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\mode<presentation>

\title{More Linear Algebra\footnote{See Chapter 2 of \emph{Linear models in statistics} for more detail. This slide show is an open-source document. See last slide for copyright information.}}
\subtitle{STA 302: Fall 2013}
\date{} % To suppress date

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
\frametitle{Overview}
\tableofcontents
\end{frame}

\section{Things you already know}

\begin{frame}
\frametitle{You already know about}

  \begin{itemize}
    \item Matrices $\mathbf{A} = [a_{ij}]$
    \item Matrix addition and subtraction $\mathbf{A}+\mathbf{B} = [a_{ij}+b_{ij}]$
    \item Scalar multiplication $a \mathbf{B} = [a\,b_{ij}]$
    \item Matrix multiplication 
          $\mathbf{AB}  = \left[ \displaystyle{\sum_k}a_{ik}b_{kj} \right]$
    \item Inverse $\mathbf{A}^{-1}\mathbf{A} = \mathbf{AA}^{-1} = \mathbf{I}$
    \item Transpose $\mathbf{A}^\prime = [a_{ji}]$
    \item Symmetric matrices $\mathbf{A} = \mathbf{A}^\prime$
    \item Determinants
    \item Linear independence
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Linear independence}
%\framesubtitle{} 
$\mathbf{X}$ be an $n \times p$ matrix of constants.The columns of 
$\mathbf{X}$ are said to be \emph{linearly dependent} if there exists $\mathbf{v} \neq \mathbf{0}$ with $\mathbf{Xv} = \mathbf{0}$. We will say that the columns of $\mathbf{X}$ are linearly \emph{independent} if $\mathbf{Xv} = \mathbf{0}$ implies $\mathbf{v} = \mathbf{0}$.

\vspace{10mm}

For example, show that $\mathbf{A}^{-1}$ exists implies that the columns of $\mathbf{A}$ are linearly independent.
\begin{displaymath}
    \mathbf{Av} = \mathbf{0} \Rightarrow \mathbf{A}^{-1}\mathbf{Av} = \mathbf{A}^{-1}\mathbf{0} \Rightarrow  \mathbf{v} = \mathbf{0}
\end{displaymath}
\end{frame}


\begin{frame}
\frametitle{How to show $\mathbf{A}^{-1\prime}= \mathbf{A}^{\prime-1}$} 
Suppose $\mathbf{B} = \mathbf{A}^{-1}$, meaning $\mathbf{AB} = \mathbf{BA} = \mathbf{I}$. Must show two things: $\mathbf{B}^\prime\mathbf{A}^\prime = \mathbf{I}$ and $\mathbf{A}^\prime\mathbf{B}^\prime = \mathbf{I}$. 

\begin{eqnarray*}
    \mathbf{AB} = \mathbf{I} & \Rightarrow &  \mathbf{B}^\prime\mathbf{A}^\prime = \mathbf{I}^\prime = \mathbf{I} \\
    \mathbf{BA} = \mathbf{I} & \Rightarrow &  \mathbf{A}^\prime\mathbf{B}^\prime  = \mathbf{I}^\prime = \mathbf{I}
\end{eqnarray*} 
$\blacksquare$
\end{frame}

% \section{Trace} deleted: See STA431s13 for omitted material

\begin{frame}
\frametitle{Extras}
\framesubtitle{You may not know about these, but we may use them occasionally} 
  \begin{itemize}
    \item Trace
    \item Rank
    \item Partitioned matrices
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Trace of a square matrix}
%\framesubtitle{} 
  \begin{itemize}
    \item Sum of diagonal elements
    \item Obvious: $tr(\mathbf{A}+\mathbf{B}) = tr(\mathbf{A})+tr(\mathbf{B})$
    \item Not obvious: $tr(\mathbf{AB}) = tr(\mathbf{BA})$
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Rank}
%\framesubtitle{} 
  \begin{itemize}
    \item Row rank is the number of linearly independent rows 
    \item Column rank is the number of linearly independent columns
    \item Rank of a matrix is the minimum of row rank and column rank
    \item $rank(\mathbf{AB}) = \min\left(rank(\mathbf{A}),rank(\mathbf{B})\right)$
  \end{itemize}
\end{frame}

\begin{frame}
\frametitle{Partitioned matrix}
%\framesubtitle{} 

  \begin{itemize}
    \item A matrix of matrices
    {\LARGE
    \begin{displaymath} \left[ \begin{array}{c|c}
        \mathbf{A} & \mathbf{B} \\ \hline
        \mathbf{C} & \mathbf{D}    
    \end{array} \right] 
    \end{displaymath} 
    } % End size
    \item Row by column (matrix) multiplication works, provided the matrices are the right sizes.
  \end{itemize}
\end{frame}


