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) Let $x\in\mathbb{R} \ $. Then the matrix

$\begin{pmatrix} 0 & x \\ -x & x \end{pmatrix}$

has all-real entries, but complex eigenvalues $\frac{x\pm ix\sqrt{3}}{2} \ $ with eigenvectors

$\begin{pmatrix} y \\ \frac{1+i\sqrt{3}}{2}y \end{pmatrix}, \ \begin{pmatrix} z \\ \frac{1-i\sqrt{3}}{2}z \end{pmatrix} \ $.

For the matrix we've described, observe that the discriminant of the determinant polynomial is $ -3x^2 \ $.

This means 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} $,

leading to

$xv_2 = x\frac{1+i\sqrt{3}}{2}v_1 \ $ and

$x(v_2-v_1)=x\frac{1+i\sqrt{3}}{2}v_2 \ $.

The first equation gives us $v_2 = \frac{1+i\sqrt{3}}{2}v_1 \ $; we can plug this into the second to get

$(\frac{1+i\sqrt{3}}{2}v_1-v_1)=\frac{1+i\sqrt{3}}{2}\frac{1+i\sqrt{3}}{2}v_1 \ $,

which is just

$\frac{-1+i\sqrt{3}}{2}v_1 = \frac{1}{4}\left({ 1+2i\sqrt{3}-3 }\right)v_1 \ $,

a true statement. So this eigenvector works.

An eigenvector for the other eigenvalue can be found as well:

$\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} $

leads to

$u_2 = u_1\frac{1- i\sqrt{3}}{2}, \ -u_1+u_2=u_2\frac{1- i\sqrt{3}}{2} \ $

which are consistent, since

$-u_1+u_1\frac{1- i\sqrt{3}}{2}=u_1\frac{1- i\sqrt{3}}{2}\frac{1- i\sqrt{3}}{2} \ $

becomes a true statement,

$\frac{-1- i\sqrt{3}}{2}= \frac{1}{4}\left({ 1-2i\sqrt{3}-3 }\right) $

3) Let $U_1, U_2 \ $ be vector spaces, and set $V=U_1 \oplus U_2, \ T:V\to V \ $ as $(\vec{u_1},\vec{u_2})\mapsto (\vec{u_1},\vec{0}) \ $.

Let $\phi_j :\mathbb{F}^{\text{dim}(U_j)} \to U_j, \ j=1,2 $ be any isomorphisms. Then we can define bases for $U_1, U_2 \ $ as $B_j = \left\{{\phi_j(\vec{e_{k,j}})}\right\}_{k=1}^{\text{dim}(U_j)} \ $, where

$\vec{e}_{k,j}=(0,0,\dots,\underbrace{1}_{k^{th} \ \text{place}},0, \dots, \underbrace{0}_{\text{dim}(U_j)^{th} \ \text{place}}) \ $

are the standard bases for $U_1, U_2 \ $.

Then $\phi_1(B_1)\oplus \phi_2(B_2) \ $ forms a basis for $V \ $. Call this basis $B \ $. Observe that $(\vec{u_1},\vec{u_2})\mapsto(\phi_1(\vec{u_1}),\phi_2(\vec{u_2})) \ $ is an isomorphism $\mathbb{F}^{\text{dim}(U_1)+\text{dim}(U_2)} \ $. Call this isomorphism $\phi \ $. Define the transformation $S=\phi^{-1}T\phi \ $.

Note that

$\mathfrak{M}_B^B (S) = \begin{pmatrix}

1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \dots \\ 0 & \dots & 1 & \dots \\ 0 & \dots & \dots & 0 \\ 0 & \dots & \dots & 0 \end{pmatrix} $,

that is, $\mathfrak{M}_B^B (S) = (m_{ij})$, where $m_{ij}= \begin{cases} 1, & \mbox{if } i=j\leq \text{dim}(U_1) \\ 0, & \mbox{if } i\neq j \ \text{or} \ j>\text{dim}(U_1) \end{cases} \ $.

Hence, the eigenvalues of $S \ $ are precisely the solutions to the equation

$(1-\lambda)^{\text{dim}(U_1)}\lambda^{\text{dim}(U_2)} = 0 \ $.

Obviously, $\lambda = 0 \ $ is a solution with multiplicity $\text{dim}(U_2) \ $ and $\lambda=1 \ $ is a solution with multiplicity $\text{dim}(U_1) \ $.

The eigenspace corresponding to the nonzero eigenvalues is of course just $T(U_1)\cong U_1 \ $.