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We need to find an explicit conjugacy between the flows of these 2 systems

1st system $X'$ = $AX$ and second system $Y'$ = $BY$

A = $$\begin{bmatrix} -1 & 1 \\ 0 &2\end{bmatrix}$$

B= $$\begin{bmatrix} 1 & 0 \\ 1 &-2 \end{bmatrix}$$

I have tried by finding the 2 eigen values of both and solving both systems. then applying a Mapping H such that $X(0)$ = $X_{0}$= $(x_{0},y_{0})$ so that $HAX$ = $ BY(h_{1},h_{2})$

we want $Y(0)$ = $Y_{0}$= $(h_{1} x_{0}, h_{2}y_{0})$

my text book has no examples. their should be a capital Theta showing the flow of theta A and Flow of theta B such that their must some mapping $H$ that acts differently on y then x to that they are the same.

Any anything welcome especially considering i don't expect you to understand what i am talking about cause i don't.

for $X'$ eigenvalues $a_{1}$=-1 and $a_{2}$=2 and vectors $v_{1}$= <1,0> $v_{2}$= <1,3>

for $Y'$ eigenvalues $b_{1}$=-2 and $b_{2}$=1 and vectors $w_{1}$= <0,1> $w_{2}$= <3,1>

we have $C_{a1}$ = $x_{0}$ - $y_{0}/3$ and $C_{a2}$ = $y_{0}/3$

$C_{b1}$ = $y_{0}$ - $x_{0}/3$ and $C_{b2}$ = $x_{0}/3$

$h_{1}$ $C_{a1}$ + $h_{1}$ $C_{a2}$ + $h_{2}$ 3$C_{a2}$

needs to equal $h_{1}$ $C_{b1}$ + $h_{2}$ [3 $C_{b1}$ + $C_{b1}$]

and we need to make the 2 equal by guessing h1 and h2

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up vote 3 down vote accepted

Some preliminaries.

Theorem Let $A$ and $B$ be conjugate, or similar matrices, that is, there exists $S$ invertible such that $A = S \cdot J \cdot S^{-1}$. Then the flows defined by $A$ and $B$ are conjugate, that is $\phi^{A}_t = S \circ \phi^{B}_t \circ S^{-1}$ for all $t$. In particular, the flow of $A$ is related to the flow of its canonical form in the same exact manner that $A$ is related to its canonical form, namely, via multiplying on both sides with a transformation $S$ and its inverse.

Proof. We could reduce everything to canonical form, or observe that, since $\phi^{A}_t = e^{At}$, the matrix exponential, we have:

$$S \phi^{A}_t S^{-1} = S e^{At} S^{-1} = e^{t S A S^{-1}} = e^{B t} = \phi^{B}_t.$$

We have the two matrices:

$$A = \begin{bmatrix} -1 & 1 \\ 0 &2\end{bmatrix}$$

$$B = \begin{bmatrix} 1 & 0 \\ 1 &-2 \end{bmatrix}$$

For the matrix $A$, we have the Jordan Normal Form (JNF) as (you found the eigenvalues and eigenvectors already and I am making use of them):

$$\tag 1 A = \begin{bmatrix} -1 & 1 \\ 0 &2\end{bmatrix} = S_A \cdot J_A \cdot S^{-1}_A = \begin{bmatrix} 1 & 1 \\ 0 & 3\end{bmatrix} \cdot \begin{bmatrix} -1 & 0 \\ 0 & 2\end{bmatrix} \cdot \begin{bmatrix} 1 & \frac{-1}{3} \\ 0 & \frac{1}{3}\end{bmatrix}$$

Similarly, for the matrix $B$, we have:

$$\tag 2 B = \begin{bmatrix} 1 & 0 \\ 1 & -2\end{bmatrix} = S_B \cdot J_B \cdot S^{-1}_B = \begin{bmatrix} 0 & 3 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix} -2 & 0 \\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix} -\frac{1}{3} & 1 \\ \frac{1}{3} & 0 \end{bmatrix}$$

Now, we want to write the flow for each of these systems (I am assuming you know where this comes from).

$$\phi^{A}(x_0, y_0, t) = \frac{1}{\lambda_2 - \lambda_1} \begin{pmatrix} (\lambda_2 x_0 - y_0)e^{\lambda_1 t} \begin{pmatrix} 1 \\ \lambda_1 \end{pmatrix} + (y_0 -\lambda_1x_0)e^{\lambda_2 t} \begin{pmatrix} 1 \\ \lambda_2 \end{pmatrix} \end{pmatrix}.$$

The position is given by: $$x(t) = \frac{1}{\lambda_2 - \lambda_1} \begin{pmatrix} (\lambda_2 x_0 - y_0)e^{\lambda_1 t} + (y_0 -\lambda_1x_0)e^{\lambda_2 t} \end{pmatrix}.$$

Next, what do you notice about $J$ in both cases? The eigenvalues are the same, except for a different sign. This is a very important observation! It allows us to write for $A$ and $B$ (you should verify this!), that:

$H(x, y) = \large (sgn(x)|x|^{\frac{\lambda_{1B}}{\lambda_{1A}}} = sgn(y)|y|^{\frac{\lambda_{2B}}{\lambda_{2A}}}) = (sgn(x)|x|^{\frac{-2}{-1}}, sgn(y)|y|^{\frac{1}{2}}) = (sgn(x)|x|^{2}, sgn(y)|y|^{\frac{1}{2}})$,

where sgn($x$) is the signum function, $\frac{x}{|x|}$, which gives the sign of $x$, $-1$ if negative and $1$ if positive, and $0$ if $0$.

$H(x, y)$ is the congugacy between the canonical forms of the two systems defined by $A$ and $B$.

To get our conjugacy, then, we must first take initial values $(x_0, y_0)$ in the $A$ domain, and apply $S^{-1}$ in order to get corresponding coordinates for the canonical form of $A$. Then the (nonlinear) transformation $H$ to change the canonical form of $B$. Finally, one applies $S$ in order to make $H \circ T^{-1}(x_0, y_0)$ fit for consumption by the flow $\phi^{B}_t$. That is:

$$\tag 3 G(x, y) = S_B \circ H \circ S^{-1}_A(x,y)$$

is the required conjugacy. In full detail, multiplying out $(3)$, yields:

$$G(x, y) = (2sgn(x-\frac{y}{3})|x-\frac{y}{3}|^{2}, -3sgn(x-\frac{y}{3})|x - \frac{y}{3}|^{2})$$

This satisfies:

$$G \circ \phi^{A}(x, y, t) = \phi^{B}(G(x,y), t)$$

You might find these notes helpful.

Here are some additional references:


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You know its bad when i can't even read the answer to my own question, thank you for being so detailed i especially appreciate the notes and references. More then one person in my class will be reading this. – Faust7 Feb 11 '13 at 20:38
+1 complete Amzoti. I know, how much time it took you to be arranged like an answer. – Babak S. Feb 15 '13 at 19:50
@Amzoti: I knew that and that's why it deserved much pluses. Thanks for the edit. I will keep it signed by you. Best wishes. ;-) – Babak S. Feb 15 '13 at 19:55
+ 1 for the time-commitment, effort, and quality! – amWhy May 3 '13 at 0:19
Those kinds of problems are wonderful to "play" with! – amWhy May 3 '13 at 0:23

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