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\section*{Lecture 1: The geometry of linear equations}

The fundamental problem of linear algebra is to solve $n$ linear
equations in $n$ unknowns; for example:
\begin{eqnarray*}
2x - y &=& 0\\
-x + 2y &=& 3.
\end{eqnarray*}
In this first lecture on linear algebra we view this problem in three
ways.

The system above is two dimensional ($n=2$).  By adding a third variable $z$ we
could expand it to three dimensions.

\subsection*{Row Picture}


Plot the points that satisfy each equation.  The
intersection of the plots (if they do intersect) represents the solution to
the system of equations.
Looking at Figure~\ref{fig:l1_g1} we see that the solution to this system of equations is $x=1$, $y=2$.

\begin{figure}[h]
\centering
  \mypic{../Images/lec1fig1}\\
  \caption{The lines $2x-y=0$ and $-x+2y =3$ intersect at the point $(1,2)$.}\label{fig:l1_g1}
\end{figure}

We plug this solution in to the original system of equations to check our work:
\begin{eqnarray*}
2\cdot 1 - 2 &=& 0\\
-1 + 2\cdot2 &=& 3.
\end{eqnarray*}

The solution to a three dimensional system of equations is the common
point of intersection of three planes (if there is one).

\subsection*{Column Picture}


In the column picture we rewrite the system of linear equations as a
single equation by turning the coefficients in the columns of the
system into vectors:
$$x \twobyone{2}{-1} + y \twobyone{-1}{2} = \twobyone 0 3.$$

Given two vectors $\vb c$ and $\vb d$ and scalars $x$ and $y$, the sum
$x \vb c + y \vb d$ is called a {\em linear combination} of $\vb c$
and $\vb d$.  Linear combinations are important throughout this
course.

Geometrically, we want to find numbers $x$ and $y$ so that $x$ copies of vector $\twobyone{2}{-1}$ added to $y$ copies of vector $\twobyone{-1}{2}$ equals the vector $\twobyone 0 3$.  As we see from Figure~\ref{fig:l1_g2}, $x=1$ and $y=2$, agreeing with the row picture in Figure~\ref{fig:l1_g2}.
\begin{figure}[h]
\centering
  \mypic{../Images/lec1fig2}\\
  \caption{A linear combination of the column vectors equals the vector $\vb b$.}\label{fig:l1_g2}
\end{figure}

In three dimensions, the column picture requires us to find a linear
combination of three 3-dimensional vectors that equals the vector $\vb b$.

\subsection*{Matrix Picture}

We write the system of equations
\begin{eqnarray*}
2x - y &=& 0\\
-x + 2y &=& 3\\
\end{eqnarray*}
as a single equation by using matrices and vectors:
$$\twobytwo{2}{-1}{-1}{2} \twobyone x y = \twobyone 0 3.$$

The matrix ${\mb A} =
\twobytwo{2}{-1}{-1}{2}$ is called the {\em coefficient matrix}.  The vector $\vb x = \twobyone x y$ is the vector of unknowns.  The
values on the right hand side of the equations form the vector ${\vb
  b}$:
$$\mb A \vb x = \vb b.$$

The three dimensional matrix picture is very like the two dimensional
one, except that the vectors and matrices increase in size.

\subsubsection*{Matrix Multiplication}

How do we multiply a matrix $\mb A$ by a vector $\vb x$?

$$\twobytwo 2 5 1 3 \twobyone 1 2 = \quad ?$$

One method is to think of the entries of $\vb x$ as
the coefficients of a linear combination of the column vectors of the
matrix:
$$\twobytwo 2 5 1 3 \twobyone 1 2 = 1 \twobyone 2 1 + 2 \twobyone 5 3
= \twobyone {12}{7}.$$ This technique shows that $\mb A \vb
x$ is a linear combination of the columns of $\mb A$.

You may also calculate the product $\mb A \vb x$ by taking the dot
product of each row of $\mb A$ with the vector $\vb x$:
$$\twobytwo 2 5 1 3 \twobyone 1 2 = \twobyone{2\cdot 1 + 5\cdot
  2}{1\cdot 1 + 3\cdot 2} = \twobyone {12}{7}.$$

\subsection*{Linear Independence}

In the column and matrix pictures, the right hand side of the equation
is a vector $\vb b$.  Given a matrix $\mb A$, can we solve:
$$\mb A \vb x = \vb b$$ for every possible vector $\vb b$?  In other
words, do the linear combinations of the column vectors fill the
$xy$-plane (or space, in the three dimensional case)?

If the answer is ``no'', we say that $\mb A$ is a {\em singular
  matrix}.  In this singular case its column vectors are {\em linearly
  dependent}; all linear combinations of those vectors lie on a point
or line (in two dimensions) or on a point, line or plane (in three
dimensions).  The combinations don't fill the whole space.

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