Eigenvalues,charateristic polynomial

[stefan1988]

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 going to 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 I am 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

 

[Deveno]

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).

 

[Mark44]

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 going to 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 I am 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

 

[Deveno]

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).

 

[stefan1988]

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 I am not sure what I am supposed to do with it though

 

[Mark44]

(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 I am not sure what I am supposed to do with it though

 

[Mark44]

that's what i get for not reading closely enough *facepalm*.

Happens to us all (well, it happens to me). Don't feel bad.