# Matrix exponential problem

1. Oct 20, 2011

### JamesGoh

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

Given matrix A= [1,1;0,0] and matrix B = [1,-1;0,0]

n.b. the semicolon separates the matrix rows

find exp(A)exp(B)

2. Relevant equations

$exp^{A}=\sum\frac{A^{n}}{n!}=I + A + \frac{A^{2}}{2} + \frac{A^{3}}{6} + ...... + \frac{A^{n}}{n!}$

for n= 0 to infinity

3. The attempt at a solution

The answer the tutor give is the following matrix

exp(A).exp(B) = [e^2,-(e-1)^2 ; 0,1]

this would imply that A and B are diagonalisable (since they need to be in order to get the form of the answer)

However, if I worked out the characteristic ppolynomial of A and B, I would only get one eigenvalue, not two distinct eigenvalues which rules out any chance of diagaonlisation

Is there a way to diagaonlise A and B ?

2. Oct 20, 2011

### Staff: Mentor

I get two eigenvalues for each matrix, so I don't know what you did that you came up with only one apiece.

3. Oct 20, 2011

### Staff: Mentor

You can also do this problem without diagonalizing either matrix. Use your relevant equation, which should be slightly amended. It's not a finite sum.
$$exp^{A}=\sum\frac{A^{n}}{n!}=I + A + \frac{A^{2}}{2} + \frac{A^{3}}{6} + ...... + \frac{A^{n}}{n!} + ...$$

4. Oct 20, 2011

### Ray Vickson

A standard result for an nxn matrix with all eigenvalues different is: if r_1, r_2,...,r_n are the eigenvalues, there exist matrices E_1,E_2,...,E_n such that for any analytic function f(x) = sum c_n x^n (with radius of convergence containing all the r_j) then f(A) = c_0*I + c_1*A + c_2*A^2 + ... = sum_j f(r_j)* E_j . Here the E_j are the same for any f, so cn be found, for example, by looking at f(x) = 1 = x^0, f(x) = x, f(x) = x^2, etc. Having the E_j, now use f(x) = exp(x) to get exp(A).

For a 2x2 matrix with eigenvalues 'a' and 'b' we have I = E1 + E2, which comes from using f(x) = 1. Similarly, A = a*E1 + b*E2 (using f(x) = x).

RGV

5. Oct 20, 2011

### JamesGoh

What was your characteristic polynomial for A ? What eigenvalues did you get for A ?

Also, what char. polynomial and eigenvalues did you get for B ?

6. Oct 20, 2011

### HallsofIvy

Those are both "Jordan form" matrices so the numbers on the diagonal are the eigenvalues.

7. Oct 20, 2011

### vela

Staff Emeritus
What did you get?

8. Oct 20, 2011

### JamesGoh

I got 0 and 1 as the eigenvalues for both matrix A and Matrix B.

The characteristic polynomial I got for A was $λ^{2}-λ$

Similary, my char. polynomial for B is $λ^{2}-λ$

9. Oct 20, 2011

### vela

Staff Emeritus
So you got two distinct eigenvalues, and you can diagonalize them both.

As Mark mentioned, you can calculate eA and eB without diagonalizing though. A matrix satisfies its characteristic polynomial, so A2-A=0. Using that fact, you can easily simplify the series.

10. Oct 21, 2011

### JamesGoh

Thanks for that. I was asking in the first placecause I knew the diagonal form of both A and B would be

[ 0, 1 ; 0, 0]

which is nowhere close to the tutor's answer

11. Oct 21, 2011

### vela

Staff Emeritus
That's not the diagonal form of either A or B.

Also, your tutor calculated exp(A)exp(B), which isn't the diagonalization of A or B.

12. Oct 21, 2011

### JamesGoh

but doesn't the Diagonal matrix contain the eigenvalues down the main diagonal and 0's elsewhere ?

what don't I understand ?

13. Oct 21, 2011

### vela

Staff Emeritus
The matrix you wrote was
\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}which isn't diagonal.