Given the transformations A_1: x'=3x-y, y'=x+y and A_2…

Given the transformations $A_1:\ x'=3x-y,\ y'=x+y$ and $A_2:\ x'=-x+3y,\ y'=x+y$, find $A_1A_2$ and $A_2A_1$, and the invariant lines through the origin of $A_1$ and of $A_2$.

Explanation of Method

Products of two linear maps in both orders, and the invariant lines of each map.

Step 1: Products

\[ A_1=\begin{pmatrix}3&-1\\1&1\end{pmatrix},\qquad A_2=\begin{pmatrix}-1&3\\1&1\end{pmatrix}. \]

$A_1A_2$ means “first $A_2$, then $A_1$”: \[A_1A_2=\begin{pmatrix}-4&8\\0&4\end{pmatrix},\qquad A_2A_1=\begin{pmatrix}0&4\\4&0\end{pmatrix}.\]

\[A_1A_2:\ x'=-4x+8y,\ y'=4y;\qquad A_2A_1:\ x'=4y,\ y'=4x;\qquad A_1A_2\ne A_2A_1\]

Step 2: Invariant lines

A line $y=kx$ through $O$ is invariant iff its direction is an eigenvector: $A\vec v=\lambda\vec v$.

$A_1$: $\det(A_1-\lambda I)=\lambda^2-4\lambda+4=(\lambda-2)^2$. From $(A_1-2I)\vec v=0$: $x-y=0$. There is only one eigen-direction.

\[A_1:\ \text{only } y=x\quad(\lambda=2)\]

$A_2$: $\lambda^2-4=0$, $\lambda=\pm2$.

$\lambda=2$: $-3x+3y=0$, which gives $y=x$.
$\lambda=-2$: $x+3y=0$, which gives $y=-\tfrac13x$.

\[A_2:\ y=x\ (\lambda=2),\qquad y=-\tfrac13x\ (\lambda=-2)\]

$\vec u$ $A_1\vec u=2\vec u$ $\vec e_2$ $A_1\vec e_2$ $y=x$

$A_1$: the only invariant line through $O$ is $y=x$ (double eigenvalue $2$); other directions, e.g. $\vec e_2$, are turned.

$A_2\vec u=2\vec u$ $\vec v$ $A_2\vec v=-2\vec v$ $y=-\tfrac x3$

$A_2$: two invariant lines, $y=x$ (eigenvalue $2$) and $y=-x/3$ (eigenvalue $-2$, each point reflected through $O$ and doubled).

Result

\[\boxed{\,A_1A_2:\ x'=-4x+8y,\ y'=4y;\qquad A_2A_1:\ x'=4y,\ y'=4x;\qquad A_1A_2\ne A_2A_1\,}\]

\[\boxed{\,A_1:\ \text{only } y=x\quad(\lambda=2)\,}\]

\[\boxed{\,A_2:\ y=x\ (\lambda=2),\qquad y=-\tfrac13x\ (\lambda=-2)\,}\]

Remark

Since the maps are linear, a line $y=kx+n$ with $n\ne0$ and an eigen-slope $k$ goes to a parallel line $y=kx+\lambda n$. So it is invariant only if $\lambda=1$, which does not happen here. The only invariant lines are those through $O$.