How Do Eigenvalues and Eigenvectors Change for Matrix B = exp(3A) + 5I?

[Physgeek64]

Homework Statement

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?

Homework Equations

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

 

[PeroK]

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

 

[Physgeek64]

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

They are the exponentials of the eigenvalues of A

 

[PeroK]

They are the exponentials of the eigenvalues of A

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

 

[Physgeek64]

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

$e^{3\lambda}$?

 

[PeroK]

$e^{3\lambda}$?

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

 

[Physgeek64]

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

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 $e^{3a}$

 

[PeroK]

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 $e^{3a}$

Yes. Although, I would start with something like:

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

 

[Physgeek64]

Yes. Although, I would start with something like:

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

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

 

[PeroK]

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

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

 

[PeroK]

Yes. Although, I would start with something like:

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

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

Does that help?

 

[Ray Vickson]

Homework Statement

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?

Homework Equations

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

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.

 

[Physgeek64]

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

Does that help?

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

Thank you for your help

 

[PeroK]

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

Thank you for your help

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

 

[Physgeek64]

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

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