Explicit conjugacy on 2 linear systems involving flow. 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 
 A: 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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*Notes
Regards
A: i tried to do this exercise step by step:
First i find that the matrices obtained by Jordan form $ J_{a} $ and $ J_{B} $ are not similar! so how can we find the conjugation between $ \phi^{A} $ and $ \phi^{B} $
second : i can't understand how can we obtained
$$\phi^{A}(x_0, y_0, t) = \frac{1}{\lambda_2 - \lambda_1} \begin{pmatrix}
(\lambda_2 x_0 - y_0)e^{\lambda_1 t} \begin{pmatrix} 1 \\ \lambda_1 \end{pmatrix}
 + (y_0 -\lambda_1x_0)e^{\lambda_2 t} \begin{pmatrix} 1 \\ \lambda_2 \end{pmatrix} \end{pmatrix}$$ 
