Find the eigenvalues of this matrix

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Homework Help Overview

The discussion revolves around finding the eigenvalues of a given 2x2 matrix associated with a system of differential equations. Participants are exploring the classification of the equilibrium point at the origin based on these eigenvalues.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts to calculate the eigenvalues and eigenvectors, with some providing results from MATLAB. There are questions about the accuracy of arithmetic and the process of row reduction to find eigenvectors. Some participants express frustration with the matrix representation and the tools used for calculations.

Discussion Status

There is an ongoing exploration of the eigenvalue calculation process, with some participants confirming the correctness of the eigenvalues found. Guidance has been offered regarding the approach to finding eigenvectors, and there is recognition of the challenges faced in the calculations.

Contextual Notes

Participants mention issues with formatting in LaTeX and the complexity of the problem, which may be affecting their ability to communicate their work clearly. There is also a reference to a related problem with a different matrix setup, indicating a broader context of learning.

zfolwick
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Homework Statement


consider the system
[tex]x' = \left[-1 & -1\\<br /> <br /> -.5 & -1\right]x[/tex]
(I'm sorry I can't seem to get a new row in! the second line is [-.5 -1]


solve the system. What are the eigenvalues of the coefficient matrix? Classify the equilibrium point at the origin as to type.

Homework Equations





The Attempt at a Solution



I found the eigenvalues to be [tex]-1 \pm \sqrt{2}/2[/tex]

but solving the matrix for the eigenvalues is proving to be very difficult. MATLAB produces a result: v =

881/1079 881/1079
-780/1351 780/1351

for the command >> [v,d]=eig(A)

any help in an analytical solution would be appreciated.
 
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Use the bmatrix environment to write a matrix in LaTeX. You can click on the matrix below to see the code.

[tex]x' = \begin{bmatrix}-1 & -1 \\ -0.5 & -1\end{bmatrix}[/tex]

Show us your work from when you tried to solve for the eigenvectors analytically.
 
wow... this tex editor is stupid. I give up on that. I hope this is intelligible to everybody.

r1 <- eigenvalue 1 := -1+sqrt(2)/2

A-r1*I -->

[-2+sqrt(2)/2 ...-1 ]
[-.5 ...... -2+sqrt(2)/2]

where A is the 2x2 matrix:
-1 -1
-.5 -1I've tried everything I know to solve this... I haven't a clue, but I know that one eigenvector is [-sqrt(2);1]
 
I also have another problem where they change the .5 to a 2 and ask us to do the same thing, but since I can't figure this out, I won't be able to figure the other out. Any help would be excellent.
 
zfolwick said:
wow... this tex editor is stupid. I give up on that. I hope this is intelligible to everybody.
Keep in mind you are asking others for help. TeX isn't rocket science, so you should make some effort into making your request easy to read. All you had to do was click on the matrix I wrote above, copy and paste, and tweak two entries.
r1 <- eigenvalue 1 := -1+sqrt(2)/2

A-r1*I -->

[-2+sqrt(2)/2 ...-1 ]
[-.5 ...... -2+sqrt(2)/2]

where A is the 2x2 matrix:
-1 -1
-.5 -1

I've tried everything I know to solve this... I haven't a clue, but I know that one eigenvector is [-sqrt(2);1]
Start by checking your arithmetic.
 
copying and pasting your code:

[tex]x' = \begin{bmatrix}-1-r & -1 \\ -0.5 & -1-r\end{bmatrix}[/tex]

which gives the characteristic polynomial
[tex](-1 - r)^2 - (-.5)(-1) = 0[/tex]

I work this out to be
[tex]r^2 +2r +1 - .5 = r^2 +2r +.5[/tex]

pluggin into the quadratic formula:
[tex] r = -2/2 \pm \sqrt{4 - 4(1)(1/2)}/2[/tex]

which agrees with the book's answer of [tex]-1 \pm \sqrt{2}/2[/tex]

this is eigenvalue number one. The negativity of both eigenvalues tells me that this will be a node, which agrees with the books' answer.

To find the eigenvector, I will again copy the TeX code above and do the initial subtraction:

[tex]\begin{bmatrix}-2+$\sqrt{2}$/2 & -1 \\ -0.5 & -2+$\sqrt{2}/2$\end{bmatrix}[/tex]

the question I have is how do I row reduce this to find the eigenvectors? I can only assume my arithmetic is correct, since that's what the book tells me, but in trying to row reduce them, I'm only coming up with trivial answers, or the terms aren't reducing... This is getting increasingly frustrating as this should be a relatively simple problem to solve.

**EDIT: For some reason the square root signs are now messed up... I've tried to fix but it should read: "The square root of 2 divided by 2". Sorry.

**EDIT 2: I've fixed the matrix for A-rI
 
Sorry, I meant for you to check your arithmetic when calculating A-rI. Your eigenvalues are correct.
 
sooo... any ideas for row reduction? Or is there an alternative?
 
To find the eigenvectors, you need to solve the equation [itex](A - rI)x = 0[/itex]. You can do row reduction, but you should be able to just solve this by inspection.

The eigenvectors that MATLAB returns are normalized so that their length is 1. Any constant multiple of an eigenvector is also an eigenvector, of course.
 
  • #10
There's no need to do a row reduction with only a 2x2 matrix.

What do you get for [itex]-1-(-1\pm\sqrt{2}/2)[/itex]? I suspect you didn't bother to write in the parentheses, which is causing your confusion.
 
  • #11
vela said:
There's no need to do a row reduction with only a 2x2 matrix.

What do you get for [itex]-1-(-1\pm\sqrt{2}/2)[/itex]? I suspect you didn't bother to write in the parentheses, which is causing your confusion.

That'd be (at least one of) the reasons why I'm so torqued up over this problem.

All afternoon on a SIGN error...wow...

I'll grind through and see what comes out. Thanks.
 
  • #12
It happens to all of us some time. While it's obviously frustrating to realize how much time you spent because of a simple mistake, you have to admit it's relieving to finally figure out what the problem was. :smile:
 
  • #13
vela said:
It happens to all of us some time. While it's obviously frustrating to realize how much time you spent because of a simple mistake, you have to admit it's relieving to finally figure out what the problem was. :smile:

no kidding.

Now on to complex eigenvalues and shenanigans...
 

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