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#1 2011-04-05 05:25:06

Reuel
Member
Registered: 2010-11-28
Posts: 178

Eigenvalues, Eigenvectors

Hello.

I am an undergraduate differential equations student. My teacher and my textbook seem both incapable of saying in plain English how eigenvalues affect eigenvectors. Could someone who understands differential equations please help me to understand their nature?

How do positive eigenvalues affect a vector field? That is, does the vector field spiral sink, spiral source, rotate around the center?

How do negative eigenvalues affect a vector field?

How do complex eigenvalues affect it? Does it matter whether it is positive or negative real parts? How do such details affect the system?

I do not understand and I thank the helper who can explain it. smile

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#2 2011-04-05 05:48:37

Bob
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Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi Reuel,

Haven't heard from you for a while.  How are you?

From my understanding of the topic, your questions don't make sense to me.  What have they been teaching you?

Have a look at

http://en.wikipedia.org/wiki/Eigenvector

and say whether this helps you at all.

Or post a specific question so I can see more clearly where your teachers are taking you.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2011-04-05 06:00:17

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Right now we are dealing with linear solutions. Supposing you have a linear system in terms of dy/dt and dx/dt you are asked to judge, just by the eigenvalues, how the system would act if plotted as a direction field.

For example, say you have the system


and you compute the eigenvalue to be -4, if I am even doing that part correctly (I don't know how to show my work with matricies on this forum) then you find the system, if plotted in Maple, rotates clockwise as a wheel would, rather than spiraling toward or away from the origin.

Here is an image from Maple to demonstrate my point (attached).

I am required to be able to tell, just from calculating the eigenvalues (and without Maple) what the system will do. Will it rotate? Will it go toward the center? away from the origin? Both?

Thanks. And I am good, thank you. How are you?



Edit: This won't let me upload my picture. Or I am unable to see my uploaded picture.

Last edited by Reuel (2011-04-05 06:03:27)

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#4 2011-04-05 06:11:07

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi,

I've got a cold but then that's normal at this time of year.  Otherwise great.

I'm not seeing the pic either and I don't have Maple.

Can you do a 'screen shot' and post as a GIF ?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2011-04-05 06:14:17

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Test, test...

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#6 2011-04-05 06:16:46

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Some of these problems take forever and the concept of eigenvalues seems really vague. I was hoping someone had a list of clear-cut, straight-forward rules for how eigenvalues affect a system. I do not like it when math is sketchy. tongue

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#7 2011-04-05 06:28:33

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

Hi

Right.  Got the picture now.

Confession:  I've never used eigenvectors in connection with DEs so I'm having to learn fast myself here.

I've got to:

Is that what you did?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2011-04-05 06:33:53

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Eigenvalues, Eigenvectors

Hi Bob and Reuel;

I do not want to add my input because I never did much of that type of work. I was more interested in the computation of eigenvalues than on what they did.

Here is a little list, see if any of this is useful.

The real part of an eigenvalue tells whether its term increases or
decreases in magnitude.

If the real part is negative, the term dampens in magnitude toward 0.
If the real part is zero, the term has constant
magnitude.

If the real part is positive, the term increases in magnitude toward infinity.

The imaginary part of an eigenvalue tells whether its term is
oscillatory or not.

If the imaginary part is zero, the term is not oscillatory.

If the imaginary part is not zero, the term is oscillatory.

This is for a linear dynamical system. Also hoping this info is correct.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2011-04-05 06:38:16

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi

Thanks bobbym.

I'm still a way from understanding all of that but I remain hopeful.

Using what I've learnt so far  I seem to get an eigenvector of

with eigenvalue = -4i

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#10 2011-04-05 06:38:25

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Yeah. Then you subtract λ from both zeroes and find the determinant or whatever it is called. In many cases you will get what is called a characteristic polynomial which you solve to find one or two eigenvalues which may or may not be complex.

So in this case I guess we find that λ is ±4i.

Last edited by Reuel (2011-04-05 06:42:29)

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#11 2011-04-05 06:47:13

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi

Yes, I've got + or - 4i now.  I didn't use determinants though.  Just matrix multiplication.

Do you know what to do with these values or shall I try to learn some more?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#12 2011-04-05 06:57:03

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

The idea is to compute eigenvectors by substituting the eigenvalues in for λ one at a time. Then you enter the two into a general solution of the form

where L1 and L2 are the eigenvalues since the forum won't let me use λ in math text. V1 and V2 are the eigenvectors.

The initial problem, however, was not in finding a general solution but in recognizing how eigenvalues affect direction fields. It seems like such a nebulous business.

Last edited by Reuel (2011-04-05 06:58:00)

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#13 2011-04-05 07:04:33

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

I have


with

note \lambda is Latex

and


with

Have a look at

http://www.math.purdue.edu/~wilker/Math266S01/complex.pdf

Bob

Last edited by Bob (2011-04-05 07:33:34)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#14 2011-04-05 08:15:36

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Bob,

That is pretty great. I am going to try their exercises. Thanks for the link!

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#15 2011-04-05 08:33:47

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Here is my attempt at exercise 1 on page 3 of that pdf you sent. How does this look?




Which yields that λ is 1 ± i.

So that


And by the same logic it can be found that


If I am doing this correctly.

How does that look?


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#16 2011-04-05 08:36:46

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Hmmm... do I need to remove the complex part?

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#17 2011-04-05 08:49:12

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Here is a problem from a handout I tried.