\section{Spectral decomposition}

\begin{frame}
\frametitle{Eigenvalues and eigenvectors}
%\framesubtitle{} 
Let $\mathbf{A} = [a_{i,j}]$ be an $n \times n$ matrix, so that the following applies to square matrices. $\mathbf{A}$ is said to have an \emph{eigenvalue} $\lambda$ and  (non-zero) \emph{eigenvector} $\mathbf{x}$ corresponding to $\lambda$ if
\begin{displaymath} 
    \mathbf{Ax} =  \lambda\mathbf{x}.
\end{displaymath}
  \begin{itemize}
    \item Eigenvalues are the $\lambda$ values that solve the determinantal equation $|\mathbf{A}-\lambda\mathbf{I}| = 0$.
    \item The determinant is the product of the eigenvalues: $|\mathbf{A}| = \prod_{i=1}^n \lambda_i$
  \end{itemize}
\end{frame}


\begin{frame}
\frametitle{Spectral decomposition of symmetric matrices}
%\framesubtitle{} 
The \emph{Spectral decomposition theorem} says that every square and symmetric matrix $\mathbf{A} = [a_{i,j}]$ may be written
\begin{equation*} 
   \mathbf{A} = \mathbf{CDC}^\prime,
\end{equation*}
where the columns of $\mathbf{C}$ (which may also be denoted $\mathbf{x}_1, \ldots, \mathbf{x}_n$) are the eigenvectors of $\mathbf{A}$, and the diagonal matrix $\mathbf{D}$ contains the corresponding eigenvalues.  
\begin{displaymath} 
   \mathbf{D} = 
    \left[ \begin{array}{c c c c }
        \lambda_1 & 0          & \cdots  & 0            \\
            0     & \lambda_2  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \lambda_n    \\ 
\end{array} \right] 
\end{displaymath}
The eigenvectors may be chosen to be orthonormal, so that $\mathbf{C}$ is an orthogonal matrix. That is, 
$\mathbf{CC}^\prime = \mathbf{C}^\prime\mathbf{C} = \mathbf{I}$.
\end{frame}

\section{Positive definite matrices}

\begin{frame}
\frametitle{Positive definite matrices}
%\framesubtitle{} 
The $n \times n$ matrix $\mathbf{A}$ is said to be \emph{positive definite} if
\begin{displaymath} 
   \mathbf{y}^\prime \mathbf{A} \mathbf{y} > 0
\end{displaymath}
for \emph{all} $n \times 1$ vectors $\mathbf{y} \neq \mathbf{0}$. It is called \emph{non-negative definite} (or sometimes positive semi-definite) if 
$\mathbf{y}^\prime \mathbf{A} \mathbf{y} \geq 0$.
\end{frame}

\begin{frame}
\frametitle{Example: Show $\mathbf{X}^\prime\mathbf{X}$ non-negative definite}
Let $\mathbf{X}$ be an $n \times p$ matrix of real constants and $\mathbf{y}$ be $p \times 1$. Then $\mathbf{Z} = \mathbf{Xy}$ is $n \times 1$, and 
\begin{eqnarray*}
     &   & \mathbf{y}^\prime \, (\mathbf{X}^\prime\mathbf{X}) \, \mathbf{y} \\
     & = & (\mathbf{Xy})^\prime (\mathbf{Xy}) \\
     & = & \mathbf{Z}^\prime \mathbf{Z} \\
     & = & \sum_{i=1}^n Z_i^2 \geq 0
\end{eqnarray*} 
\end{frame}


