### A Two Dimensional Rotation

The matrix that represents an anticlockwise rotation by $45$ degrees is $$\frac{\sqrt{2}}{2}\left( \begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right)$$ with respect to the standard basis. If $$v=\left(\begin{array}{cc} 1\\ 0 \end{array}\right)$$ then $$\alpha(v)=\frac{\sqrt{2}}{2}\left(\begin{array}{cc} 1\\ 1 \end{array}\right)$$ and $$\alpha^2(v) = \frac{\sqrt{2}}{2}\left(\begin{array}{cc} \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} \end{array}\right)+ \frac{\sqrt{2}}{2}\left(\begin{array}{cc} -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} \end{array}\right)= \left(\begin{array}{cc} \frac{1}{2}\\ \frac{1}{2} \end{array}\right)+ \left(\begin{array}{cc} -\frac{1}{2}\\ \frac{1}{2} \end{array}\right) = \left(\begin{array}{cc} 0\\ 1 \end{array}\right)$$ Therefore $$0 = \alpha^2(v)-\sqrt{2}\alpha(v)+v = (\alpha^2 -\sqrt{2}\alpha + I)v$$ which we can factorise as $$\left(\alpha -\frac{\sqrt{2}}{2}(1+i)I\right)\left(\alpha -\frac{\sqrt{2}}{2}(1-i)I\right)=0$$ and hence $$\left(\alpha - \frac{\sqrt{2}}{2}(1+i)I\right)v=\alpha(v)-\frac{\sqrt{2}}{2}(1+i)v$$ $$= \frac{\sqrt{2}}{2} \left(\begin{array}{cc} 1\\ 1 \end{array}\right) - \frac{\sqrt{2}}{2}(1+i) \left(\begin{array}{cc} 1\\ 0 \end{array}\right)$$ $$=\frac{\sqrt{2}}{2} \left(\begin{array}{cc} -i\\ 1 \end{array}\right)$$ is an eigenvector with eigenvalue $\frac{\sqrt{2}}{2}(1-i)$ and $$\left(\alpha - \frac{\sqrt{2}}{2}(1-i)I\right)v=\alpha(v)-\frac{\sqrt{2}}{2}(1-i)v$$ $$= \frac{\sqrt{2}}{2} \left(\begin{array}{cc} 1\\ 1 \end{array}\right) - \frac{\sqrt{2}}{2}(1-i) \left(\begin{array}{cc} 1\\ 0 \end{array}\right)$$ $$=\frac{\sqrt{2}}{2} \left(\begin{array}{cc} i\\ 1 \end{array}\right)$$ is an eigenvector with eigenvalue $\frac{\sqrt{2}}{2}(1+i)$.

### A Two Dimensional Rotation then Magnification

The matrix that represents an anticlockwise rotation by $90$ degrees followed by a magnification of $2$ is $$\left( \begin{array}{cc} 0 & -2\\ 2 & 0 \end{array}\right)$$ with respect to the standard basis. If $$v=\left(\begin{array}{cc} 1\\ 0 \end{array}\right)$$ then $$\alpha(v)=\left(\begin{array}{cc} 0\\ 2 \end{array}\right)$$ and $$\alpha^2(v)=\left(\begin{array}{cc} -4\\ 0 \end{array}\right)$$ so $$\alpha^2(v)+4v = 0$$ $$(\alpha^2+4)v=0$$ which factorises as $$(\alpha -2i)(\alpha +2i)v=0$$ Hence $$(\alpha-2i)v=\alpha(v)-2iv=\left(\begin{array}{cc} 0\\ 2 \end{array}\right)-2i \left(\begin{array}{cc} 1\\ 0 \end{array}\right)= \left(\begin{array}{cc} -2i\\ 2 \end{array}\right)$$ is an eigenvector with eigenvalue $-2i$ and $$(\alpha+2i)v=\alpha(v)+2iv=\left(\begin{array}{cc} 0\\ 2 \end{array}\right)+2i \left(\begin{array}{cc} 1\\ 0 \end{array}\right)= \left(\begin{array}{cc} 2i\\ 2 \end{array}\right)$$ is an eigenvector with eigenvalue $2i$.

### A Three Dimensional Rotation

The matrix that represents an anticlockwise rotation by $90$ degrees around the vector $$v=\left(\begin{array}{cc} 0\\ \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} \end{array}\right)$$ is $$\left( \begin{array}{cc} 0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} &\frac{1}{2} & \frac{1}{2}\\ -\frac{\sqrt{2}}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right)$$ with respect to the standard basis. Since we spot that the sum of the second and third columns has a particularly nice form we try the initial vector $$v=\left(\begin{array}{cc} 0\\ 1\\ 1 \end{array}\right)$$ and see that $$\alpha(v)=\left(\begin{array}{cc} 0\\ 1\\ 1 \end{array}\right)$$ and so $$\alpha(v)=\left(\begin{array}{cc} 0\\ 1\\ 1 \end{array}\right)$$ is an eigenvector with eigenvalue $1$. (This shows us the axis of rotation.) If $$v=\left(\begin{array}{cc} 1\\ 0\\ 0 \end{array}\right)$$ then $$\alpha(v)=\left(\begin{array}{cc} 0\\ \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} \end{array}\right)$$ and $$\alpha^2(v)=\frac{\sqrt{2}}{2} \left(\begin{array}{cc} -\frac{\sqrt{2}}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{array}\right) -\frac{\sqrt{2}}{2}\left(\begin{array}{cc} \frac{\sqrt{2}}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{array}\right)=\left(\begin{array}{cc} -1\\ 0\\ 0 \end{array}\right)$$ hence $$\alpha^2(v)+v = 0$$ $$(\alpha^2+I)v=0$$ factorising gives $$(\alpha+iI)(\alpha-iI)v=0$$ so $$(\alpha-iI)v=\alpha(v)-iv=\left(\begin{array}{cc} 0\\ \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} \end{array}\right)-i\left(\begin{array}{cc} 1\\ 0\\ 0 \end{array}\right)= \left(\begin{array}{cc} -i\\ \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} \end{array}\right)$$ is an eigenvector with eigenvalue $-i$. Also $$(\alpha+iI)v=\alpha(v)+iv=\left(\begin{array}{cc} 0\\ \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} \end{array}\right)+i\left(\begin{array}{cc} 1\\ 0\\ 0 \end{array}\right)= \left(\begin{array}{cc} i\\ \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} \end{array}\right)$$ is an eigenvector with eigenvalue $i$.