# Need help finding exponential matrix e^AT

1. Apr 22, 2009

### bengaltiger14

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

Given the matrix: A=[-2 0 0;4 -2 0;1 0 -2], find e^AT.

I found the eigenvalues to be -2,-2,-2.

How do I solve this problem with 3 identical eigenvalues? Do I cube to matrix after plugging in my eigenvalue -2 and then solve it that way?

After plugging in the eigenvalue -2, the resulting matrix is: [0 0 0;4 0 0;1 0 0]

I could eliminate the 1 from row 3,col 1 and end up with two free variables. I am unsure after this.

2. Apr 22, 2009

### Dick

If you can split the matrix A=B+C in such a way that B and C commute (i.e. BC=CB) then you can use e^A=e^B*e^C. Can you figure out how to do this in such a way that e^B and e^C are easy to compute?

3. Apr 22, 2009

### bengaltiger14

splitting the original matrix or the matrix with the -2 eigenvalue?

4. Apr 22, 2009

### Dick

The original matrix. If you can split it into easy commuting parts, you don't need the eigenvectors.

5. Apr 22, 2009

### bengaltiger14

I think my prof. wants us to use the Eigenvectors, but i do not know how to split it the way you are talking of.

6. Apr 22, 2009

### Dick

Do it the easy way first. Split it into a diagonal part and an off-diagonal part. Do they commute? The off-diagonal part is nilpotent, it's easy to find both exponentials.

7. Apr 22, 2009

### bengaltiger14

I am not sure how to split as you are talking. But the diagonal would just be a diagonal of e^-2t? And everything else just 0?

8. Apr 22, 2009

### Dick

Yes, the exponential of the diagonal part (call it B) is the diagonal matrix with entries e^(-2). 'Everything else' is the matrix C=[0,0,0;4,0,0;-2,0,0]. e^C isn't 0. What is it? Notice that C^2=0 and put it into the series definition of the matrix exponential.

9. Apr 22, 2009

### bengaltiger14

e^C would be: C=[0,0,0;e^4t,0,0;e^-2t,0,0]

10. Apr 22, 2009

### Dick

Not at ALL. e^C=I+C+C^2/2!+C^3/3!+... Isn't it? And don't forget before you can say e^(B+C)=e^B*e^C you HAVE to show B and C commute.

11. Apr 22, 2009

### bengaltiger14

So, the resulting matrix would be: [e^-2t,0,0;4e^-2t,e^-2t,0;e^-2t,0,e^-2t]

Does this look right?

12. Apr 22, 2009

### Dick

No, not yet. Concentrate on e^(Ct) first. It's I+Ct+(Ct)^2/2!+... What is that? I just noticed that t floating around. You'll have to add it to some of my previous posts.

13. Apr 22, 2009

### bengaltiger14

That is: [1,0,0;4t,1,0;1t,0,1]

The e^C*eB = [e^-2t,0,0;4te^-2t,e^-2t,0;te^-2t,0,e^-2t]

14. Apr 22, 2009

### Dick

That looks much better. You did check B and C commute, right? It's necessary.