Math! Distinct Real Eigenvalues Help :D

[Zopyrus]

:--sunny:

I need help! I'm reading Differential Equations with Boundary-value problems and on one of its problems about EIGENVECTORS:

Question was to solve dx/dt= 2x+3y ----- dy/dt= 2x+y

From a characterstic equation i took out: ¤1 = -1 and ¤2 = 4 ;

And for ¤1=-1 is equivalent to; 3k1 + 3k2=0 & 2k1 + 2k2=0

How do they get the eigenvector K1 =
(1)
(-1)

and for ¤2=4 they get eigenvector K2 =
(3)
(2)

How? :kiss:

 

[Real Madrid]

:--sunny:

I need help! I'm reading Differential Equations with Boundary-value problems and on one of its problems about EIGENVECTORS:

Question was to solve dx/dt= 2x+3y ----- dy/dt= 2x+y

From a characterstic equation i took out: ¤1 = -1 and ¤2 = 4 ;

And for ¤1=-1 is equivalent to; 3k1 + 3k2=0 & 2k1 + 2k2=0

How do they get the eigenvector K1 =
(1)
(-1)

and for ¤2=4 they get eigenvector K2 =
(3)
(2)

How? :kiss:

Here you go . hope this helps :smile:

10 + 10 = 20 -5 = 15

 

[Zopyrus]

Here you go . hope this helps :smile:

10 + 10 = 20 -5 = 15

Thank you mate! 10 + 10 = 15? :friends: You need to stop being a fan of Real Madrid dude

 

[westwienmaskulin]

can you somehow post the problem, if you have it on pdf or something?

basically, you get the eigenvectors by A*x = lambda*x

so if you have eigenvalue -1, you need to find a vector where A*vector = -1*vector.

if you do that, you will see that A*vector = eigenvalue*vector.

 

[Payandeh Iran]

I used to be one of the best in university at Diff. Eq. (about 18-19 years ago). Damn I hardly remember anything now!!!

 

[Agha Shojaa]

I used to be one of the best in university at Diff. Eq. (about 18-19 years ago). Damn I hardly remember anything now!!!

And I have a masters' in math and I don't remember $hit....so don't feel bad!

 

[westwienmaskulin]

And I have a masters' in math and I don't remember $hit....so don't feel bad!

LOOOOOOOOOL! You have masters in math?? Then I shouldn't feel that bad at all, that I had to pick up a book again to make sure about it.

 

[Zopyrus]

Thank you for helping me out here guys =)

Here is two part picture

Part One

Part Two

Sorry for the blurriness! Took it with my celly.. I hope it's readable

 

[Agha Shojaa]

LOOOOOOOOOL! You have masters in math?? Then I shouldn't feel that bad at all, that I had to pick up a book again to make sure about it.

And what's worse is I took every single ODE and PDE course that you could find out there and it still does not ring a bell...lol
I know you were getting your math degree westi...are you done with it?

 

[westwienmaskulin]

LOOOL..dude..that 3k1+3k2 thing was really confusing..

anyway..it's good

You just have to take the vector in a way, that A*x = EV*x where x is the eigenvector.

Generally, you take the easiest option and for EV = -1, the Eigenvector is (1,-1). You do the same step for EV=4.

 

[westwienmaskulin]

And what's worse is I took every single ODE and PDE course that you could find out there...
I know you were getting your math degree westi...are you done with it?

It should be the last year hopefully. I need 2 bigger exams, 4 smaller ones and a masters thesis to finish it.

 

[Agha Shojaa]

It should be the last year hopefully. I need 2 bigger exams, 4 smaller ones and a masters thesis to finish it.

Nice!

 

[Babr]

well i dont know if this can help but
if you know how to get the equation of Lambda at first place , then if you solve the equation it give u lambda = 4 and -1 ...
(to solve the problem u have to find 2 numbers that multiple is -4 and addition is -3 which are 4 and -1 ..
(1)(-4)=-4
1+(-4)=-3 )

then u replace the value in det .. (add K1 and k2) and if you replace
-1 in det you will find 3k1+3k2=0 same for 4 and second equation ..
after that it would be easy as the result of equation is equal to zero that mean k1=-k2 and in this case will be 1 and -1..same way for second one ..

 

[Zopyrus]

Thanks for the help guys! I'm really grateful to how I always get help when asking for it!!!! And Westie jan good luck to you!

 

[Babr]

LOOOL..dude..that 3k1+3k2 thing was really confusing...

Mehdi jan if you replace -1 in the Det you will find this equation but you have to add k1 and k2 ( or anyother Par to find the 2 values i guess )

 

[IPride]

k1 and k2 are the two components of the eigen vector.
You will have two independent eigen vectors because you have two distinct eigen values.

The equation
3K1+3K2=-0
2K1+2K2= 0
will give you the eigen vector corresponding to Lamda=1 as your eigen value.
which is of the form v= [S -S] < transpose usually S is taken to be the lowest possible value... hence [1 -1]

and the equation
-2K1+3K2=0
2K1-3K2=0

will give you the other eigen-vector.
V=[1 3/2]

 

[mashtnaghi]

Baz mashghatono avordin inja?

Agha Shoja jaan, I did not know that about you. I knew there was a reason I liked you. Now, you only need to give up supporting Taj.

Okay Zopyrus jaan, I will help you this time I can justify it with providing additional info, with the hope of a deeper understanding and not helping you solve a problem.

Eigen values and their associated eigenvectors come from the solutions to the homogenous equations with nontrivial solutions (trivial solution would be zero). The following are equivalent:

det(A) does not equal 0.
AX=b is consistent.

To have nontrivial solution (infinitely many solutions) we want the opposite, se we set det(A-LI)=0


Matrix A is
[2 3]
[2 1]

I will use L for Lambda.

Characteristic matrix is [A-LI]

A-LI=
[2-L 3]
[2 1-L]

det(A-LI)= (2-L)(1-L)-6

det(A-LI)= 0, so:

(2-L)(1-L)-6=0
L^2-3L-4=0

L=4 and L=-1

Substitute L=4 and L=-1 into the characteristic matrix to get the characteristic equations.

L=4

[-2 3]
[2 -3]

I will use x and y to avoid subscripts.
-2x+3y=0
2x-3y=0

This systems of equation has infinitely many solutions and needs to be parameterized. This can be done through row-reduction, but for 2 by 2 systems you can pick any of the variables and solve it in terms of the other.

-2x=-3y
x=3/2y

Let y=t, then x=3/2t.

(Note the choice is arbitrary. You can let x=t and solve for y.)


Eigenvector for L=4 thus is:

[3/2t]
[t ]

If you don’t want fractions you can multiply by 2 and get

[3t]
[t]

Let t=1 to get the answer they have.


L=-1

[3 3]
[2 2]

3x+3y=0
2x+2y=0

3x=-3y
x=-y

Let y=t, then x=-t.

Eigenvector for L=-1 thus is:

[-t]
[t ]

and

[t]
[-t]

if you parameterize x. Again both produce the same eigenvectors. Let t=1 and you will get the answers they have.

 

[Agha Shojaa]

Agha Shoja jaan, I did not know that about you. I knew there was a reason I liked you. Now, you only need to give up supporting Taj..

koochikeh aghayeh Prof.

 

[mashtnaghi]

koochikeh aghayeh Prof.

Ma chakerim do barabar.

 

[spinhead]

i love linear algebra. wish i had a chance 2 teach it.