% math 211 - 02/03 - Tutorial (Linear Algebra Review)
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\begin{titlepage}
\vspace*{-30pt}        
\begin{center} {\huge\bf Math 211} \end{center} \vspace{0pt}
\begin{center} {\Large\bf Linear Algebra Review} \end{center} 

\newcommand{\rank}{\mbox{rank}}
\newcommand{\Lra}{\Leftrightarrow}
            
\begin{list}{\arabic{number}. }{\usecounter{number}}

\item[\bf 1.] {\bf Solving Systems of Linear Equations} $A\vec{x} = \vec{b}$
\begin{itemize}
\item[(a)] {\em Row reduction}, {\em row echelon form} (REF) of $A$, the {\em augmented matrix} $\tilde{A} = (A|\vec{b})$.

\item[(b)] General solution of $A\vec{x} = \vec{b}$ 
via {\em back substitution} (with parameters).

\item[(c)] For {\em homogeneous systems} (i.e.\ those for which $\vec{b} = \vec{0}$) we have the formula: 
$$
\mbox{\# parameters} = n - \mbox{rank}(A)
$$
where \begin{minipage}[t]{8cm} 
$A$ is an $m\times n$ matrix ($m$ rows, $n$ columns), \\
$\rank(A)$ = \# non-zero rows of a REF of $A$.
\end{minipage}

\item[(d)] A general {\em inhomogeneous}  system $A\vec{x}=\vec{b}$ is {\em consistent} (i.e.\ has a solution) $\Lra \rank(A) = \rank(\tilde{A})$. 
(Otherwise, the system has no solution, i.e.\ is {\em inconsistent.})

\item[(e)] Invertible matrices: $A$ is {\em invertible} if $A^{-1}$ exists, i.e.\ if there is a matrix $B$ such that $AB = BA =I$.

{\bf Theorem.} If $A$ is an $n\times n$ matrix, then the following conditions are equivalent: \\
\ind (1) $A$ is invertible. \\
\ind (2) The homogeneous system $A\vec{x} = \vec{0}$ only has the trivial solution $\vec{x} = \vec{0}$.  \\ 
\ind (3) The inhomogeneous system $A\vec{x} = \vec{b}$ has a solution for every $\vec{b}$. \\  
\ind (4) $\mbox{rank}(A) = n$. \\
\ind (5) The rows of $A$ are linearly independent. \\
\ind (6) The columns of $A$ are linearly independent. \\
\ind (7) The rows (columns) span $\R^n$. \\ 
\ind (8) $\det(A) \ne 0$. 
\end{itemize} 

\item[\bf 2.] {\bf Subspaces} 
\begin{itemize}
\item[(a)] Definition of a {\em subspace} $V$ (of $\R^n$) and of a {\em linearly independent} set.
 
\item[(b)] For a subspace $V$, definition of a {\em spanning set} of $V$ and of a {\em basis} of $V$. 

{\bf Theorem.} All bases of a subspace $V$ have the same number of elements. 

Definition: the {\em dimension} of $V$ is $\dim V = \#B$, $B$ any basis of $V$.

\item[(c)] Subspaces attached to an $m\times n$ Matrix $A$: the {\em solution space} $\mbox{Solsp}(A) \subset \R^n$, the {\em row space} $\mbox{Rowsp}(A) \subset \R^n$ and the {\em column space} $\mbox{Colsp}(A)\subset \R^m$. 

Bases for these spaces may be determined by row reducing $A$ to $R$, an REF of $A$. (Note that $\mbox{Solsp}(A) = \mbox{Solsp}(R)$ and $\mbox{Rowsp}(A) = \mbox{Rowsp}(R)$, but that $\mbox{Colsp}(A)\ne \mbox{Colsp}(R)$ (in general).) Thus we have:
$$
\dim \mbox{Solsp}(A) = n -\rank(A),\quad\dim \mbox{Rowsp}(A) = \dim \mbox{Colsp}(A) = \rank(A).
$$
In particular: $\rank(A) = \rank(A^t)$, where $A^t$ = {\em transpose} of $A$.
\end{itemize}

\end{list}
\end{titlepage}        
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