gistfile1.tex

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\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}


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