# Solution by Characteristic Value

• Jun 9th 2008, 11:54 AM
Kim Nu
Solution by Characteristic Value
Heres a problem that I have solved:

Find the general solution of the equation x' = Ax, where A is the given matrix:

[-2 4]
[1 1]

Using characteristic values, this matrix turns into:

[-2-λ 4]
[1 1-λ]

Simplifies to (λ + 3)(λ - 2) = 0

λ1 = 2
λ2 = -3

For λ1, (A - λI)k = 0
[-4 4][k1]=[0]
[1 -1][k2]=[0]

Solving: c1e^(2t)[1]
[1]

Set this up the same way with λ2:

[1 4][k1]=[0]
[1 4][k2]=[0]

Solving: c2e^(-3t)[4 ]
[-1]

Thus the solution is x(t) = c1c1e^(2t)[1] + c2e^(-3t)[4 ]
[1] [-1]

How do I go about solving for the general solution of the matrix:

[-2 0]
[0 -2]

I run into a brick wall pretty quick on this one.
• Jun 9th 2008, 03:21 PM
albi
In this case we have multiple eigenvalue. There is the method which deals with such cases. Generally it relays on solving equation (A-lambda*I)^n x = 0

Maybe I will write something tomorrow about it....

I guess you don't speak polish, but equations are international.
• Jun 9th 2008, 04:13 PM
Kim Nu
Hey thanks for the response. I did take a look at the "polish" document, but I don't think it addressed my pickle:

[-2 0]
[0 -2]

Using characteristic values this turns into:

[-2-λ 0]
[0 -2-λ]

Thus:

(-2-λ)^2 = 0

λ1 = -2, λ2 = -2

So:

[-2--2 0]
[0 -2--2]

[0 0][k1]=[0]
[0 0][k2]=[0]

So this is where I'm stuck.
• Jun 10th 2008, 12:46 PM
albi
This case is very simple. We have equation:

$\dot{\mathbf{x}} = \mathbf{A}\mathbf{x}$

for
$\mathbf{A} = \left[\begin{array}{lr} -2 & 0 \\ 0 & -2 \end{array}\right]$

The case is very simple. If $\mathbf{x} = [x_1, x_2]^T$ than the equation reduces to:
$\begin{array}{l} \dot{x_1} = -2 x_1 \\ \dot{x_2} = -2 x_2 \end{array}$

For which obvious solution is
$\begin{array}{l} x_1 = C_1 e^{-2t} \\ x_2 = C_2 e^{-2t} \end{array}$

Hence $\mathbf{x} = [C_1, C_2]^T e^{-2t}$
• Jun 10th 2008, 01:31 PM
albi
Of course you want to use eigenvalues at all costs. Than the case complicates.

The problem is we have multiple eigenvalue. The general formula for fundamental matrix I mentioned before is:
(*) $
e^{\mathbf{A}t} = \sum_{i = 1}^{n} \sum_{j = 0}^{k_i} e^{\lambda_i t} \frac{t^j}{j!}(\mathbf{A} - \lambda_i \mathbf{I} )^{j} \mathbf{E}_{\lambda_i}
$

Where n is eigenvalues number, k_i is multiplicity of i-th eigenvalue and $\mathbf{E}_{\lambda_i}$ is projection matrix from $\mathbb{R}^m$ to

$span\{v_{\lambda_i}^{(k)}\}_{k=1}^{k_i}$

Where (Wink) $\{v_{\lambda_i}^{(k)}\}_{k=1}^{k_i}$ are generalised eigenvectors which are solution of the equation:
$(\mathbf{A} - \lambda_i \mathbf{I} )^{k_i} \mathbf{x} = 0$

In our case $span\{v_{-2}^{(k)}\}_{k=1}^{2} = \mathbb{R}^2$ (up to isomorphism). Hence projection matrix $\mathbf{E}_{-2} = \mathbf{I}$

Using formula (*) we should assume $0^0 = 1$ otherwise it does not make sense... We get (only one part in sum):

$e^{\mathbf{A}t} = \mathbf{I} e^{-2t}$

Thus:

$\mathbf{x} = [C_1, C_2]^T e^{\mathbf{A}t} = [C_1, C_2]^T\mathbf{I} e^{-2t} = [C_1, C_2]^T e^{-2t}$

We got the previous result!