# System of equations solution description

• Apr 8th 2009, 12:10 PM
Oijl
System of equations solution description
If I have

x' = Ax

where A is the matrix below

(1, -2
3, -4)

then the direction field of the solutions would have everything moving towards (0, 0) as t --> infinity. My question is: how would I describe the behavior as t --> infinity in words? Is it just "stable"?

EDIT:

A much more important question is: if I have a 2X2 matrix A, and the eigenvalues of that A are both greater than zero, then what would the direction field look like?

Edit:

Again, an even MORE important question is: if I have A as a 3X3 matrix, and so have three eigenvalues for it, and two of them are the same.... writing the general solution wouldn't be just

x = (constant1)(eigenvector1)(e^eigenvalue1t) + (constant2)(eigenvector2)(e^eigenvalue2t) + (constant3)(eigenvector3)(e^eigenvalue3t)

would it? That just doesn't seem right, and I can't get my answer doing it that way to match the book's answer.

Thanks.

ANOTHER EDIT:

While I'm at it, let's put in another question here. I'm sure it's a simple question, but my class is at eight in the morning, and lasts forever, so... I drift.

Anyway, how could I write an eigenvector in terms of its eigenvalue? What I'm trying to do is show that, considering the 2X2 system x' = Ax,
(A - r1I)E1 = (r1 - r2)E1
where r1 and r2 are the eigenvalues and E1 is the eigenvector corresponding to r1.
• Apr 8th 2009, 05:26 PM
Ill throw in a bit of help.

Here's how you classify linear systems...

Eigenvalues ------------------- Condition ------------------------Type of system
real, opposite sign ------------ $det A < 0$ ------------------------ saddle

real, positive ----------------- $0 < det A < \frac{(tr A)^2}{4}, tr A > 0$ ----- unstable node

real, negative ---------------- $0 < det A < \frac{(tr A)^2}{4}, tr A < 0$ ------- stable node

complex, positive real part ----- $det A > \frac{(tr A)^2}{4}, tr A > 0$ -------- unstable focus

complex, negative real part ----- $det A > \frac{(tr A)^2}{4}, tr A < 0$ -------- stable focus

complex, zero real part --------- $det A > 0, tr A = 0$ -------------- centre

Where det = determinant and tr = trace of a matrix

Sorry im no good at doing latex tables!