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Eigenvalues,charateristic polynomial

  • Thread starter stefan1988
  • Start date
  • #1

Homework Statement


http://img213.imageshack.us/img213/7867/problemeigen.png [Broken]

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Homework Equations


http://img339.imageshack.us/img339/771/equationh.jpg [Broken]

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The Attempt at a Solution



so i tried doing the problem i gonna use L for lambda
i got the matrix
[1-L, 3]
[K , 4-L]

then i do determinant to get the characteristic polynomial
(1-L)(4-L)-3K=0
L^2-5L+4=3k
K=(L^2-5L+4)/(3)

the other way i did it but im not sure is [(L^2-5L+4)/(K)] -3

i also tried doing this in matlab by doing
k=-100:100
a=[1 3; k 4;]
but that i think would just put a bunch of numbers where k is i think
not really sure how to proceed from here
 
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Answers and Replies

  • #2
Deveno
Science Advisor
906
6
i think what you are being asked to do is plot the function:

k(L) = (1/3)(L2-5L+4)

limiting the range (the "y" axis, or in this case, the "k" axis) to [-100,100]

(so you can fit all L values that give output values within your range on the plot).
 
  • #3
33,265
4,965

Homework Statement


http://img213.imageshack.us/img213/7867/problemeigen.png [Broken]

Uploaded with ImageShack.us

Homework Equations


http://img339.imageshack.us/img339/771/equationh.jpg [Broken]

Uploaded with ImageShack.us


The Attempt at a Solution



so i tried doing the problem i gonna use L for lambda
i got the matrix
[1-L, 3]
[K , 4]
Missing part of the lower right element.
then i do determinant to get the characteristic polynomial
(1-L)(4-L)-3K=0
L^2-5L+4=3k
K=(L^2-5L+4)/(3)
No, the idea is to solve for lambda (L here). k is just a constant.

Solve this equation: λ2 - 5λ + 4 - 3k = 0. This is a quadratic in λ.
the other way i did it but im not sure is [(L^2-5L+4)/(K)] -3

i also tried doing this in matlab by doing
k=-100:100
a=[1 3; k 4;]
but that i think would just put a bunch of numbers where k is i think
not really sure how to proceed from here
 
Last edited by a moderator:
  • #4
Deveno
Science Advisor
906
6
i think what you are being asked to do is plot the function:

k(L) = (1/3)(L2-5L+4)

limiting the range (the "y" axis, or in this case, the "k" axis) to [-100,100]

(so you can fit all L values that give output values within your range on the plot).
that's what i get for not reading closely enough *facepalm*.

right it should be L in terms of f(k).
 
  • #5
Solve this equation: λ2 - 5λ + 4 - 3k = 0. This is a quadratic in λ.
(λ-4)(λ-1)-3k
λ=4
λ=1

i solved it but im not sure what im supposed to do with it though
 
  • #6
33,265
4,965
(λ-4)(λ-1)-3k
λ=4
λ=1
You're ignoring the -3k. And the above should be an equation; namely λ2 - 5λ + 4 - 3k = 0.

Do you know how to solve quadratic equations?
i solved it but im not sure what im supposed to do with it though
 
  • #7
33,265
4,965
that's what i get for not reading closely enough *facepalm*.
Happens to us all (well, it happens to me). Don't feel bad.
 

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