Homework 5 Solutions

Due Friday, February 25, 2005.
Post Reply
goodwine
Site Admin
Posts: 1596
Joined: Tue Aug 24, 2004 4:54 pm
Location: 376 Fitzpatrick
Contact:

Homework 5 Solutions

Post by goodwine »

Here they are.

For problem 6 C and FORTRAN are available.
Last edited by goodwine on Thu Jun 15, 2006 4:22 pm, edited 2 times in total.
Bill Goodwine, 376 Fitzpatrick
sstarch1

Post by sstarch1 »

These solutions arent there, they are all blank white pages.
goodwine
Site Admin
Posts: 1596
Joined: Tue Aug 24, 2004 4:54 pm
Location: 376 Fitzpatrick
Contact:

Post by goodwine »

sstarch1 wrote:These solutions arent there, they are all blank white pages.
Yes they are there and it works for sure in linux. Maybe there is an encoding problem and something doesn't work too well with various versions of Acrobat. If lots of people have problems then I'll have them re-scanned tomorrow. For the time being, I'd suggest trying another computer platform.
Bill Goodwine, 376 Fitzpatrick
goodwine
Site Admin
Posts: 1596
Joined: Tue Aug 24, 2004 4:54 pm
Location: 376 Fitzpatrick
Contact:

Post by goodwine »

In the solution to homework number 5, in problem number 5, we have a repeated
root of zero. I understand the first solution, but not how he got the second
solution.

For second solution: A * eta = Xi where eta = {{(1+2k)/4} , {k}}.

Where did this eta come from?
It's confusing because of the zeros, but two approaches are equally valid and equally easy.
  1. xi satisfies (A - lambda I) xi = 0 and then eta is a generalized eigenvector that satisfies
    (A - lambda I) eta = xi. This approach works just fine since
    dim(N(A - lambda I)) = 1, so there is one regular eigenvector. Since there is only one eigenvector, regardless of how you scale it,
    (A - lambda I) eta = xi
    will have a solution (you won't have to consider linear combinations of regular eigenvectors when trying to compute the generalized eigenvector).
  2. Alternatively, find two vectors in the null space of (A - lambda I)^2 = A^2 (since lambda=0). Then substitute each of these into
    • Image
    The exp(lambda t) term is 1 since lambda=0.
This is just a repeated eigenvalues problem with repeated zero eigenvalues; however, the TA did it using the book's method, which is fine since A is only 2x2, which I can see is a bit confusing.
Bill Goodwine, 376 Fitzpatrick
Post Reply

Return to “AME 302 Homework 5”