# Homework Help: Find the eigenvalues of this matrix

1. Jul 18, 2010

### zfolwick

1. The problem statement, all variables and given/known data
consider the system
$$x' = \left[-1 & -1\\ -.5 & -1\right]x$$
(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.

2. Relevant equations

3. The attempt at a solution

I found the eigenvalues to be $$-1 \pm \sqrt{2}/2$$

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.

2. Jul 18, 2010

### vela

Staff Emeritus
Use the bmatrix environment to write a matrix in LaTeX. You can click on the matrix below to see the code.

$$x' = \begin{bmatrix}-1 & -1 \\ -0.5 & -1\end{bmatrix}$$

Show us your work from when you tried to solve for the eigenvectors analytically.

3. Jul 18, 2010

### zfolwick

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 -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]

4. Jul 18, 2010

### zfolwick

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.

5. Jul 18, 2010

### vela

Staff Emeritus
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.

6. Jul 18, 2010

### zfolwick

$$x' = \begin{bmatrix}-1-r & -1 \\ -0.5 & -1-r\end{bmatrix}$$

which gives the characteristic polynomial
$$(-1 - r)^2 - (-.5)(-1) = 0$$

I work this out to be
$$r^2 +2r +1 - .5 = r^2 +2r +.5$$

pluggin in to the quadratic formula:
$$r = -2/2 \pm \sqrt{4 - 4(1)(1/2)}/2$$

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

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:

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

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

7. Jul 18, 2010

### vela

Staff Emeritus
Sorry, I meant for you to check your arithmetic when calculating A-rI. Your eigenvalues are correct.

8. Jul 18, 2010

### zfolwick

sooo... any ideas for row reduction? Or is there an alternative?

9. Jul 18, 2010

### hgfalling

To find the eigenvectors, you need to solve the equation $(A - rI)x = 0$. 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. Jul 18, 2010

### vela

Staff Emeritus
There's no need to do a row reduction with only a 2x2 matrix.

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

11. Jul 18, 2010

### zfolwick

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. Jul 18, 2010

### vela

Staff Emeritus
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.

13. Jul 18, 2010

### zfolwick

no kidding.

Now on to complex eigenvalues and shenanigans...