# Eigenvalue Problem

1. Jul 16, 2009

### mlarson9000

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

2. Relevant equations

3. The attempt at a solution
I think I should be doing something like this:
det(A-$$\lambda$$I), and

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

The eigenvalues would be $$\lambda$$ where the det(A-$$\lambda$$I)=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?

Last edited by a moderator: Jul 16, 2009
2. Jul 16, 2009

### Staff: Mentor

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?

3. Jul 17, 2009

### mlarson9000

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

4. Jul 17, 2009

### Staff: Mentor

So now put together what I wrote and what you wrote.
An eigenvalue of A is a number $\lambda$ such that (A - $\lambda$I)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?

5. Jul 18, 2009

### 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.

Last edited: Jul 18, 2009
6. Jul 18, 2009

### 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?

7. Jul 18, 2009

### mlarson9000

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

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

8. Jul 19, 2009

Right.