Question

find the fundamental matrix e^At for the linear system:
x' = [1 1 0] x
[0 1 0]
[1 0 1]

Answer 1

i believe that the given matrix is the matrix A in x = e^At , i.e no work required.

Answer 2

no.. there is some work that needs to be done.

Answer 3

Did you find the eigenvalues of the given matrix?

Answer 4

\[\begin{vmatrix}
1-\lambda&1&0\\
0&1-\lambda&0\\
1&0&1-\lambda
\end{vmatrix}
=(1-\lambda)^3=0~\Rightarrow~\lambda_1,\lambda_2,\lambda_3=1\]
The eigenvalue \(\lambda=1\) has multiplicity 3, so you'll have to find a generalized eigenvector, if I recall correctly.

Answer 5

i got to this part.. i don't know how to go further.

Answer 6

Do you know how to do it using putzer's algorithm? Using e^At?

Answer 7

I haven't heard of that algorithm, no. But I have learned a thing or two about matrix exponentials.

Answer 8

okay... can you help me go further please?

Answer 9

Just a sec... I need to see if I can work this out.

Answer 10

First eigenvector: Solve for \(\vec{a}\) such that \(A-\lambda I=0\), or equivalently,
\[\begin{pmatrix}
0&1&0\\0&0&0\\1&0&0\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
Solving yields \(a_1=0,a_2=0\) and \(a_3\) can be any real number. Pick any non-zero real for \(a_3\); I'll pick \(1\).

So your first eigenvector is
\[\vec{a}_1=\begin{pmatrix}0\\0\\1\end{pmatrix}\]

Answer 11

Have you covered generalized eigenvectors yet? I don't want to present anything new because it's been a while since I've learned all this and I doubt I'd be able to introduce it with ease.

Answer 12

no we haven't... so using the same method, i find out a2 and a3?

Answer 13

Not the exact same method, no. It's a bit different.

The first generalized eigenvector is obtained by solving for \(\vec{a}_2\) in \((A-\lambda I)\vec{a}_2=\vec{a}_1\).
\[\begin{pmatrix}
0&1&0\\0&0&0\\1&0&0\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=\begin{pmatrix}0\\0\\1\end{pmatrix}\]
This time, you get \(a_1=1,a_2=0\) and \(a_3\) can be any real. Pick \(a_3=0\) for convenience. You want something that makes \(\vec{a}_1\) and \(\vec{a}_2\) linearly independent.
\[\vec{a}_2=\begin{pmatrix}1\\0\\0\end{pmatrix}\]

Answer 14

The second generalized eigenvector is found similarly. Solve for \(\vec{a}_3\) such that \((A-\lambda I)\vec{a}_3=\vec{a}_2\).
\[\begin{pmatrix}
0&1&0\\0&0&0\\1&0&0\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\]
This time you get \(a_1=0\) and \(a_2=1\), and \(a_3\) can be any real. Let \(a_3=0\) for ease:
\[\vec{a}_3=\begin{pmatrix}0\\1\\0\end{pmatrix}\]

Answer 15

so is a2 (001)?

Answer 16

No that's the first eigenvector.
\[\vec{a}_1=\begin{pmatrix}0\\0\\1\end{pmatrix},~\vec{a}_2=\begin{pmatrix}1\\0\\0\end{pmatrix},~\vec{a}_3=\begin{pmatrix}0\\1\\0\end{pmatrix}\]

Answer 17

yeah.. that's what i meant!

Answer 18

So the linearly independent solutions are
\[x_1=e^t\begin{pmatrix}0\\0\\1\end{pmatrix}\]
\[x_2=e^t\begin{pmatrix}1\\0\\0\end{pmatrix}\]
\[x_3=e^t\begin{pmatrix}0\\1\\0\end{pmatrix}\]
giving you the fundamental matrix solution of
\[X(t)=\begin{pmatrix}0&e^t&0\\0&0&e^t\\e^t&0&0\end{pmatrix}\]
(The eigenvectors multiplied by \(e^t\) make up the columns of the fundamental matrix solution.)

Lastly, to find \(e^{At}\), you make use of this theorem:
If \(X\) is a fundamental matrix solution of the system \(x'=Ax\), then \(e^{At}=X(t)X^{-1}(0)\).

Answer 19

thank you so much!!

Answer 20

how do you find out a1 again?

Answer 21

Given some matrix \(A\), you first find the eigenvalues, which are the values of \(\lambda\) such that \(det(A-\lambda I)=0\).

Most of the time, you'll you be given a matrix that has non-repeating eigenvalues. In this case, you plug in each of the values of \(\lambda\) and solve the following equation for the corresponding eigenvector: \((A-\lambda I)\vec{v}=\vec{0}\), where \(\vec{v}\) is the corr. eigenvector.

Here's a link for more practice: http://tutorial.math.lamar.edu/Classes/DE/LA_Eigen.aspx