\begin{frame}
\frametitle{Some properties of symmetric positive definite matrices}
\framesubtitle{Variance-covariance matrices are often assumed positive definite.} 
For a symmetric matrix,
\vspace{5mm}
  \begin{itemize}
    \item[] Positive definite
    \item[]~~~~~~~~$\Downarrow$
    \item[] All eigenvalues positive 
    \item[]~~~~~~~~$\Downarrow$
    \item[] Inverse exists $\Leftrightarrow$ Columns (rows) linearly independent
  \end{itemize}
\vspace{5mm}
If a real symmetric matrix is also non-negative definite, as a variance-covariance matrix \emph{must} be, Inverse exists $\Rightarrow$ Positive definite
\end{frame}

\begin{frame}
\frametitle{Showing Positive definite $\Rightarrow$ Eigenvalues positive}
\framesubtitle{For example}
Let $\mathbf{A}$ be square and symmetric as well as positive definite. 
  \begin{itemize}
    \item Spectral decomposition says $\mathbf{A} = \mathbf{CDC}^\prime$.
    \item Using $\mathbf{y}^\prime \mathbf{A} \mathbf{y} > 0$, let $\mathbf{y}$ be an eigenvector, say the third one.
    \item Because eigenvectors are orthonormal,
  \end{itemize}
\begin{eqnarray*}
   \mathbf{y}^\prime \mathbf{A} \mathbf{y} 
    & = & \mathbf{y}^\prime \mathbf{CDC}^\prime \mathbf{y} \\
    & = &  ( \begin{array}{ccccc}
              0 & 0 & 1 & \cdots & 0
              \end{array} )
    \left( \begin{array}{c c c c }
        \lambda_1 & 0          & \cdots  & 0            \\
            0     & \lambda_2  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \lambda_n    \\ 
\end{array} \right)
             \left( \begin{array}{c}
              0 \\ 0 \\ 1 \\ \vdots \\ 0
              \end{array} \right) \\
    & = &  \lambda_3 \\
    & > & 0
\end{eqnarray*} 
\end{frame}


\begin{frame}
\frametitle{Inverse of a diagonal matrix}
%\framesubtitle{}
Suppose $\mathbf{D} = [d_{i,j}]$ is a diagonal matrix with non-zero diagonal elements. It is easy to verify that
\begin{displaymath}
       \left( \begin{array}{c c c c }
        d_{1,1} & 0          & \cdots  & 0            \\
            0     & d_{2,2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & d_{n,n}    \\ 
\end{array} \right)
       \left( \begin{array}{c c c c }
        1/d_{1,1} & 0          & \cdots  & 0            \\
            0     & 1/d_{2,2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & 1/d_{n,n}    \\ 
\end{array} \right) = \mathbf{I}
\end{displaymath} 
And
\begin{displaymath}
       \left( \begin{array}{c c c c }
        1/d_{1,1} & 0          & \cdots  & 0            \\
            0     & 1/d_{2,2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & 1/d_{n,n}    \\ 
\end{array} \right)
       \left( \begin{array}{c c c c }
        d_{1,1} & 0          & \cdots  & 0            \\
            0     & d_{2,2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & d_{n,n}    \\ 
\end{array} \right) = \mathbf{I}
\end{displaymath} 
\end{frame}

\begin{frame}
\frametitle{Showing Eigenvalues positive $\Rightarrow$ Inverse exists}
\framesubtitle{For a symmetric, positive definite matrix} 

Let $\mathbf{A}$ be symmetric and positive definite. Then $\mathbf{A} = \mathbf{CDC}^\prime$ and its eigenvalues are positive.

\vspace{2mm}

Let $\mathbf{B} = \mathbf{C}\mathbf{D}^{-1}\mathbf{C}^\prime$

\vspace{5mm}

Showing $\mathbf{B} = \mathbf{A}^{-1}$:
\begin{eqnarray*}
    \mathbf{AB} & = & \mathbf{CDC}^\prime \, \mathbf{C}\mathbf{D}^{-1}\mathbf{C}^\prime =  \mathbf{I} \\
    \mathbf{BA} & = & \mathbf{C}\mathbf{D}^{-1}\mathbf{C}^\prime \, \mathbf{CDC}^\prime =  \mathbf{I}
\end{eqnarray*}
So
{\LARGE
\begin{displaymath}
    \mathbf{A}^{-1} = \mathbf{C}\mathbf{D}^{-1}\mathbf{C}^\prime
\end{displaymath} 
} % End size
\end{frame}

