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\begin{document}
	
	\begin{frame}
	
	\color{yellow}
	
	Show that 
	\[
	\left\| \left(\begin{array}{c c}
	\delta_0 & 0 \\
	0 & \delta_1  \end{array} \right)
	\right\|_2 = \max( \vert \delta_0 \vert, \vert \delta_1 \vert ).
	\]
\end{frame}

	\begin{frame}

	\color{yellow}
		
	\begin{eqnarray*}
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
			\end{array} \right) \right\|_2^2
		\pause
			& = & 
			\max_{\| x \|_2 = 1}
			\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) 
		\left( \begin{array}{c}
			\chi_0 \\
			\chi_1
		\end{array}
		\right)
		\right\|_2^2 \\
				\pause
		& = & 
		\max_{\| x \|_2 = 1}
		\left\| 
		\left( \begin{array}{c}
			\delta_0 \chi_0 \\
			\delta_1 \chi_1
		\end{array}
		\right)
		\right\|_2^2 \\
				\pause
		& = & 
		\max_{\| x \|_2 = 1}
		\left[ 
			\vert \delta_0 \chi_0 \vert^2 +
			\vert \delta_1 \chi_1 \vert^2
		\right] \\	
				\pause
		& = & 
		\max_{\| x \|_2 = 1}
		\left[ 
		\vert \delta_0 \vert^2 \vert\chi_0\vert^2 +
		\vert \delta_1 \vert^2 \vert\chi_1\vert^2
		\right] \\
				\pause
		& \leq & 
		\max_{\| x \|_2 = 1}
		\left[
		\max( \vert \delta_0 \vert, \vert\delta_1\vert )^2 \left[ 
		\vert\chi_0\vert^2 +
		\vert\chi_1\vert^2
		\right] \right] \\
				\pause
		& = & 
		\max( \vert \delta_0 \vert, \vert\delta_1\vert )^2 				
	\end{eqnarray*}
		\pause
	Conclude: $ \left\| \left( \begin{array}{c c}
	\delta_0 & 0 \\
	0 & \delta_1
	\end{array} \right) \right\|_2 \leq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
\end{frame}

\begin{frame}

	\color{yellow}
	
	\begin{eqnarray*}
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) \right\|_2
			\pause
		& = & 
		\max_{\| x \|_2 = 1}
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) 
		\left( \begin{array}{c}
			\chi_0 \\
			\chi_1
		\end{array}
		\right)
		\right\|_2 \\
				\pause
		& \geq & 
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) 
		\left( \begin{array}{c}
			0 \\
			1
		\end{array}
		\right)
		\right\|_2 \\
				\pause
		& = & 
		\left\| 
		\left( \begin{array}{c}
			0 \\
			\delta_1 
		\end{array}
		\right)
		\right\|_2 \\
				\pause
				& = & 
				\left\| 
				\delta_1
				\left( \begin{array}{c}
					0 \\
					1 
				\end{array}
				\right)
				\right\|_2 \\
						\pause
		& = & 
		\vert \delta_1 \vert
		\left\| 
		\left( \begin{array}{c}
			0 \\
			1 
		\end{array}
		\right)
		\right\|_2
				\pause
=
		\vert \delta_1 \vert 
	\end{eqnarray*}
			\pause
	Conclude: $ \left\| \left( \begin{array}{c c}
	\delta_0 & 0 \\
	0 & \delta_1
	\end{array} \right) \right\|_2 \geq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
		\end{frame}
	
	
	\begin{frame}
	
	\color{yellow}
	
	\frametitle{\color{yellow} More precisely:}
	
	Choose $ j \in \{ 0, 1 \} $ such that $ \vert \delta_j \vert = \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
	
			\pause
	\begin{eqnarray*}
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) \right\|_2
			\pause
		& = & 
		\max_{\| x \|_2 = 1}
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) 
		\left( \begin{array}{c}
			\chi_0 \\
			\chi_1
		\end{array}
		\right)
		\right\|_2 \\
				\pause
		& \geq & 
		\left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
		\end{array} \right) 
		e_j
		\right\|_2 \\
				\pause
		& = & 
		\left\| 
		\delta_j e_j 
		\right\|_2 \\
				\pause
		& = & 
		\vert \delta_j \vert
		\left\| 
		e_j
		\right\|_2
		\\
				\pause
		& = & 
		\vert \delta_j \vert 
		\\
		\pause& = & 
		\max ( \vert \delta_0 \vert, \vert \delta_1 \vert ).					
	\end{eqnarray*}
			\pause
	Conclude: $ \left\| \left( \begin{array}{c c}
	\delta_0 & 0 \\
	0 & \delta_1
	\end{array} \right) \right\|_2 \geq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
\end{frame}

		\begin{frame}
		\color{yellow}
		
			\[ \left\| \left( \begin{array}{c c}
			\delta_0 & 0 \\
			0 & \delta_1
			\end{array} \right) \right\|_2 \leq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) \]			
					\pause
			and
			\[
		 \left\| \left( \begin{array}{c c}
		\delta_0 & 0 \\
		0 & \delta_1
		\end{array} \right) \right\|_2 \geq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) .
		\]
				\pause
				
