# Basic Eigenvalue Proof

maherelharake

## Homework Statement

If v is an eigenvector of A with corresponding eigenvalue $$\lambda$$ and c is a scalar, show that v is an eigenvector of A-cI with corresponding eigenvalue $$\lambda$$-c.

## The Attempt at a Solution

I started out thinking that I have to figure out how to go from:
Ax=$$\lambda$$x
to
(A-cI)x=($$\lambda$$-c)x

Is this the right start?

Looks like a good start to me. Keep going.....

maherelharake
I don't see how I'm supposed to continue. It just seems obvious to me and I can't think of any intermediate steps.

Try expanding (A-cI) x.

maherelharake
Ok I expanded the first step to...
Ax-(CI)x=$$\lambda$$x-cx

maherelharake
I also know that CI is similar to cx, except CI is a matrix and cx is not. But I'm stuck...

I don't see your problem. You showed that $$(A-cI)x = (\lambda-c)x$$, which is what you were supposed to show.

Homework Helper
No, he didn't show that, he simply wrote it and asked how to go from $Ax= \lamda x$ to that.

maherelharake, From (A- cI)x you get Ax- (cI)x= Ax- c(Ix). Since x is an eigenvector of A corresponding to eigenvalue $\lambda$, $Ax= \lambda x$. I is the identity operator so Ix= x. Put those together.

maherelharake
This is what I did step by step. How does it look?
(A-cI)x=($$\lambda$$-c)x
Ax-(cI)x=$$\lambda$$x-cx
Ax-c(Ix)=$$\lambda$$x-cx
Ax-cx=$$\lambda$$x-cx
Ax=$$\lambda$$x

maherelharake
Is that sufficient?