\documentclass[12pt]{article}

\title{Matrices and Least Square Fit}

\author{FOSSEE}

\usepackage{listings}

\lstset{language=Python,

    basicstyle=\ttfamily,

  commentstyle=\itshape\bfseries,

  showstringspaces=false,

}

\newcommand{\typ}[1]{\lstinline{#1}}

\usepackage[english]{babel}

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\usepackage{times}

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\begin{document}

\date{}

\vspace{-1in}

\begin{center}

\LARGE{Matrices and Least Square Fit}\\

\large{FOSSEE}

\end{center}

\section{Matrices}

\subsection{Basics}

Matrix Creation\\

\typ{In []: C = array([[1,1,2], [2,4,1], [-1,3,7]])}\\

It creates C matrix of shape 3x3\\

Shape is dimensions of given array.

\begin{lstlisting}

In []: C.shape 

Out[]: (3, 3)

In []: shape([[1,2],[4,5],[3,0]])

Out[]: (3, 2)

\end{lstlisting}

\typ{In []: B = ones_like(C)} \\

B would be array of ones with the same shape and type as C.\\

\typ{In []: A = ones((3,2))} \\

A would be new matrix of given shape(arguments), filled with ones.\\ 

\typ{In []: I = identity(3)}\\

I would be identity matrix of shape 3x3

\subsection{Accessing Elements}

\begin{lstlisting}

In []: C

Out[]: 

array([[ 1,  1,  2],

       [ 2,  4,  1],

       [-1,  3,  7]])

In []: C[1,2]

Out[]: 1

\end{lstlisting}

Two indexes seperated by \typ{','} specifies [row, column]. So \typ{C[1,2]} gets third element of second row(indices starts from 0).

\newpage

\begin{lstlisting}

In []: C[1]

Out[]: array([2, 4, 1])

\end{lstlisting}

Single index implies complete row.

\subsection{Changing elements}

\begin{lstlisting}

In []: C[1,1] = -2

In []: C

Out[]: 

array([[ 1,  1,  2],

       [ 2, -2,  1],

       [-1,  3,  7]])

In []: C[1] = [0,0,0]

In []: C

Out[]: 

array([[ 1,  1,  2],

       [ 0,  0,  0],

       [-1,  3,  7]])

\end{lstlisting}

\subsection{Slicing}

Accessing rows with Matrices is straightforward. But If one wants to access particular Column, or want a sub-matrix, Slicing is the way to go.

\begin{lstlisting}

In []: C[:,1]

Out[]: array([1, 0, 3])

\end{lstlisting}

First index(:) specifies row(':' implies all the rows) and second index(1) specifies column(second column).

\begin{lstlisting}

In []: C[1,:]

Out[]: array([0, 0, 0])

\end{lstlisting}

Here we get second row(1), all columns(':') of C matrix.

\newpage

\begin{lstlisting}

In []: C[0:2,:]

Out[]: 

array([[1, 1, 2],

       [0, 0, 0]])

\end{lstlisting}

Result is sub-matrix with first and second row(endpoint is excluded), and all columns from C.

\begin{lstlisting}

In []: C[1:3,:]

Out[]: 

array([[ 0,  0,  0],

       [-1,  3,  7]])

In []: C[:2,:]

Out[]: 

array([[1, 1, 2],

       [0, 0, 0]])

\end{lstlisting}

\typ{':2'} => start from first row, till and excluding third row.

\begin{lstlisting}

In []: C[1:,:]

Out[]: 

array([[ 0,  0,  0],

       [-1,  3,  7]])

In []: C[1:,:2]

Out[]: 

array([[ 0,  0],

       [-1,  3]])

\end{lstlisting}

\typ{'1:'} => Start from second row, till last row\\

\typ{':2'} => Start from first column, till and excluding third column.

\newpage

\subsection{Striding}

Often apart from submatrix, one needs to get some mechanism to jump a step. For example, how can we have all alternate rows of a Matrix. \\

Following method will return Matrix with alternate rows.

\begin{lstlisting}

In []: C[::2,:]

Out[]: 

array([[ 1,  1,  2],

       [-1,  3,  7]])

\end{lstlisting}

\typ{C[startR:stopR:stepR,startC:stopC:stepC]} => Syntax of mentioning starting index, ending index, and step to jump.\\

In above mentioned case, \typ{'::2'} means, start from first row, till last row(both are blank), with step of 2, that is, skipping alternate row. After first row, C[startR], next row would be C[startR+stepR] and so on.

\begin{lstlisting}

In []: C[:,::2]

Out[]: 

xarray([[ 1,  2],

       [ 0,  0],

       [-1,  7]])

\end{lstlisting}

Same as above, just that here we get matrix with each alternate column and all rows.

\begin{lstlisting}

In []: C[::2,::2]

Out[]: 

array([[ 1,  2],

       [-1,  7]])

\end{lstlisting}

\section{Matrix Operations}

For a Matrix A and B of equal shapes.

\begin{lstlisting}

In []: A.T # Transpose

In []: sum(A) # Sum of all elements

In []: A+B # Addition

In []: A*B # Element wise product

In []: dot(A,b) #Matrix multiplication

In []: inv(A) # Inverse

In []: det(A) # Determinant

\end{lstlisting}

Eigen Values and Eigen Vectors

\begin{lstlisting}

In []: eig(A) #Eigen Values and Vectors

In []: eigvals(A) #Eigen Values 

\end{lstlisting}

%% Norm

%% \begin{lstlisting}

%% In []: norm(A)

%% \end{lstlisting}

%% Single Value Decomposition

%% \begin{lstlisting}

%% In []: svd(A)

%% \end{lstlisting}

\section{Least Square Fit Line}

\begin{lstlisting}

L = []

T = []

for line in open('pendulum.txt'):

    point = line.split()

    L.append(float(point[0]))

    T.append(float(point[1]))

Tsq = []

for time in T:

    Tsq.append(time*time)

plot(L, Tsq, '.')

\end{lstlisting}

This is exact curve we get from L Vs Tsq from data.This relation among L and Tsq is not of straight line. For getting Least Square Fit line, we have to solve the relations:\\

$L=m*Tsq+c$ (something similar to $y=m*x+c$)\\

For present scenario, we have L and corresponding Tsq values. For finding m and c at given points we use \typ{lstlq} function provided by pylab. It returns the least-squares solution to an equation. \\

For finding Least Square Fit line for this particular data we have to do following steps:\\

\typ{In []: A = array([L, ones\_like(L)])}\\

A is 2x(Length of array L) array.

\begin{lstlisting}

In []: A = A.T #now A.shape = (Length of array L)x2

In []: result = lstsq(A,TSq)

In []: coef = result[0]

In []: Tline = coef[0]*L + coef[1]

\end{lstlisting}

\typ{coef[0]} is array with all $m$ values, and \typ{coef[1]} contains $c$.\\

To get the final plot.\\

\typ{In []: plot(L, Tline)}

\end{document}