# Eigenvalues and Eigenvectors of exponential matrix

1. Jan 14, 2014

### unscientific

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

Part (a): Find the eigenvalues and eigenvectors of matrix A:

$$\left( \begin{array}{cc} 2 & 0 & -1\\ 0 & 2 & -1\\ -1 & -1 & 3 \\ \end{array} \right)$$

Part(b): Find the eigenvalues and eigenvectors of matrix $B = e^{3A} + 5I$.
2. Relevant equations

3. The attempt at a solution

Part (a)
$$\lambda = 1, 2, 4$$
$$u_1 = \frac{1}{\sqrt3}(1,1,1)$$
$$u_2 = \frac{1}{\sqrt 2}(1,-1,0)$$
$$u_3 = \frac{1}{\sqrt 5}(1,1,-2)]$$

Part(b)

Realize A is a hermitian matrix.
Diagonalize A:
$$A'= \left( \begin{array}{cc} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 4 \\ \end{array} \right)$$

$$B' = exp(3A') + 5I$$

Therefore, eigenvalues of $B'= e+5, e^2 + 5, e^4+5$. Also, eigenvalues of B = B'.

How do I find the eigenvectors of B? Do I need to undiagonalize B' using the transformation matrix made up of eigenvectors of A?

2. Jan 14, 2014

### Office_Shredder

Staff Emeritus
You can, but it's easier than that. You found the eigenvectors of A, then observed that doing a change of basis to the set of eigenvectors of A diagonalizes A (which is a general fact). You then got B', which is diagonal after doing a change of basis using the eigenvectors of A. That should instantly tip you off about what the eigenvectors of B are.

3. Jan 15, 2014

### unscientific

I can instantly find out what the eigenvalues of B are, but I don't see the trick/link to find eigenvectors of B. I know it's related to eigenvectors of A.

I know the relation between B and A. I know how A is diagonalized using its eigenvectors. But it tells us nothing about how B is diagonalized!

4. Jan 15, 2014

### unscientific

Can I simply say that since eigenvectors of A also diagonalize B, eigenvectors of B is also eigenvectors of A?

5. Jan 15, 2014

### haruspex

Are eigenvectors of A also eigenvectors of A2, A3, ...? Of a power series in A?

6. Jan 15, 2014

### unscientific

Ah, for a $A^n$, to completely diagonalize it, we must apply matrix $U$ n times. Following this line of thought, I would say the eigenvectors of B are $exp(v_1)$, $exp(v_2)$ and $exp(v_3)$.

7. Jan 15, 2014

### haruspex

Why?
U-1A2U = U-1AUU-1AU.

8. Jan 15, 2014

### unscientific

Ok, then I have no idea..

9. Jan 15, 2014

### haruspex

U-1A2U = U-1AU U-1AU. (Agreed?)
If U-1AU = D is diagonal, then what does that tell you for U-1A2U? For U-1AnU?

10. Jan 16, 2014

### unscientific

They are all diagonal! So each term in the $exp(3A)$ is diagonal. Meaning B is made diagonal by eigenvectors of A. Therefore, eigenvectors of B = eigenvectors of A. (The factor of 3 cancels out, because (3A)u = (3λ)u.

11. Jan 16, 2014

### Office_Shredder

Staff Emeritus
This is the statement I was going for, which is correct. More generally it's good to notice that for exponentiating, along with many other power series operations, you get the same eigenvectors as when you started. The only thing that can screw things up is the constant term of the power series.