Solution: Show ui=vi if u0=v0 in Difference Equations

[simmonj7]

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]

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 w_k = u_k - v_k, for all k.

 

[simmonj7]

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]

But there's only one eigenvalue here (and it's A).

 

[simmonj7]

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]

The eigenvalues of A don't matter …

all that matters is the roots of the characteristic equation of this recurrence relation which in this case is the single root, A.

Try using the w_k I mentioned earlier. ​