I got


Is anyone getting similar answers? smile

P. S. Thanks for the help.

Last edited by Reuel (2011-04-05 08:49:53)

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#18 2011-04-05 19:19:41

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi Reuel,

Post #15.  I agree with all of your working here.   smile

post #16  I don't know.

post #17  I agree with you here as well.  smile

I'm almost ready for a longer post on directional fields and eigenvectors.  I've just got one element to get sorted.  Hopefully I'll do this evening (It's just after 8 am at the moment here.)

Bob

Last edited by Bob (2011-04-05 19:35:30)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#19 2011-04-06 08:21:29

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi Reuel,

Drawing directional fields without software:
In the x-y plane, a directional field is a diagram showing how typical slopes (dy/dx) for different points (x,y).

You do not have to have software to make these.

Example:

We want the slope dy/dx so divide the second by the first:

Now use this to calculate the slope at certain points.

(i) On the line y = x the gradient will be -1.  On the line y = -x the slope will be +1.

I have shown some lines for this in red on my diagram below.

(ii) When x = 0, the slope will be zero.  When y = 0, the slope will be vertical.

I have shown some lines for this in green on my diagram below.

(iii) If y > x with both positive, the slope will be less than 1.  If y < x with both positive the slope will be more than 1.  You can similarly work out slopes for points in the other quadrants.

I have shown these in blue on my diagram below.

Thus we get the direction field diagram.

It looks like solutions to the DEs will be circles.  That is not surprising as this question can also be done by separation of variables.

Rearrange and put a new constant:

But not all DEs can be simplified in this way.

Why does finding eigenvectors and eigenvalues help?
This was the bit I had not met before.

The explanation on http://www.sosmath.com/diffeq/system/linear/eigenvalue/eigenvalue.html

is very good and I don't think I could improve upon it.  It shows why the DE converts to a matrix calculation involving eigenvectors and values and also why a solution will have the form

And then the general solution comes from constructing a generalised linear combination of all the solutions found.

Because any linear combination of eigenvectors is also an eigenvector.

proof

as A is a linear transformation.

(I think this proof may need a bit of tidying up but it should be enough to justify why the method works.)

Bob

Last edited by Bob (2011-04-06 09:00:59)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#20 2011-04-18 03:33:00

Puneeth B C
Guest

Re: Eigenvalues, Eigenvectors

The proof for "any linear combination of eigenvectors is also an eigenvector" is a false proof.
    A*(k1*V1+k2*V2)
=  k1*A*V1 + k2*A*V2
A*V1 = L1*V1 because L1(lambda 1) is the eigen value for the corresponding eigen vector V1 and
A*V2 = L2*V2 because L2(lambda 2) is the eigen value for the corresponding eigen vector V2.

The two lambdas need not be same.
=k1*L1*V1 + k2*L2*V2
unless L1 and L2 are the same, the proof is not valid.

In general a linear combination of eigen vectors is not an eigen vector. An nxn matrix with distinct eigen values will have linearly independent eigen vectors which can span the whole of nD space. If the above proof were to be true there was no need to calculate the eigen vectors as any vector could serve as one.

#21 2011-04-18 05:41:09

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Eigenvalues, Eigenvectors

hi Puneeth B C

Thanks for that.  I did suspect that bit myself for that reason but assumed there was a way round it as the property seems to be used extensively in this area of differential equations.

Perhaps you can enlighten me about the correct way to do this as I can find nothing to explain the method beyond what I have said.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#22 2011-04-18 08:25:39

Reuel
Member
Registered: 2010-11-28
Posts: 178

Re: Eigenvalues, Eigenvectors

Thanks, everyone, for your input.

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#23 2011-04-18 19:15:18

Puneeth B C
Guest

Re: Eigenvalues, Eigenvectors

As I said earlier, suppose I have two linearly independent vectors (real), then they are capable of spanning the whole of 2D space. Assume we have a matrix A = [2 1;1 2] (Ref: Wiki) has 2 LI(linearly independent) EVs(eigen vectors); (-1,1) with ev(eigen value) 1 and (1,1) with ev 3.
Now,
    A*(-1,1) = (-1,1)   -  is in the same direction as (-1,1)
    A*(1,1) = (3,3)      -  is in the same direction as (1,1)
Suppose now that I want to get the transformation at a point (1,3) which is not an eigen vector, I can either multiply it directly with A to get
A*(1,3) = (5,7)   -  Note that this is not in the same direction as (1,3) because (1,3) is not an eigen vector ------------------->    (a)
Since the two LI EVs can span the entire 2D space I should be able to span the point (1,3) as a linear combination(LC) of the EVs.
(1,3) = 2*(1,1) + 1*(-1,1)
A*(1,3) = A*2*(1,1) + A*1*(-1,1)
            = 2*A*(1,1) + 1*A*(-1,1)
            = 2*3*(1,1) + 1*1*(-1,1)
            = (6,6) + (-1,1)
            = (5,7)    ----------------------------->       (b)
        (a) = (b)

So for a nxn invertible matrix A any vector in the nD space can be represented as a LC of its n LI EVs. Any vector in the nD space then can be represented as a LC of the EVs and individually linearly transformed by A. The vector represented as a LC does not become an EV.

#24 2011-04-18 20:42:20

Puneeth B C
Guest

Re: Eigenvalues, Eigenvectors

You can read this for more info:
http://www.jirka.org/diffyqs/htmlver/diffyqsse20.html

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