# Another Eigenvalue proof

1. Apr 7, 2009

### sana2476

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

Let λ be an eigenvalue of A. Then λ+σ is an eigenvalue of A+σI

2. Relevant equations

3. The attempt at a solution

I'm guessing I need to use the fact that λ is an e.v of A to start with. But then when I add σ to both sides somehow I feel like I'm begging the question..

2. Apr 7, 2009

### Cyosis

You know that if $\lambda$ is an eigenvalue of A, then $A\boldsymbol{v}=\lambda \boldsymbol{v}$ is the appropriate eigenvalue equation with v an eigenvector. So if $\lambda+\sigma$ is an eigenvalue of$A+\sigma I$,then what would the appropriate eigenvalue equation be?

Last edited: Apr 7, 2009
3. Apr 7, 2009

### sana2476

The appropriate e.v equation would then be: Ax=(λ+σ)x..is that right?

4. Apr 7, 2009

### Cyosis

No that eigenvalue does not belong to A but to $A+\sigma I$. So replace A by $A+\sigma I$ then write out the left side of the equation and see if it checks out.

5. Apr 7, 2009

### sana2476

But wouldnt that be begging the question?

6. Apr 7, 2009

### aostraff

Yeah, Try adding sigma*x to both sides of equation if you're still stuck.

7. Apr 7, 2009

### Cyosis

You want to solve the following equation $(A+\sigma I) \boldsymbol v= \eta \boldsymbol v$ for $\eta$. Do you understand why this is the relevant equation? Now write out the left hand side and use the information given in the exercise.

8. Apr 7, 2009

### sana2476

Thanks I did that...and here's what I have

Ax +σx=λx+σx
Then you have (A+σI)x=(λ+σ)x
and because x doesnt equal zero therefore (λ+σ) doesnt equal zero

which then means that (λ,λ+σ) is an eigenpair of A+σI

9. Apr 7, 2009

### Cyosis

While the first two lines you wrote are correct I am not entirely convinced. Why is $A \boldsymbol{ v}+\sigma v =(A+\sigma I) \boldsymbol{v}$ ?

As for the rest. You are correct to say that x is not zero for if it was it wouldn't be an eigenvector. However 0 is a valid eigenvalue so the eigenvalue of $A+\sigma I$ could be zero.

Last edited: Apr 7, 2009
10. Apr 7, 2009

### tiny-tim

Looks ok to me … distributive law.

11. Apr 7, 2009

### aostraff

$$x=Ix \Rightarrow \sigma x=\sigma (Ix) = (\sigma I)x$$