anonymous
  • anonymous
Finding eigenvectors: Can some just show me the steps to find the eigenvector of the following matrix, with (lambda)=2. Matrix is given below.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\left[\begin{matrix}-1 & 2 & -1\\ 3 & 0 & 1\\ -3 & -2 & -3\end{matrix}\right]\]
anonymous
  • anonymous
I hated linear algebra and don't remember anything, but I will suggest Paul's online notes for these (just Google is). Also, if you can get your hands on a pdf copy of 'Elementary Linear Algebra' by Larson and Falvo.
zzr0ck3r
  • zzr0ck3r
first do A - bI where b is some constant variable and I is the identity matrix

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zzr0ck3r
  • zzr0ck3r
so [-1-b,2,-1;3,-b,1;-3,-2,-3-b]
zzr0ck3r
  • zzr0ck3r
understand? b is your lamda
anonymous
  • anonymous
yes, but I thought you used A-bI=0, in order to find all the lambdas... and then use Ax=bx, in order to find the eigenvectors.
zzr0ck3r
  • zzr0ck3r
o sorry you labda is given
zzr0ck3r
  • zzr0ck3r
you have your lambda its 2
anonymous
  • anonymous
and I have all three eigenvalues, but for some reason when i am calculating for the eigenvector, i am not getting the right answer
zzr0ck3r
  • zzr0ck3r
so solve for the null space of that equation
zzr0ck3r
  • zzr0ck3r
one sec
anonymous
  • anonymous
and if i can see how to do it just for lambda = 2 , i can figure out the other two eigenvectors.
TuringTest
  • TuringTest
if I remember this right, you just plug in each lambda and solve (A-b_1*I)=0 (A-b_2*I)=0
zzr0ck3r
  • zzr0ck3r
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anonymous
  • anonymous
ahhh, got it, okay... i see where i went wrong, thank you for the help
zzr0ck3r
  • zzr0ck3r
np

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