Invariant points and linesEdexcel A-Level Further Maths: Revision notes

A point is invariant under a transformation if it is mapped to itself. For a transformation with matrix $\mathbf{M}$, the point $(x,y)$ is invariant when $\mathbf{M}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x\\ y\end{pmatrix}.$ Write out the two equations and solve them. The origin is always invariant for a matrix transformation, and sometimes there are more: if the two equations reduce to the same equation, a whole line of invariant points exists. Example: $\mathbf{A}=\begin{pmatrix}3&-2\\1&0\end{pmatrix}$ gives $3x-2y=x$ and $x=y$, which are both $y=x$. Every point on $y=x$ is invariant. For $\mathbf{M}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ the equations $2x=x$, $3y=y$ force $(0,0)$ only.

A line is invariant if every point on it is mapped to a point on the same line. The points themselves may move along the line, so an invariant line is not the same as a line of invariant points. A line of invariant points is always an invariant line; the converse is false. For $\mathbf{M}=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ the $x$-axis is invariant because $(x,0)\to(2x,0)$, but only the origin is a fixed point. Test a line by taking a general point on it, finding its image, and checking the image satisfies the same equation.

For a line $y=mx$, a general point is $(x,mx)$. If $\mathbf{M}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ its image is $\big((a+bm)x,\,(c+dm)x\big)$. The image is on $y=mx$ when $c+dm=m(a+bm).$ This is a quadratic in $m$ whose roots give the invariant lines. For $\mathbf{B}=\begin{pmatrix}4&2\\1&3\end{pmatrix}$: $1+3m=m(4+2m)$, so $2m^2+m-1=0$, giving $m=\frac12$ or $m=-1$, and the lines $y=\frac12x$ and $y=-x$. The vertical line $x=0$ cannot be written $y=mx$, so check it separately: it is invariant when the top-right entry $b=0$.

For a line $y=mx+c$, take a general point $(x,mx+c)$, find its image $(x',y')$ and require $y'=mx'+c$ for all $x$. Compare the coefficients of $x$ and the constants to find $m$ and $c$. Example: $\mathbf{C}=\begin{pmatrix}1&2\\0&3\end{pmatrix}$ maps $(x,mx+c)$ to $\big((1+2m)x+2c,\,3mx+3c\big)$. Coefficients of $x$: $3m=m(1+2m)$, so $m=0$ or $m=1$. Constants: $3c=2mc+c$, so $c(1-m)=0$. For $m=1$ this holds for every $c$: all the lines $y=x+c$ are invariant. For $m=0$ it forces $c=0$: the line $y=0$. A shortcut is to find where a point $P$ and its image $P'$ lie: if the line $PP'$ is invariant it is the line through them. For $P(1,4)$, $P'=(9,12)$ and the line is $y=x+3$.