	\vspace{0.25in}
		Hence
		\[
		 \left\| \left( \begin{array}{c c}
		\delta_0 & 0 \\
		0 & \delta_1
		\end{array} \right) \right\|_2 = \max( \vert \delta_0 \vert, \vert \delta_1 \vert ). \]
		\end{frame}
	
	
\begin{frame}
\frametitle{\color{yellow} Key insight}

\color{yellow}

Given: $ f: D \rightarrow \mathbb R $.
		\pause
		\vspace{0.2in}
		
If we want to show that
\[
\max_{x \in D} f( x ) = \alpha
\]
		\pause
then a strategy is
		\pause
\begin{itemize}
	\color{yellow}
	\item[\color{yellow} $ \star $]
	Show that 
	\[
	 \max_{x \in D} f( x ) \leq \alpha.
	 \]
		\color{yellow}
				\pause
	\item[\color{yellow} $ \star $]
	Find a specific $ y \in D $ such 
	\[
	 f( y ) = \alpha .
	 \]
	 
\end{itemize}

\end{frame}

	\begin{frame}

\frametitle{\color{yellow}  Show that $ \max_{x \in D} f( x ) \leq \alpha $:}
	
\color{yellow}


\begin{eqnarray*}
	\left\| \left( \begin{array}{c c}
		\delta_0 & 0 \\
		0 & \delta_1
	\end{array} \right) \right\|_2^2
	& = & 
	\max_{\| x \|_2 = 1}
	\left\| \left( \begin{array}{c c}
		\delta_0 & 0 \\
		0 & \delta_1
	\end{array} \right) 
	\left( \begin{array}{c}
		\chi_0 \\
		\chi_1
	\end{array}
	\right)
	\right\|_2^2 \\
	& = & 
	\max_{\| x \|_2 = 1}
	\left\| 
	\left( \begin{array}{c}
		\delta_0 \chi_0 \\
		\delta_1 \chi_1
	\end{array}
	\right)
	\right\|_2^2 \\
	& = & 
	\max_{\| x \|_2 = 1}
	\left[ 
	\vert \delta_0 \chi_0 \vert^2 +
	\vert \delta_1 \chi_1 \vert^2
	\right] \\	
	& = & 
	\max_{\| x \|_2 = 1}
	\left[ 
	\vert \delta_0 \vert^2 \vert\chi_0\vert^2 +
	\vert \delta_1 \vert^2 \vert\chi_1\vert^2
	\right] \\
	& \leq & 
	\max_{\| x \|_2 = 1}
	\left[
	\max( \vert \delta_0 \vert, \vert\delta_1\vert )^2 \left[ 
	\vert\chi_0\vert^2 +
	\vert\chi_1\vert^2
	\right] \right] \\
	& = & 
	\max( \vert \delta_0 \vert, \vert\delta_1\vert )^2 				
\end{eqnarray*}
Conclude: $ \left\| \left( \begin{array}{c c}
\delta_0 & 0 \\
0 & \delta_1
\end{array} \right) \right\|_2 \leq \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
\end{frame}

	\begin{frame}

\frametitle{\color{yellow} Find $ y $ such that $ f( y )= \alpha $:}


\color{yellow}



Choose $ j \in \{ 0, 1 \} $ such that $ \vert \delta_j \vert = \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.

\begin{eqnarray*}
	\left\| \left( \begin{array}{c c}
		\delta_0 & 0 \\
		0 & \delta_1
	\end{array} \right) 
	e_j
	\right\|_2 
	& = & 
	\left\| 
	\delta_j e_j 
	\right\|_2 \\
	& = & 
	\vert \delta_j \vert
	\left\| 
	e_j
	\right\|_2
	\\
	& = & 
	\vert \delta_j \vert 
	= \max( \vert \delta_0 \vert, \vert \delta_1 \vert ).				
\end{eqnarray*}

Conclude: $ \left\| \left( \begin{array}{c c}
\delta_0 & 0 \\
0 & \delta_1
\end{array} \right) \right\|_2 = \max( \vert \delta_0 \vert, \vert \delta_1 \vert ) $.
\end{frame}

\end{document}