# Math Help - Solution by Characteristic Value

1. ## 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.

2. 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.

3. 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.

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

5. 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 $\{v_{\lambda_i}^{(k)}\}_{k=1}^{k_i}$ 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!