I've been struggling with this exercise from Strogatz', I would appreciate any correction of what I've done so far since I've been self studying all of these topics and some (if not all) of it might be wrong. "5.2.11 Show that any matrix of the form $\mathbf{A}=\begin{pmatrix}\lambda && b \\ 0 && \lambda\end{pmatrix}$, with $b\neq 0$, has only a one dimensional eigenspace corresponding to the eigenvalue $\lambda$. Then solve the system $\dot{\mathbf{x}} = \mathbf{A} \mathbf{x}$ and sketch the phase portrait."
So given $\mathbf{A}$ I need to find an eigenvector for the unique eigenvalue $\lambda$ such that $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ where $\mathbf{v}$ is the eigenvector. $$ \mathbf{A}\mathbf{v} = \lambda\mathbf{v} \\ \mathbf{A}\mathbf{v} - \lambda\mathbf{v} = 0 \\ (\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = 0 $$
so basically I have to find the null space of $(\mathbf{A} - \lambda\mathbf{I})$ .- $$ (\mathbf{A} - \lambda\mathbf{I})\mathbf{v}= \left( \begin{pmatrix}\lambda && b \\ 0 && \lambda\end{pmatrix} - \begin{pmatrix}\lambda && 0 \\ 0 && \lambda\end{pmatrix}\right)\begin{pmatrix}\mathrm{v_1} \\\mathrm{v_2} \end{pmatrix} = \begin{pmatrix}0 && b \\ 0 && 0\end{pmatrix} \begin{pmatrix}\mathrm{v_1} \\\mathrm{v_2} \end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix} $$
which is already in its reduced form, thus the null space (kernel) is formed by just one vector, the column vector that is zero, and this will be our eigenvector $\mathbf{v} = (\mathrm{v_1}, \mathrm{v_2})$, since this vector forms the basis for the eigenspace we can conclude that the eigenspace is one dimensional. now we have $b\mathrm{v_2} = 0$ thus $\mathrm{v_2} =0$ and we have $\mathrm{v_1}$ as a free variable that we will set to 1 thus our eigenvector is.- $$ \mathbf{v}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} $$ that way $\mathbf{A}\mathbf{v} = \lambda\mathbf{v}$ holds.
WRONG STATEMENTS FROM THIS PART ON AND SOLUTION BELOW
Now to solve the system $\dot{\mathbf{x}} = \mathbf{A} \mathbf{x}$ we would like a solution of the form $\mathbf{x(t)} = e^{\lambda t} \mathbf{v}$ so $$ \dot{\mathbf{x}} = \lambda e^{\lambda t} \mathbf{v} = \mathbf{A} e^{\lambda t} \mathbf{v}\\ \mathbf{A} \mathbf{v}=\lambda \mathbf{v} $$ so $\mathbf{v}$ is our eigenvector and $\lambda$ our eigenvalue thus- $$ \mathbf{x(t)} = e^{\lambda t} \mathbf{v} =c_1e^{\lambda t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$
where $\mathbf{x(t)} = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$ so $x(t) = c_1e^{\lambda t}$ and $y(t)=0$ but I'm not sure of the result because the system is - $$ \dot{x} = \lambda x + by \\ \dot{y} = \lambda y $$ and $\dot{y}$ can be solved as $y(t)=y_0e^{\lambda t}$ then substitution in $\dot{x}$ gives- $$ \dot{x} = \lambda x +b y_o e^{\lambda t} $$ which I'm no longer now how to solve that's the part in which I'm completely lost, I don't now if my assumption of solving $y$ by separation of variables is fine or the result with the eigenvectors is the correct, I would appreciate any help.
END OF WRONG STATEMENTS
I made the phase portrait though, using mathematica and this is what I got -
With the help of Amzoti who send me this link http://www.math.vt.edu/people/afkhamis/class_home/notes/F08W12.pdf I was able to find the answer, now I'll post it just in case someone else is having problem with this.
Examples and a deeper exposition of the topic is given in the link.
Basically what we would like to have is a "generalized vector of $\mathbf{ A}$" such that $$ \mathbf{x} = te^{\lambda t}\mathbf{v} +e^{\lambda t}\mathbf{w} $$
so that at the time we substitute in $\dot{\mathbf{x}} = \mathbf{A}\mathbf{x}$ what we get is $$ e^{\lambda t}\mathbf{v} +\lambda te^{\lambda t}\mathbf{v}+\lambda e^{\lambda t}\mathbf{w} = te^{\lambda t}\mathbf{A}\mathbf{v} +e^{\lambda t}\mathbf{A}\mathbf{w} $$ And equating terms and eliminating we get.- $$ \mathbf{v} +\lambda\mathbf{w} = \mathbf{A}\mathbf{w}\\ \lambda \mathbf{v}=\mathbf{A}\mathbf{v} $$ we have already worked out $(\mathbf{A}-\lambda\mathbf{I})\mathbf{v} =0$ we're left out with just $(\mathbf{A}-\lambda\mathbf{I})\mathbf{w} =\mathbf{v}$ which we will obtain in the following way.- $$ (\mathbf{A} - \lambda\mathbf{I})\mathbf{w}=\mathbf{v} \\ \left( \begin{pmatrix}\lambda && b \\ 0 && \lambda\end{pmatrix} - \begin{pmatrix}\lambda && 0 \\ 0 && \lambda\end{pmatrix}\right)\begin{pmatrix}\mathrm{w_1} \\\mathrm{w_2} \end{pmatrix} = \begin{pmatrix}\mathrm{v_1} \\\mathrm{v_2} \end{pmatrix}\\ \begin{pmatrix}0 && b \\ 0 && 0\end{pmatrix} \begin{pmatrix}\mathrm{w_1} \\\mathrm{w_2} \end{pmatrix}= \begin{pmatrix}1 \\ 0 \end{pmatrix} $$ We have $b\mathrm{w_2} =1$ so $\mathrm{w_2}=\frac{1}{b}$ and $\mathrm{w_1}$ is a free variable that we will set to 1 thus $$ \mathbf{w} = \begin{pmatrix}1 \\ \frac{1}{b} \end{pmatrix} $$
Now the solution of the system is the following.- $$ \mathbf{x(t)}=c_1e^{\lambda t}\mathbf{v}+ c_2e^{\lambda t}(t\mathbf{v}+\mathbf{w})\\ \begin{pmatrix}x(t) \\ y(t) \end{pmatrix} =c_1e^{\lambda t}\begin{pmatrix}1 \\ 0 \end{pmatrix} +c_2e^{\lambda t} \left(t \begin{pmatrix}1 \\ 0 \end{pmatrix} + \begin{pmatrix}1 \\\frac{1}{b} \end{pmatrix} \right) \\ x(t) =c_1e^{\lambda t} +c_2e^{\lambda t} +c_2te^{\lambda t}\\ y(t)=\frac{c_2}{b}e^{\lambda t} $$ and that's pretty much all you can plot some solutions with this code in mathematica
fx[t_, c1_, c2_] := c1 Exp[t] + c2 t Exp[t] + c2 Exp[t]
fy[t_, c2_] := c2 Exp[t]
Manipulate[ ParametricPlot[{fx[t, c1, c2], fy[t, c2]}, {t, -5, 5}], {c1, -1, 1}, {c2, -1, 1}]