# How to obtain the state matrix of this trajectory?

Continuous-time LTI case.
I have a problem getting the state matrix of this trajectory.

One element of the state matrix is known. $$A = \begin{pmatrix} a & 4 \\c & d \end{pmatrix}$$

I know that one Eigenvalue is 0, $s_1 = 0$, and one Eigenvector is $p_1=\begin{pmatrix} 1\\1 \end{pmatrix}$.
With the equation $s_1 - a*p_{11} + s - 4*p_{12} = 0$, i get $a = -4$.
I also found out that $c=-d$.
But I am stuck now. I don't know how to get the other elements.

Help appreciated...

## German version:

Für ein freies System 2. Ordnung der Form $\frac{dx}{dt} = A*x$ ist das dazugehörige Trajektorienbild gegeben.
Außerdem ist die zugehörige Systemmatrix A (teilweise) bekannt: $$A = \begin{pmatrix} a & 4 \\c & d \end{pmatrix}$$ Bestimmen Sie die fehlenden Elemente der Systemmatrix A . (HINWEIS: Benützen Sie das angegebene Trajektorienbild!)

## Translation:

The trajectory image associated with a free system of second order of the form $\frac{dx}{dt} = A*x$ is given. Further, the associated system matrix $A$ is (partially) known: $$A = \begin{pmatrix} a & 4 \\c & d \end{pmatrix}$$ Determine the missing elements of the system matrix $A$. (HINT: Use the given trajectory image!)

• A Google search seems to indicate that "system matrix" and "trajectory image" are terms of art in optical imaging. Neither of them has a Wikipedia entry. I suspect you'll have a much better chance of getting an answer if you define these terms and explain what it is you're trying to do. – joriki Oct 4 '11 at 16:03
• I tried to explain it better. All my studies are in german so it it not that easy to translate the math specific expressions. – madmax Oct 4 '11 at 16:12
• You can try writing in German, and then somebody else can translate for you here. – J. M. is a poor mathematician Oct 4 '11 at 16:14
• German version added. Topic is control engineering. – madmax Oct 4 '11 at 16:19
• I added a translation to English. Are you sure it says "2. Ordnung"? The differential equation given is of first order. – joriki Oct 4 '11 at 16:31

If ruhezone is the set of the equilibrium points, then $\dot x=0$ when $x_1(t)=x_2(t)$, and if those are arrows pointing towards the $x=y$ line, the elements of the eq. point set are stable. The question, as far as I understand, wants you to make a qualitative assesment of how this line is the attractor of the trajectories in its neighborhood.
From the first hint, $A_{12} = 4$ we obtain $$\pmatrix{\dot x_1\\\dot x_2} = \pmatrix{-4 &4\\ c&d}\pmatrix{x_1\\ x_2}$$ Just test for yourself, whenever $x_1>x_2$, $\dot x_1<0$ and $x_1<x_2$, $\dot x_1>0$.
Now, this is where I might mistaken: if those arrows are for vector field annotation and horizontal arrow means the vector field has no $x_2$ contribution, that means $\dot x_2$ is zero everywhere no matter what. Hence $c=d=0$.
$$\pmatrix{\dot x_1\\\dot x_2} = \pmatrix{-4 &4\\ 0&0}\pmatrix{x_1\\ x_2}$$
Indeed, 0 is an eigenvalue and $\begin{pmatrix} 1\\1 \end{pmatrix}$ is the eigenvector together with -4 and $\begin{pmatrix} 1\\0 \end{pmatrix}$.