# Difference equations

## Homework Statement

Suppose that {uk} and {vk} are sequences satisfying uk = Auk-1 k = 1, 2, ... and vk = Avk-1 k = 1, 2,... Show that if u0 = v0 then ui = vi for all values of i.

## Homework Equations

uk --> is u subscript k
u0 --> is u subscript 0
uk-1 --> is u subscript k-1
ui --> is u subscript i

## The Attempt at a Solution

Well so far I have...
uk = A^k(u0)
= A^k(a1u1 + a2u2 +...+anun)
= a1(A^k)u1 + a2(A^k)u2 +...+ an(A^k)un
= a1(lambda1^k)u1 + a2(lambda2^k) +...+ an(lambdan^k)un

But since u0 = v0 A^k(u0) = A^k(v0)

But after there I get uncertain cause I think my next steps would be:
= A^k(a1v1 + a2v2 +...+ anvn)
= a1(A^k)v1 + a2(A^k)v2 +...+ an(A^k)vn
= a1(lambda1^k)v1 + a2(lambda2^k)v2 +...+ an(lambdan^k)vn)

Then conclude uk = vk?
Is this correct?

Thanks! :)

tiny-tim
Homework Helper
Hi simmonj7! (try using the X2 tag just above the Reply box )
Well so far I have...
uk = A^k(u0)
= A^k(a1u1 + a2u2 +...+anun)
= a1(A^k)u1 + a2(A^k)u2 +...+ an(A^k)un
= a1(lambda1^k)u1 + a2(lambda2^k) +...+ an(lambdan^k)un

(what are the as and the lambdas? anyway …)

Hint: put wk = uk - vk, for all k. When solving a difference equation all the way out you have to find the eigen values, then find the corresponding eigen vectors and then have to find a relationship between those eigen vectors and the vector x0. Thus x0 = a1u1 + a2u2 +...+ anun is the relationship between all the eigen vectors.

Then you just plug in that relationship for x0 into the equation xk = (A^k)x0. From there you distribute the A^k through and then (because (A)x = (lambda)x then (A^k)x = (lambda^k)x where lambda is an eigen vector of A) you substitute that back into the equation and that is what all the a's and lambda's are.

tiny-tim
Homework Helper
But there's only one eigenvalue here (and it's A). Where are you getting that there is only one eigen value? There is no way to determine how many eigen values there are of A...

tiny-tim
Try using the wk I mentioned earlier. 