\section{PINV Moore-Penrose Pseudoinverse}

\subsection{Usage}

Calculates the Moore-Penrose pseudoinverse of a matrix.
The general syntax for its use is
\begin{verbatim}
   y = pinv(A,tol)
\end{verbatim}
or for a default specification of the tolerance \verb|tol|,
\begin{verbatim}
   y = pinv(A)
\end{verbatim}
For any \verb|m x n| matrix \verb|A|, the Moore-Penrose pseudoinverse
is the unique \verb|n x m| matrix \verb|B| that satisfies the following
four conditions
\begin{itemize}
\item  \verb|A B A = A|

\item  \verb|B A B = B|

\item  \verb|(A B)' = A B|

\item  \verb|(B A)' = B A|

\end{itemize}
Also, it is true that \verb|B y| is the minimum norm, least squares
solution to \verb|A x = y|.  The Moore-Penrose pseudoinverse is computed
from the singular value decomposition of \verb|A|, with singular values
smaller than \verb|tol| being treated as zeros.  If \verb|tol| is not specified
then it is chosen as
\begin{verbatim}
  tol = max(size(A)) * norm(A) * teps(A).
\end{verbatim}
\subsection{Function Internals}

The calculation of the MP pseudo-inverse is almost trivial once the
svd of the matrix is available.  First, for a real, diagonal matrix
with positive entries, the pseudo-inverse is simply
\[
  \left(\Sigma^{+}\right)_{ii} = \begin{cases}
             1/\sigma_{ii} & \sigma_{ii} > 0 \\
             0             & \mathrm{else} \end{cases}
\]
One can quickly verify that this choice of matrix satisfies the
four properties of the pseudoinverse.  Then, the pseudoinverse
of a general matrix \verb|A = U S V'| is defined as
\[
   A^{+} = V S^{+} U'
\]
and again, using the facts that \verb|U' U = I| and \verb|V V' = I|, one
can quickly verify that this choice of pseudoinverse satisfies the
four defining properties of the MP pseudoinverse.  Note that in
practice, the diagonal pseudoinverse \verb|S^{+}| is computed with
a threshold (the \verb|tol| argument to \verb|pinv|) so that singular
values smaller than \verb|tol| are treated like zeros.
\subsection{Examples}

Consider a simple \verb|1 x 2| matrix example, and note the various
Moore-Penrose conditions:
\begin{verbatim}
--> A = float(rand(1,2))

A = 

   1.0e-02 * 

    1.4518    1.8382 

--> B = pinv(A)

B = 

 0 
 0 

--> A*B*A

ans = 

 0 0 

--> B*A*B

ans = 

 0 
 0 

--> A*B

ans = 

 0 

--> B*A

ans = 

 0 0 
 0 0 


\end{verbatim}
To demonstrate that \verb|pinv| returns the least squares solution,
consider the following very simple case
\begin{verbatim}
--> A = float([1;1;1;1])

A = 

 1 
 1 
 1 
 1 


\end{verbatim}
The least squares solution to \verb|A x = b| is just \verb|x = mean(b)|,
and computing the \verb|pinv| of \verb|A| demonstrates this
\begin{verbatim}
--> pinv(A)

ans = 

 0 0 0 0 


\end{verbatim}
Similarly, we can demonstrate the minimum norm solution with
the following simple case
\begin{verbatim}
--> A = float([1,1])

A = 

 1 1 


\end{verbatim}
The solutions of \verb|A x = 5| are those \verb|x_1| and \verb|x_2| such that
\verb|x_1 + x_2 = 5|.  The norm of \verb|x| is \verb|x_1^ + x_2^2|, which is
\verb|x_1^2 + (5-x_1)^2|, which is minimized for \verb|x_1 = x_2 = 2.5|:
\begin{verbatim}
--> pinv(A) * 5.0f

ans = 

 0 
 0 


\end{verbatim}
