Eigenvalues & Eigenvectors of A & A+rI

[mlarson9000]

Homework Statement

Let A be an nxn matrix and let I be the nxn identity matrix. Compare the eigenvectors and eigenvalues of A with those of A+rI for a scalar r.

Homework Equations

The Attempt at a Solution

I think I should be doing something like this:
det(A-\lambdaI), and

det((A+rI)-\lambdaI)=det(A-(\lambda-r)I).

The eigenvalues would be \lambda where the det(A-\lambdaI)=0
and det(A-(\lambda-r)I)=0.

So does that mean the eigen values for the first matrix are \lambda =n and the second will be n+r?

 

[Mark44]

An eigenvector of A is a nonzero vector x such that (A - \lambda)x = 0, and \lambda is the eigenvalue for that eigenvector.

Can you say something in this vein for the eigenvectors of A + rI?

 

[mlarson9000]

[/B][/B]

An eigenvector of A is a nonzero vector x such that (A - \lambda)x = 0, and \lambda is the eigenvalue for that eigenvector.

Can you say something in this vein for the eigenvectors of A + rI?

The eigenvectors are the solution to:
(A-(\lambda-r)I)x=0
What do you suppose is meant by "compare the eigenvectors and eigenvalues?"

 

[Mark44]

So now put together what I wrote and what you wrote.
An eigenvalue of A is a number \lambda such that (A - \lambdaI)x = 0.
An eigenvalue of A + rI is a number ? such that (A - (\lambda - r)I)x = 0.

(Fill in the question mark.)
What can you say about the values of x in either case?

 

[mlarson9000]

I don't see why I should be assuming that \lambda in the first equation should be equal to \lambda in the second equation. The same goes for x. It's really giving me trouble as I work with this.

 

[Dick]

lambda is an eigenvalue corresponding to eigenvector x if Ax=lambda*x, right? Then (A+rI)x=Ax+rx=(lambda+r)x, also right? I haven't changed any lambda's or x's. What does (A+rI)x=(lambda+r)x tell you about eigenstuff of A+rI?

 

[mlarson9000]

(A+rI)x=(\lambda+r)x
(A+rI)x-(\lambda+r)x=0
(A+rI-\lambdaI-rI)x=0
(A-\lambdaI)x=0

So the eigenvectors of A and A+rI are the same, right?

 

[Dick]

(A+rI)x=(\lambda+r)x
(A+rI)x-(\lambda+r)x=0
(A+rI-\lambdaI-rI)x=0
(A-\lambdaI)x=0

So the eigenvectors of A and A+rI are the same, right?

Right.