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Basic Eigenvalue Proof

  • #1

Homework Statement


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


Homework Equations





The Attempt at a Solution


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

Is this the right start?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
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Looks like a good start to me. Keep going.....
 
  • #3
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.
 
  • #4
phyzguy
Science Advisor
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Try expanding (A-cI) x.
 
  • #5
Ok I expanded the first step to...
Ax-(CI)x=[tex]\lambda[/tex]x-cx
 
  • #6
I also know that CI is similar to cx, except CI is a matrix and cx is not. But I'm stuck...
 
  • #7
phyzguy
Science Advisor
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I don't see your problem. You showed that [tex] (A-cI)x = (\lambda-c)x[/tex], which is what you were supposed to show.
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
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No, he didn't show that, he simply wrote it and asked how to go from [itex]Ax= \lamda x[/itex] 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 [itex]\lambda[/itex], [itex]Ax= \lambda x[/itex]. I is the identity operator so Ix= x. Put those together.
 
  • #9
This is what I did step by step. How does it look?
(A-cI)x=([tex]\lambda[/tex]-c)x
Ax-(cI)x=[tex]\lambda[/tex]x-cx
Ax-c(Ix)=[tex]\lambda[/tex]x-cx
Ax-cx=[tex]\lambda[/tex]x-cx
Ax=[tex]\lambda[/tex]x
 
  • #10
Is that sufficient?
 
  • #11
phyzguy
Science Advisor
4,505
1,447
It seems like you are confused with how to go about demonstrating or proving something. You need to start with something you know, then apply known operations until you arrive at what you are trying to prove. Here, the first line of your proof is what you are trying to prove, and you don't yet know it to be true, so if I were grading this proof I would object to the very first line. Actually if you just reverse the order of steps, starting from the bottom and working up, you have pretty well shown what you are trying to prove
 
  • #12
I see what you are saying. So just reverse it?
 
  • #13
Well I guess I will just go with the reverse of what I had...
 

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