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

A point $P$ is invariant under a transformation with matrix $\mathbf{M}$ if its image is itself: $\mathbf{M}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x\\ y\end{pmatrix}$. Writing this as two equations and solving them finds every invariant point. The origin is always invariant, because $\mathbf{M}\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$. Example: $\mathbf{M}=\begin{pmatrix}1&2\\0&3\end{pmatrix}$ gives $x+2y=x$ and $3y=y$. Both say $y=0$, so every point $(x,0)$ is invariant: the $x$-axis is a line of invariant points. (Check: $(4,0)\mapsto(4,0)$.) If the two equations are independent, the origin is the only invariant point. For $\mathbf{T}=\begin{pmatrix}4&-2\\1&1\end{pmatrix}$: $4x-2y=x$ and $x+y=y$ give $x=0$ and then $y=0$.

A line is invariant if every point on it is mapped to a point on the same line. The individual points may move along the line. This is different from a line of invariant points, where no point moves.

• A line of invariant points is always an invariant line, but an invariant line need not be a line of invariant points.
• Example: under reflection in the $y$-axis, the line $y=5$ is invariant (the point $(x,5)$ goes to $(-x,5)$), but its points move. To test a line, take a general point on it, find its image, and check that the image satisfies the same equation.

For a line $y=mx$, a general point is $(x,mx)$. With $\mathbf{M}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ the image is $\big((a+bm)x,\;(c+dm)x\big)$. This is on $y=mx$ when $c+dm=m(a+bm)\quad\Longrightarrow\quad bm^2+(a-d)m-c=0.$ Solve this quadratic for $m$. Example: $\mathbf{T}=\begin{pmatrix}4&-2\\1&1\end{pmatrix}$ gives $(x,mx)\mapsto\big((4-2m)x,(1+m)x\big)$, so $1+m=m(4-2m)$ and $2m^2-3m+1=0$, giving $m=1$ and $m=\frac12$. The invariant lines are $y=x$ and $y=\frac12x$. Do not forget the line $x=0$ (the $y$-axis). It is invariant when the image of $(0,y)$ has zero $x$-coordinate, that is when $b=0$.

For $y=mx+k$ with $k\neq0$, a general point is $(x,mx+k)$. Find its image $(x',y')$ and require $y'=mx'+k$ for all $x$. Compare coefficients of $x$ and the constant terms. Example: for $\mathbf{S}=\begin{pmatrix}2&-1\\1&0\end{pmatrix}$ and the line $y=x+c$, the image of $(x,x+c)$ is $(x-c,x)$. Then $x'+c=x=y'$ for all $x$, so every line $y=x+c$ is invariant. This is typical of a shear: lines parallel to the line of invariant points are invariant. For $\mathbf{T}$ above, the coefficient condition gives $m=1$ or $\frac12$, but the constant terms give $0=-2mk$. Since $m\neq0$, $k=0$, so there are no invariant lines with $k\neq0$.

• Reflection in a line through the origin: the mirror line is a line of invariant points; every line perpendicular to the mirror is invariant.
• Rotation through $180^\circ$ about the origin: every line through the origin is invariant (no line of invariant points apart from the origin).
• Enlargement, centre the origin: every line through the origin is invariant.
• Stretch parallel to the $x$-axis, $\begin{pmatrix}a&0\\0&1\end{pmatrix}$: the $y$-axis is a line of invariant points, and every line $y=c$ (including the $x$-axis) is invariant.
• Shear $\begin{pmatrix}1&\lambda\\0&1\end{pmatrix}$: the $x$-axis is a line of invariant points and every line $y=c$ is invariant. Use these as a check on an answer from calculation.