Setting up linear systems in a matrix

[Zem]

I am getting stuck on the very beginning of these homework problems.

Solve the linear system to determine whether the critical point (0,0) is stable, asymtotically stable, or unstable.

dx/dt = -2x, dy/dt = -2y

The book uses separation of variables, but the professor has instructed us to use matrices to do these homework problems.
x= \left(\begin{array}{c} -2 \ 0\\ 0 \ -2 \end{array}\right)
\lambda_1 = 0, \lambda_2 = 4

Is this the right matrix? When I use lambda = 0 in the eigenvalue method (A - (lambda)I)=0 on that matrix, the vector entries a and b both = 0 because it's just a diagonal matrix. I need to find a general solution such that
x = c1v1e^((lambda)t) + c2v2e^((lambda)t). Then determine the stability of x. But first I need to know how to set it up.

 

[AKG]

You have your function x equal to a 2x2 matrix? That matrix does equal something important, but its certainly not x. And the numbers you found are not the eigenvalues of that matrix. Looking at the matrix, it should be very obvious what the eigenvalue(s) is/are. To help you get started:

\left(\begin{array}{c}x'\\y'\end{array}\right)=\left(\begin{array}{cc}-2&0\\0&-2\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right)

 

[Zem]

I got the eigenvalues with the characteristic equation.
(-2 - \lambda)(-2 - \lambda) -4 = 0
\lambda^2 + 4\lambda = 0
\lambda_1 = 0, \lambda_2 = -4

Case 1:
(A - \lambda I = 0)
-2x + 0y = 0
0x - 2y = 0
x = 0, y =0
v_1 = [0,0]

Case 2:
2x + 0y = 0
0x + 2y = 0
v_2 = [0,0]

Both eigenvectors equal 0? How can I set up a general solution without eigenvectors = 0? The general solution should be of the form x(t) = c_1v_1e^(lambda_1 t) + c_2v_2e^(lambda_2 t)

\lambda t should be in the exponent of e in the equation above, but I don't know how to make the code do that.

 

[AKG]

That's not the characteristic equation.

 

[Zem]

Oops.
\lambda ^2 + 4\lambda + 4 = 0
\lambda_1 = \lambda_2 = -2

Find v_1: (A - \lambda I = 0)
\left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right)

How can I get an eigenvector when (A - lambda) = 0?

Since the eigenvalues are equal and less than zero, I know it is a stable proper node. So the graphing part is done. But I need a general solution.

 

[J77]

substitute back in your eigenvalues (this doesn't give the null matrix) and satisfy:

A{\bf v}=\lambda {\bf v}

or

(A-\lambda I){\bf v}=0

where the eigenvector {\bf v}=(v_1; v_2)

 

[HallsofIvy]

It might be worth your time to notice that if the matrix is
\left(\begin{array}{c} a \ 0\\ b \ c \end{array}\right)
then the eigenvalue equation is
\left|\begin{array}{c}a-\lambda \ 0\\ b\ c-\lambda\end{array}\right|= (a- \lambda)(c-\lambda)= 0
In other words for any triangular (which include diagonal) matrix, the numbers on the main diagonal are the eigenvalues.

 

[Zem]

(A-\lambda I){\bf v}=0
\left(\begin{array}{c} -2 \ 0\\ 0 \ -2 \end{array}\right)
\left(\begin{array}{c} -2-(-2) \ 0\\ 0 \ -2-(-2) \end{array}\right)
\left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right)

or

(-2 - (-2)){\bf v}=0
0v = 0

What am I missing?

 

[Zem]

After asking the professor about this, it turns out the matrix is
\left(\begin{array}{c} 0 \ 0\\ 0 \ 0 \end{array}\right)
after
(A-\lambda I){\bf v}=0

This means any vector will work.
v_1 = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)
v_2 = \left(\begin{array}{c} 0 \\ 1 \end{array}\right)

General Solution:
x(t) = c_1(1)e^-2t + c_2(0)e^-2t = c_1e^-2t
y(t) = c_1(0)e^-2t + c_2(1)e^-2t = c_2e^-2t

 

[AKG]

That's correct. Note you should be able to look at your matrix, and recognize that it's -2I, where I is the identity matrix. If you know what eigenvectors and eigenvalues are (and not just how to compute them) then you should be able to see immediately that the only eigenvalue is -2, and that every vector is an eigenvector.

 

[Zem]

The linear algebra explanation of this is the easiest way for me to understand it. Since (A - \lambda I = 0) is -2a - 0b = 0, a is a free variable. So a could be anything, which leads to an eigenvector that could be anything. Thanks everyone!