Eigenvalue/vector Proof

  • #1

Homework Statement


Let A be n x n, λ ∈ ℂ, let v be n x 1, and suppose that A ⋅ v = λ ⋅ v. Show that A^jv =
mathImg.gif
^j ⋅ v
for each positive integer j.

Homework Equations




The Attempt at a Solution


I haven't been able to get very far but,

mathImg.gif
^j ⋅ v
- A^jv = 0n x 1
v(
mathImg.gif
^j -
A^j) = 0n x 1

Not sure how to prove that for every positive j that this is true. Any thought would be appreciated. Thanks.
 

Answers and Replies

  • #2
RUber
Homework Helper
1,687
344
What if you tried induction?
 
  • #3
Where j = 0

v(
mathImg.gif
^j -
A^j) = 0n x 1

v(
mathImg.gif
^0 -
A^0) = 0n x 1

v(In-1) = 0n x 1

v ⋅ In
- v= 0n x 1

v - v = 0n x 1


Since v is n x 1

0n x 1
= 0n x 1


Because this is true I can assume this for all j?
 
  • #4
RUber
Homework Helper
1,687
344
I don't think so, since it would be true for just about anything raised to the zero.
Start with the original form ##Av=\lambda v ##. This is your base case.
##A^2v=A \lambda v ##.
And for any integer...
 
  • #5
I see what you're saying about
##A^2v=A \lambda v ##.
And for any integer...
and vice-versa for
mathImg.gif
but I don't see how that helps?
My skills with induction are sub par to say the least.
 
  • #6
RUber
Homework Helper
1,687
344
Assume ##A^n v = \lambda^n v## show that ##A^{n+1}v=A A^n v = \lambda^{n+1}v##.
The argument for induction is: It works for j=1, assume it works for some j and show it is true for the next j.
 

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