Question

Find the fundamental matrix of the linear system x'=Ax where A is the 3x3 matrix [{1,1,0},{0,1,1},{0,0,1}.

I know I should use the Putzer's Method but I keep getting stuck

Answer 1

Finding the fundamental matrix is a matter of computing the eigenvalues and -vectors for the given matrix.

Solve for all \(\lambda\):
\[\begin{vmatrix}
1-\lambda&1&0\\
0&1-\lambda&1\\
0&0&1-\lambda
\end{vmatrix}=0\]

Answer 2

To get you started... With a cofactor expansion along the first column, you'll find the determinant to be
\[\begin{vmatrix}
\color{red}{1-\lambda}&1&0\\
\color{red}0&1-\lambda&1\\
\color{red}0&0&1-\lambda
\end{vmatrix}=\color{red}{(1-\lambda)}\begin{vmatrix}1-\lambda&1\\0&1-\lambda\end{vmatrix}-\color{red}0+\color{red}0=(1-\lambda)^3\]

Answer 3

right. I found that, but I was told to use the Putzer method since all of the eigenvalues are the same and will give the same eigenvectors

Answer 4

I'm not familiar with the Putzer method, at least not by name, but to deal with repeated eigenvalues, the way I learned to do it was by finding the eigenvalues \(\vec{\eta}_i\) that satisfy
\[(A-\lambda I)~\vec{\eta}_1=\vec{0}\\
(A-\lambda I)^2~\vec{\eta}_2=\vec{0}\\
(A-\lambda I)^3~\vec{\eta}_3=\vec{0}\]
and I vaguely recall that you should get the same result if you do the following:
\[(A-\lambda I)~\vec{\eta}_1=\vec{0}\\
(A-\lambda I)~\vec{\eta}_2=\vec{\eta}_1\\
(A-\lambda I)~\vec{\eta}_3=\vec{\eta}_2\]
but I might be mistaken about that...

Answer 5

hmmm I've never seen that but if it will help me solve the question I will try it. So am I trying to find when that matrix times the eigenvalue n will give 0 or is it when the determinant of that matrix times the eigenvalue is zero?

Answer 6

You're multiplying matrices, not determinants. So the first eigenvalue can be found as follows:
\[(A-I)~\vec{\eta}_1=\vec{0}~~\iff~~\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix}\eta_{1,1}\\\eta_{1,2}\\\eta_{1,3}\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\]
for which you'll find that \(\eta_{1,1}\) can take on any number. For simplicity, we'd let \(\eta_{1,1}=1\), so the first eigenvector would be
\[\vec{\eta}_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\]

Answer 7

this looks pretty similar to the putzer method. just different letters. I got stuck trying to figure out what n1, n2, and n3. I got n1=t^2/2(e^t)+1 n2=te^t snf n3=e^t-1 but I feel like these are wrong.

Answer 8

By "n1,n2,n3", if you mean \(\vec{\eta}_1,~\vec{\eta}_2,~\vec{\eta}_3\), you should be getting vectors with constant entries... I'm not sure what you're saying you're getting.
\[\begin{cases}(A-\lambda I)~\vec{\eta}_1=\vec{0}&(1)\\
(A-\lambda I)~\vec{\eta}_2=\vec{\eta}_1&(2)\\
(A-\lambda I)~\vec{\eta}_3=\vec{\eta}_2&(3)\end{cases}\]
We already solved (1), so sub that into (2) to find \(\vec{\eta}_2\).
\[\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}
\begin{pmatrix}\eta_{2,1}\\\eta_{2,2}\\\eta_{2,3}\end{pmatrix}
=\begin{pmatrix}1\\0\\0\end{pmatrix}\]
which gives \(\eta_{2,2}=1\) and \(\eta_{2,3}=0\), while \(\eta_{2,1}\) can take on any real number. Again, for simplicity, we can set \(\eta_{2,1}=1\), so the second eigenvector would be
\[\vec{\eta}_2=\begin{pmatrix}1\\1\\0\end{pmatrix}\]

Answer 9

oh i see where I messed up!!! thank you so much!!!

Answer 10

You're welcome!