# Homework Help: Eigenvectors and eigenvalues

1. Dec 30, 2017

### Physgeek64

1. The problem statement, all variables and given/known data
Find the eigenvalues and eigenvectors of the matrix
$A=\matrix{{2, 0, -1}\\{0, 2, -1}\\{-1, -1, 3} }$

What are the eigenvalues and eigenvectors of the matrix B = exp(3A) + 5I, where I is

the identity matrix?

2. Relevant equations

3. The attempt at a solution
So i've found the eigenvectors for A to be $\frac{1}{\sqrt{6}}\vec{1,1,-2}$, $\frac{1}{\sqrt{3}}\vec{1,1,1}$, $\frac{1}{\sqrt{2}}\vec{-1,1,0}$ with eigenvalues 4,1 one 2 respectively. but i don't know how to do the second part

Many thanks

2. Dec 30, 2017

### PeroK

Can you calculate the eigenvalues of $exp(A)$?

3. Dec 30, 2017

### Physgeek64

They are the exponentials of the eigenvalues of A

4. Dec 30, 2017

### PeroK

Well, that's a good start. What about $exp(3A)$?

5. Dec 30, 2017

### Physgeek64

$e^{3\lambda}$?

6. Dec 30, 2017

### PeroK

How could you prove that if you are not sure? Hint: it's not hard. Try letting $B = 3A$

7. Dec 30, 2017

### Physgeek64

the eigenvalues of $exp(B)$ are $e^b$ but $b=3a$ where a are the eigenvalues of A for $B=3A$. Hence the eigenvalues are $e^{3a}$

8. Dec 30, 2017

### PeroK

Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda \dots$

9. Dec 30, 2017

### Physgeek64

Okay. But how do you find the eigenvalues of $exp(3A)+5I$?

10. Dec 30, 2017

### PeroK

I thought you had worked it out. Where do you think you are stuck?

11. Dec 30, 2017

### PeroK

$\dots Bv = (\exp(3A) + 5I)v = \dots$

Does that help?

12. Dec 30, 2017

### Ray Vickson

For future reference: you can format a matrix nicely as
$$A = \pmatrix{2 & 0 & -1\\0 & 2 & -1 \\ -1 & -1 & 3}$$
The instructions that do that are "\pmatrix{2 & 0 & -1\\0 & 2 & -1 \\ -1 & -1 & 3}". Note the use of '&' as a separator, not a comma, and there is only one pair of curly brackets "{ }".

Also, your eigenvalues read as $\langle \frac{1}{\sqrt{6}} 1 , 1, -2 \rangle$, but you might have meant $\frac{1}{\sqrt{6}} \langle 1,1,-2 \rangle$, which is very different. Using $\vec{\mbox{ }}$ does not work well for an array of more than about two characters in length, so $\vec{v_1}$ looks OK but $\vec{v_1, v_2, v_3,v_4}$ does not.

13. Dec 30, 2017

### Physgeek64

Will have eigenvalues $e^{3a+5}$ with the same eigenvectors

14. Dec 30, 2017

### PeroK

Is that $exp(3a+5)$ or $exp(3a) + 5$?

15. Dec 30, 2017

### Physgeek64

Oops sorry its meant to be $exp(3a) + 5$