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December 12, 2013 00:25
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\documentclass[12pt]{article} | |
\usepackage{amsmath} | |
\usepackage{amssymb} | |
\begin{document} | |
\title{Mathematics Work Sample (Stanford OHSx)} | |
\author{Dhaivat Pandya} | |
\maketitle | |
\textbf{Problem: } Let $\theta$ be a real number. Prove that the following two matrices are similar over the field of complex numbers: | |
\begin{equation} | |
M_1 | |
\begin{bmatrix} | |
\cos{\theta} & -\sin{\theta} \\ \sin{\theta} &\cos{\theta} | |
\end{bmatrix}, | |
M_2 = | |
\begin{bmatrix} | |
e^{i\theta} & 0 \\ 0 & e^{-i\theta} | |
\end{bmatrix} | |
\end{equation} | |
\textbf{Solution: } | |
By definition, two matrices are similar if both represent the same linear transformation. Let T be the linear transformation represented by $M_1$ under the standard basis for $\mathbb{C}^2$. | |
Let $B=\{b_1, b_2\}$ be a basis of $\mathbb{C}^2$. Considering the columns of $M_2$, we note that if we can find a $B$ such that: | |
\begin{eqnarray} | |
T(b_1) &=& \begin{bmatrix} e^{i\theta} \\ 0 \end{bmatrix}_B = e^{i\theta}b_1 \\ | |
T(b_2) &=& \begin{bmatrix} 0 \\ e^{-i\theta}\end{bmatrix}_B = e^{-i\theta}b_2 | |
\end{eqnarray} | |
then $M_2$ must represent $T$ under the basis $B$. We now find a viable $B$. Let $b_1 = <x_1, x_2>$. | |
\begin{eqnarray} | |
\therefore T(b_1) = T(\begin{bmatrix}x_1 \\ x_2\end{bmatrix}) &=& e^{i\theta}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix} e^{i\theta}x_1 \\ e^{i\theta}x_2 \end{bmatrix} \\ | |
M_1 \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} &=& \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} &\cos{\theta}\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}e^{i\theta}x_1 \\ e^{i\theta}x_2 \end{bmatrix} \\ | |
\begin{bmatrix}x_1 \cos{\theta} - x_2\sin{\theta} \\ x_1 \sin{\theta} + x_2 \cos{\theta}\end{bmatrix} &=& \begin{bmatrix}e^{i\theta}x_1 \\ e^{i\theta}x_2 \end{bmatrix} | |
\end{eqnarray} | |
Equivalently, | |
\begin{eqnarray} | |
x_1 \cos{\theta} - x_2 \sin{\theta} = e^{i\theta}x_1 \label{firsteq} \\ | |
x_1 \sin{\theta} + x_2 \cos{\theta} = e^{i\theta}x_2 \label{secondeq} | |
\end{eqnarray} | |
From (\ref{firsteq}), using Euler's Formula, | |
\begin{eqnarray} | |
x_1\cos{\theta} - x_2\sin{\theta} = (\cos{\theta} + i\sin{\theta})x_1 = x_1\cos{\theta} + ix_1\sin{\theta} | |
\end{eqnarray} | |
\end{document} |
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