal matrix and transform back to the original basis. Ie. undo the diagonalization.

9. Apr 25, 2017

### Vitani11

Okay. After I do this then I can use the formula P-1DP=B to transform this back to B.

10. Apr 25, 2017

### Dick

Right.

11. Apr 25, 2017

### Vitani11

Wow, that is a beautiful result. I learned how to do this two different ways now :) thank you for the help!