\section{Square root matrices}

\begin{frame}
\frametitle{Square root matrices}
\framesubtitle{For symmetric, non-negative definite matrices} % Real if non-neg def 
{\small
Define
\begin{displaymath}
    \mathbf{D}^{1/2} = 
        \left( \begin{array}{c c c c }
        \sqrt{\lambda_1} & 0          & \cdots  & 0            \\
            0     & \sqrt{\lambda_2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \sqrt{\lambda_n}    \\ 
\end{array} \right)
\end{displaymath} 
So that
\begin{eqnarray*}
    \mathbf{D}^{1/2} \mathbf{D}^{1/2} 
    & = & 
        \left( \begin{array}{c c c c }
        \sqrt{\lambda_1} & 0          & \cdots  & 0            \\
            0     & \sqrt{\lambda_2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \sqrt{\lambda_n}    \\ 
\end{array} \right)
        \left( \begin{array}{c c c c }
        \sqrt{\lambda_1} & 0          & \cdots  & 0            \\
            0     & \sqrt{\lambda_2}  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \sqrt{\lambda_n}    \\ 
\end{array} \right) \\
    & = &
    \left( \begin{array}{c c c c }
        \lambda_1 & 0          & \cdots  & 0            \\
            0     & \lambda_2  & \cdots  & 0            \\
            \vdots   & \vdots  & \ddots  & \vdots       \\
            0        & 0       & \cdots  & \lambda_n    \\ 
\end{array} \right) = \mathbf{D}
\end{eqnarray*}
} % End size
\end{frame}

\begin{frame}
\frametitle{For a non-negative definite, symmetric matrix $\mathbf{A}$}
Define 
{\LARGE
\begin{displaymath}
    \mathbf{A}^{1/2} = \mathbf{C} \mathbf{D}^{1/2} \mathbf{C}^\prime
\end{displaymath}
} % End size
So that
\begin{eqnarray*}
    \mathbf{A}^{1/2}\mathbf{A}^{1/2} 
    & = & 
    \mathbf{C} \mathbf{D}^{1/2} \mathbf{C}^\prime
    \mathbf{C} \mathbf{D}^{1/2} \mathbf{C}^\prime \\
    & = & 
    \mathbf{C} \mathbf{D}^{1/2} \, \mathbf{I} \,
    \mathbf{D}^{1/2} \mathbf{C}^\prime \\
    & = & 
    \mathbf{C} \mathbf{D}^{1/2}
    \mathbf{D}^{1/2} \mathbf{C}^\prime \\
    & = & 
    \mathbf{C} \mathbf{D} \mathbf{C}^\prime \\
    & = &  \mathbf{A} 
\end{eqnarray*} 
\end{frame}

\begin{frame}
\frametitle{The square root of the inverse is the inverse of the square root}

Let $\mathbf{A}$ be symmetric and positive definite, with $\mathbf{A} = \mathbf{CDC}^\prime$.

\vspace{2mm}

Let $\mathbf{B} = \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime $. What is $\mathbf{D}^{-1/2}$?

\vspace{2mm}

Show $\mathbf{B} = \left( \mathbf{A}^{-1} \right)^{1/2}$
\begin{eqnarray*}
    \mathbf{BB}  & = & \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime 
                       \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime \\
                 & = &  \mathbf{C}\mathbf{D}^{-1} \mathbf{C}^\prime  = \mathbf{A}^{-1} 
\end{eqnarray*}

\vspace{2mm}

Show $\mathbf{B} = \left( \mathbf{A}^{1/2} \right)^{-1}$
\begin{eqnarray*}
    \mathbf{A}^{1/2}\mathbf{B}  & = & \mathbf{C}\mathbf{D}^{1/2} \mathbf{C}^\prime 
                       \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime = \mathbf{I} \\
    \mathbf{B} \mathbf{A}^{1/2} & = &  \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime
                        \mathbf{C}\mathbf{D}^{1/2} \mathbf{C}^\prime  = \mathbf{I}
\end{eqnarray*}
Just write ~~~~~~~ \LARGE{$\mathbf{A}^{-1/2} = \mathbf{C}\mathbf{D}^{-1/2} \mathbf{C}^\prime$}
\end{frame}

