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In the question, I'm asked to show that \begin{align*} F\begin{pmatrix}x\\y\end{pmatrix}=x^2+y^2 \end{align*} is an integral for the linear map \begin{align*} L(\text{x})= \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} \text{x}. \end{align*}

I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F ∘ L(x) = F(x)$.

My question is how to you compose the two in this case? In $F\begin{pmatrix}x\\y\end{pmatrix}$ Do you set both variables equal $L(x)$ or only one....

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  • $\begingroup$ The condition "$F$ is constant along orbits of $L$" is should be written as $F\circ L(x)=F(x)$ $\endgroup$
    – Miel Sharf
    Dec 15 '14 at 14:50
  • $\begingroup$ yes sorry! thanks for pointing that out $\endgroup$
    – Janet
    Dec 15 '14 at 14:54
  • $\begingroup$ The condition that $F ∘ L=F$ is offtopic. $\endgroup$
    – Did
    Dec 15 '14 at 15:00
  • $\begingroup$ The exact question in the textbook asks you to prove that $F$ is an integral for a linear map $L$ by calculating $F ∘ L(x) = F(x)$ $\endgroup$
    – Janet
    Dec 15 '14 at 15:03
  • $\begingroup$ Which textbook is this? Are we considering the (rather poor) dynamical system in discrete time $\mathbf x_{n+1}=L(\mathbf x_n)$, or the dynamical system in continuous time $\dot{\mathbf x}(t)=L(\mathbf x(t))$? $\endgroup$
    – Did
    Dec 15 '14 at 15:05
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The following is based on the assumption that one considers a dynamical system in continuous time. It appears that the object of interest is actually a dynamical system in discrete time (aka a recursive sequence...) hence the answer below does not apply. Leaving it for the picture...

Note that the question actually asks why $x^2+y^2=(-y)^2+x^2$ (?!), which makes for a rather poor question, I am sorry to say. Note also that every sequence $(\mathbf x_n)$ defined by $\mathbf x_{n+1}=L(\mathbf x_n)$ has period $4$, as an elementary computation shows.


I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F \text{ o } L(x) = L(x)$.

No. That $F$ is an invariant of the dynamical system associated to $L$ means that, if $$\dot{\mathbf x}(t)=L(\mathbf x(t)),$$ then the quantity $F(\mathbf x(t))$ does not depend on $t$. This is usually done showing that $u'(t)=0$ for every $t$, where $$u(t)=F(\mathbf x(t)),$$ which is implied by the condition that, for every $\mathbf x$, $$\nabla F(\mathbf x)\cdot L(\mathbf x)=0.$$ In your case, $$\nabla F(\mathbf x)=2\mathbf x,$$ hence one is left with $$\nabla F(\mathbf x)\cdot L(\mathbf x)=2\mathbf x\cdot L(\mathbf x)=2\begin{pmatrix}x\\y\end{pmatrix}\cdot\begin{pmatrix}y\\-x\end{pmatrix}=0.$$ Here is a rough sketch of some solutions $(\mathbf x(t))$:

$\hspace2in$enter image description here

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