Question

Can someone check my solution, please

Answer 1

of course eigenvectors are not unique

Answer 2

normally if two eigenvalues are the same you can't use e^(landa*t)*eigenvector

have you learned picart algotithm

Answer 3

on my paper, that's the way I figure out the second eigenvector from the available one(I mean eigenvector from eigenvalue 2)

Answer 4

by that way my third solution is linearly independent to the second one

Answer 5

you should have 3 different linearly indepe. sol

Answer 6

I'm not sure.. have you learned picart algotithm ?

Answer 7

the first one is linearly independent to the second one already, since they come from 2 different eigenvalues

Answer 8

no , I didn't

Answer 9

himm..

Answer 10

I'll be back

Answer 11

ok, I have to get home, I ll back when I get home. See you

Answer 12

http://www.wolframalpha.com/input/?i=x%27%3Dx%2Cy%27%3D3y%2Bz%2Cz%27%3D-y%2Bz

Answer 13

sorry the name of the method is putzer algorithm
http://www.proofwiki.org/wiki/Putzer_Algorithm_for_Matrix_Exponentials

Answer 14

do you have any example in your lecture note one has the same eigenvalue..

Answer 15

I'll check my prof's stuff and when he did on his way, his own result is different from wolfram. That's why mine stuff doesn't match to it. But as long as mine matches his, I am ok, right?

Answer 16

@e.mccormick this is one of program I wrote last spring from java course