\section{R}

\begin{frame}[fragile]
\frametitle{Matrix calculation with R}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> is.matrix(3)  # Is the number 3 a 1x1 matrix?
\end{verbatim}
} % End color
\begin{verbatim}
[1] FALSE
\end{verbatim}
{\color{blue}
\begin{verbatim}
> treecorr = cor(trees); treecorr
\end{verbatim}
} % End color
\begin{verbatim}
           Girth    Height    Volume
Girth  1.0000000 0.5192801 0.9671194
Height 0.5192801 1.0000000 0.5982497
Volume 0.9671194 0.5982497 1.0000000
\end{verbatim}
{\color{blue}
\begin{verbatim}
> is.matrix(treecorr)
\end{verbatim}
} % End color
\begin{verbatim}
[1] TRUE
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Creating matrices}
\framesubtitle{Bind rows into a matrix} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> # Bind rows of a matrix together
> A = rbind(  c(3, 2, 6,8),
+             c(2,10,-7,4),
+             c(6, 6, 9,1)   ); A
\end{verbatim}
} % End color
\begin{verbatim}
     [,1] [,2] [,3] [,4]
[1,]    3    2    6    8
[2,]    2   10   -7    4
[3,]    6    6    9    1
\end{verbatim}
{\color{blue}
\begin{verbatim}
> # Transpose
> t(A)
\end{verbatim}
} % End color
\begin{verbatim}
     [,1] [,2] [,3]
[1,]    3    2    6
[2,]    2   10    6
[3,]    6   -7    9
[4,]    8    4    1
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Matrix multiplication}
\framesubtitle{Remember, $\mathbf{A}$ is $3\times 4$} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> # U = AA' (3x3), V = A'A (4x4)
> U = A %*% t(A)
> V = t(A) %*% A; V
\end{verbatim}
} % End color
\begin{verbatim}
     [,1] [,2] [,3] [,4]
[1,]   49   62   58   38
[2,]   62  140   -4   62
[3,]   58   -4  166   29
[4,]   38   62   29   81
\end{verbatim}
} % End size
\end{frame}


\begin{frame}[fragile]
\frametitle{Determinants}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> # U = AA' (3x3), V = A'A (4x4)
> # So rank(V) cannot exceed 3 and det(V)=0
> det(U); det(V)
\end{verbatim}
} % End color
\begin{verbatim}
[1] 1490273
[1] -3.622862e-09
\end{verbatim}
} % End size

Inverse of $\mathbf{U}$ exists, but inverse of $\mathbf{V}$ does not.
\end{frame}

\begin{frame}[fragile]
\frametitle{Inverses}
%\framesubtitle{} 
  \begin{itemize}
    \item The \texttt{solve} function is for solving systems of linear equations like
$\mathbf{Mx}=\mathbf{b}$.
    \item Just typing \texttt{solve(M)} gives $\mathbf{M}^{-1}$.
  \end{itemize}

{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> solve(U)
\end{verbatim}
} % End color
\begin{verbatim}
              [,1]          [,2]          [,3]
[1,]  0.0173505123 -8.508508e-04 -1.029342e-02
[2,] -0.0008508508  5.997559e-03  2.013054e-06
[3,] -0.0102934160  2.013054e-06  1.264265e-02
\end{verbatim}
{\color{blue}
\begin{verbatim}
> solve(V)
\end{verbatim}
} % End color
{\color{red}
\begin{verbatim}
Error in solve.default(V) : 
  system is computationally singular: reciprocal condition 
  number = 6.64193e-18
\end{verbatim}
} % End color
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Eigenvalues and eigenvectors}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> eigen(U)
\end{verbatim}
} % End color
\begin{verbatim}
$values
[1] 234.01162 162.89294  39.09544

$vectors
           [,1]       [,2]        [,3]
[1,] -0.6025375  0.1592598  0.78203893
[2,] -0.2964610 -0.9544379 -0.03404605
[3,] -0.7409854  0.2523581 -0.62229894
\end{verbatim}
} % End size
\end{frame}


\begin{frame}[fragile]
\frametitle{$\mathbf{V}$ should have at least one zero eigenvalue}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> eigen(V)
\end{verbatim}
} % End color
\begin{verbatim}
$values
[1]  2.340116e+02  1.628929e+02  3.909544e+01 -1.012719e-14

