Page 1 of 1
Homework 5 Solutions
Posted: Mon Feb 28, 2005 3:57 pm
by goodwine
Here they are.
For problem 6
C and
FORTRAN are available.
Posted: Mon Feb 28, 2005 8:24 pm
by sstarch1
These solutions arent there, they are all blank white pages.
Posted: Mon Feb 28, 2005 8:55 pm
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.
Posted: Tue Mar 01, 2005 7:45 pm
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.
- 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).
- Alternatively, find two vectors in the null space of (A - lambda I)^2 = A^2 (since lambda=0). Then substitute each of these into
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.