Can Eigenvalues Be Shifted by a Scalar?

[sana2476]

Homework Statement

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

Homework Equations

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..

 

[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 ofA+\sigma I,then what would the appropriate eigenvalue equation be?

 

[sana2476]

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

 

[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.

 

[sana2476]

But wouldn't that be begging the question?

 

[aostraff]

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

 

[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.

 

[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 doesn't equal zero therefore (λ+σ) doesn't equal zero

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

 

[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.

 

[tiny-tim]

While what you wrote is correct I am not entirely convinced. Why is A \boldsymbol{v}+\sigma \boldsymbol{v} =(A+\sigma I) \boldsymbol{v} ?

Looks ok to me … distributive law.

 

[aostraff]

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