Category Archives: Co-ordinate Geometry

September 24, 2025 · 9:31 am

Eigenvalues and Eigenvectors

My Year 11 Specialist students have had an investigation which involves finding eigenvalues, eigenvectors and lines that are invariant under a particular linear transformation. This is not part of the course, but I feel for teachers who have to create new investigations every year.

Let’s find the eigenvalues and eigenvectors for matrix $T=\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}$

We want to find $\lambda$ such that

(1) $\begin{equation*}T\textbf{v}=\lambda \textbf{v}\end{equation*}$

We solve $det(T-\lambda I)=0$

$\begin{equation*}T-\lambda I=\begin{bmatrix}-\frac{1}{2}-\lambda&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}-\lambda\end{bmatrix}\end{equation}$

$det\left (T-\lambda I \right )=\left (-\frac{1}{2}-\lambda \right ) \left ( \frac{1}{2}-\lambda \right )- \left (-\frac{\sqrt{3}}{2} \right ) \left ( -\frac{\sqrt{3}}{2} \right )$

Hence $0=\lambda^2-1$ and $\lambda=\pm 1$

When $\lambda=1$ , $\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$

Hence, $-\frac{x_1}{2}-\frac{\sqrt{3}x_2}{2}=x_1$

$x_2=-\frac{3x_1}{\sqrt{3}}$ and the eigenvector is $\begin{bmatrix}1\\-\sqrt{3}\end{bmatrix}$

When $\lambda=-1$ , $\begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_1\\-x_2\end{bmatrix}$

Hence, $-\frac{x_1}{2}-\frac{\sqrt{3}x_2}{2}=-x_1$

$x_2=\frac{x_1}{\sqrt{3}}$ and the eigenvector is $\begin{bmatrix}1\\\frac{1}{\sqrt{3}}\end{bmatrix}$

Which means the invariant lines are $y=-\sqrt{3}x$ and $y=\frac{x}{\sqrt{3}}$

A quadrilateral with vertices on our lines

The vertices after they have been transformed – A and C remain in the same place (they are on the $\lambda=1$ line)

The quadrilateral (purple) after the transformation

Linear Transformation (Rotation) Question

The unit square is rotated about the origin by $45^\circ$ anti-clockwise.
(a) Find the matrix of this transformation.
(b) Draw the unit square and its image on the same set of axes.
(c) Find the area of the over lapping region.

Remember the general rotation matrix is

$\begin{equation*}\begin{bmatrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{bmatrix}\end{equation}$

Hence

$\begin{equation*}\begin{bmatrix}cos(45)&-sin(45)\\sin(45)&cos(45)\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\end{equation}$

The unit square has co-ordinates

$\begin{bmatrix}0 & 1 & 1& 0\\0&0&1&1\end{bmatrix}$

Transform the unit square

$\begin{bmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}0 & 1 & 1& 0\\0&0&1&1\end{bmatrix}=\begin{bmatrix}0&\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\0&\frac{1}{\sqrt{2}}&\frac{2}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{bmatrix}$