0

I am trying to solve the system $x'=\begin{bmatrix}1&0&1\\0&1&0\\0&0&1\end{bmatrix}x$.

The eigenvalue is $1$ with multiplicity 3. The eigenbasis is generated by $e_1,e_2$. The corresponding term in my solution (OR SO I THOUGHT!) is $(A)\\\Bigg(c_1\begin{bmatrix}1\\0\\0\end{bmatrix}+c_2\begin{bmatrix}0\\1\\0\end{bmatrix}\Bigg)e^t$.

I have to find a generalized eigenvector. Am I allowed to choose some $v_1$ in the span of $e_1,e_2$, or is it necessarily the case that $v_1=\begin{bmatrix}c_1\\c_2\\0\end{bmatrix}$?

My original idea was to choose $v_1=e_1$ ($c_1=1,c_2=0$) and I find that $v_2=e_3$. So if I proceed in this direction, should the general solution be

$$(B)\\ \Bigg(c_1\begin{bmatrix}1\\0\\0\end{bmatrix}+c_2\begin{bmatrix}0\\1\\0\end{bmatrix}\Bigg)e^t+D\Bigg(\begin{bmatrix}0\\0\\1\end{bmatrix}+t\begin{bmatrix}1\\0\\0\end{bmatrix}\Bigg)e^t$$

OR

$$(C)\\C\begin{bmatrix}1\\0\\0\end{bmatrix}e^t+D\Bigg(\begin{bmatrix}0\\0\\1\end{bmatrix}+t\begin{bmatrix}1\\0\\0\end{bmatrix}\Bigg)e^t$$

Edited: I'm confused. They both work as solutions, so (B) is the general one. But according to the outline given by Git Gud in Finding a solution basis (as well as my textbook) the general solution should be of the form given in (C).

Community
  • 1
cap
  • 2,443

1 Answers1

1

The matrices $$ A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ and $$ B = \begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$ commute, so that $$ e^{A+B} = e^A e^B = e^B e^A $$ Very similar, introducing the independent variable $t,$ it is also true that $At$ and $Bt$ commute, so that $$ e^{(A+B)t} = e^{At} e^{Bt} = e^{Bt} e^{At} $$

Will Jagy
  • 139,541