$vectors
           [,1]         [,2]       [,3]         [,4]
[1,] -0.4475551  0.006507269 -0.2328249  0.863391352
[2,] -0.5632053 -0.604226296 -0.4014589 -0.395652773
[3,] -0.5366171  0.776297432 -0.1071763 -0.312917928
[4,] -0.4410627 -0.179528649  0.8792818  0.009829883
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Spectral decomposition $\mathbf{V}= \mathbf{CDC}^\prime$}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> eigenV = eigen(V)
> C = eigenV$vectors; D = diag(eigenV$values); D
\end{verbatim}
} % End color
\begin{verbatim}
         [,1]     [,2]     [,3]          [,4]
[1,] 234.0116   0.0000  0.00000  0.000000e+00
[2,]   0.0000 162.8929  0.00000  0.000000e+00
[3,]   0.0000   0.0000 39.09544  0.000000e+00
[4,]   0.0000   0.0000  0.00000 -1.012719e-14
\end{verbatim}
{\color{blue}
\begin{verbatim}
> # C is an orthoganal matrix
> C %*% t(C)
\end{verbatim}
} % End color
\begin{verbatim}
              [,1]         [,2]         [,3]          [,4]
[1,]  1.000000e+00 5.551115e-17 0.000000e+00 -3.989864e-17
[2,]  5.551115e-17 1.000000e+00 2.636780e-16  3.556183e-17
[3,]  0.000000e+00 2.636780e-16 1.000000e+00  2.558717e-16
[4,] -3.989864e-17 3.556183e-17 2.558717e-16  1.000000e+00
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Verify $\mathbf{V}= \mathbf{CDC}^\prime$}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> V;   C %*% D %*% t(C)
\end{verbatim}
} % End color
\begin{verbatim}
     [,1] [,2] [,3] [,4]
[1,]   49   62   58   38
[2,]   62  140   -4   62
[3,]   58   -4  166   29
[4,]   38   62   29   81

     [,1] [,2] [,3] [,4]
[1,]   49   62   58   38
[2,]   62  140   -4   62
[3,]   58   -4  166   29
[4,]   38   62   29   81
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{Square root matrix $\mathbf{V}^{1/2}= \mathbf{CD}^{1/2}\mathbf{C}^\prime$}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> sqrtV = C %*% sqrt(D) %*% t(C)
\end{verbatim}
} % End color
{\color{red}
\begin{verbatim}
Warning message:
In sqrt(D) : NaNs produced
\end{verbatim}
} % End color
{\color{blue}
\begin{verbatim}
> # Multiply to get V
> sqrtV %*% sqrtV;  V
\end{verbatim}
} % End color
\begin{verbatim}
     [,1] [,2] [,3] [,4]
[1,]  NaN  NaN  NaN  NaN
[2,]  NaN  NaN  NaN  NaN
[3,]  NaN  NaN  NaN  NaN
[4,]  NaN  NaN  NaN  NaN
     [,1] [,2] [,3] [,4]
[1,]   49   62   58   38
[2,]   62  140   -4   62
[3,]   58   -4  166   29
[4,]   38   62   29   81
\end{verbatim}
} % End size
\end{frame}

\begin{frame}[fragile]
\frametitle{What happened?}
%\framesubtitle{} 
{\footnotesize % or scriptsize
{\color{blue}
\begin{verbatim}
> D; sqrt(D)
\end{verbatim}
} % End color
\begin{verbatim}
         [,1]     [,2]     [,3]          [,4]
[1,] 234.0116   0.0000  0.00000  0.000000e+00
[2,]   0.0000 162.8929  0.00000  0.000000e+00
[3,]   0.0000   0.0000 39.09544  0.000000e+00
[4,]   0.0000   0.0000  0.00000 -1.012719e-14

         [,1]     [,2]     [,3] [,4]
[1,] 15.29744  0.00000 0.000000    0
[2,]  0.00000 12.76295 0.000000    0
[3,]  0.00000  0.00000 6.252635    0
[4,]  0.00000  0.00000 0.000000  NaN
\end{verbatim}
{\color{red}
\begin{verbatim}
Warning message:
In sqrt(D) : NaNs produced
\end{verbatim}
} % End color
} % End size
\end{frame}

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\begin{frame}
\frametitle{Copyright Information}

This slide show was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistical Sciences, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/~brunner/oldclass/302f13} {\small\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/302f13}}

\end{frame}


\end{document}

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