User:J D Bowen/Math725 HW6

1) We aim to show that $$r\in\mathbb{Z}, \ T\vec{v}=\lambda \vec{v}\implies T^r\vec{v}=\lambda^r\vec{v} \ $$.

Suppose we have $$T\vec{v}=\lambda \vec{v} \ $$, and let $$r\in\mathbb{N} \ $$. Suppose $$\lambda^{r-1} \ $$ is an eigenvalue of $$T^{r-1} \ $$ with eigenvector $$\vec{v} \ $$, ie, $$T^{n-1}\vec{v}=\lambda^{r-1}\vec{v} $$. Then

$$T^r\vec{v}=T(T^{r-1}\vec{v})=T(\lambda^{r-1})\vec{v})=\lambda^{r-1}T\vec{v}=\lambda^{r-1}\lambda\vec{v}=\lambda^r\vec{v} \ $$

We have the first case, and so the theorem follows by induction for $$r\in\mathbb{N} \ $$.

Now consider the case $$r=0 \ $$. Then $$T^0 = I, \ \lambda^0=1$$, and we have $$I\vec{v}=\vec{v} \ $$.

Now suppose $$-r\in\mathbb{N} \ $$, and $$T \ $$ is invertible. Consider that

$$\vec{v}=T^{0}\vec{v}=T^{-1} (T\vec{v})= T^{-1} (\lambda\vec{v}) = \lambda T^{-1}\vec{v} \implies \lambda^{-1}\vec{v}=T^{-1}\vec{v} \ $$.

If we let $$A=T^{-1}, \psi=\lambda^{-1} \ $$, the theorem follows from the first case.

2) Observe that the matrix

$$\begin{pmatrix} w & x \\ y & z \end{pmatrix} \ $$

has eigenvalues

$$\lambda = \frac{z+w\pm\sqrt{z^2+w^2-2zw-4wz+4xy}}{2} \ $$

If we set $$w=0, z=x=-y \ $$, we have the discriminant as

$$x^2-4x^2= -3x^2 \ $$.

This means that if we set $$x \ $$ as any positive real number, we will have an invertible real matrix with complex eigenvalues

$$\lambda = \frac{x\pm ix\sqrt{3}}{2} \ $$

Some eigenvectors of this matrix then are

$$\begin{pmatrix} 0 & x \\ -x & x \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} v_1\frac{x+ ix\sqrt{3}}{2} \\ v_2\frac{x+ ix\sqrt{3}}{2} \end{pmatrix} $$

$$\begin{pmatrix} 0 & x \\ -x & x \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} u_1\frac{x- ix\sqrt{3}}{2} \\ u_2\frac{x-ix\sqrt{3}}{2} \end{pmatrix} $$