\magnification 2000
\parskip 10pt
\parindent 0pt

\hoffset -0.3truein
\hsize 7 truein
\voffset -0.4truein
\vsize 9.8 truein

\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{\vskip -8pt}
\def\v{\vskip -4pt}

Section 1.6:  $\R^n$

Thm  (Cauchy-Schwarz Inequality):  Let {\bf u} and {\bf v} $\in \R^n$.
 Then
$${\bf u} \cdot {\bf v}~ \leq ~||{\bf u} ||~||{\bf v} ||$$


Recall in $\R^2$ and in $\R^3$,  ${\bf u} \cdot {\bf v} = ||{\bf u}|| ~  ||{\bf v} || cos \theta $




Thm (Triangle Inequality):   Let {\bf u} and {\bf v} $\in \R^n$.
 Then
$$||{\bf u} + {\bf v} ||~ \leq ~||{\bf u} || + ||{\bf v} ||$$


\vskip 3pt
\hrule
\vskip -6pt

A matrix is a vector of vectors

Example of a $2 \times 3$ matrix:

$\left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6}\right]$
\hfill
$\left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6}\right]$\hfill~~~~

$\left[\matrix{a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23}}\right]$
\hfill
$\left[\matrix{a_{11} & a_{12} & a_{13} \cr a_{21} & a_{22} & a_{23}}\right]$\hfill~~



Operations on Matrices

$A = (a_{ij})$, $B = (b_{ij})$, $C = (c_{ij})$.

Defn:  Two matrices $A$ and $B$ are equal, if they have the same
size and
$a_{ij} = b_{ij}$  for all 
$i = 1, ..., n$, $j = 1, ..., m$

Defn: $A + B  = (a_{ij} + b_{ij})$.

\vfill

Defn:  $cA = (ca_{ij})$.

\vfill

Defn: $-B = (-1)B$.

\vfill

Defn:  $A - B = A + (-B)$.

\vfill

Defn:  The zero matrix = {\it 0} =  $(a_{ij})$ where $a_{ij} = 0$ for all $i,
j$.

\vfill

Defn:  The identity matrix = 
{\it I} =  $(a_{ij})$ where
$a_{ij} = 0$ for all $i \not= j$ 
{and $a_{ii} = 1$ for all $i$}
{and $I$ is a square matrix.}

\vfill
\vfill

The transpose of the $m \times n$ matrix $A = A^T =
(a_{ji})$.

\vfill

\eject

Suppose $A$ is an $n \times r$ matrix, $B$ is an $r \times n$.
$AB = C$ where 
$$c_{ij} = row(i)~ of ~A \cdot column(j) ~ of ~B$$

\centerline{$ = a_{i1}b_{1j} + a_{i2}b_{2j} + ... + 
a_{ir}b_{rj}$.}
\vskip 10pt
Ex:  $\left[\matrix{1 & 2 & 3 \cr 4 & 5 & 6}\right]$
$\left[\matrix{2 & 3 \cr 10 & 1 \cr 0 & 4}\right] =$

\vfill

\vskip 5pt
\hrule

%%\vskip -6pt

$j$th column of $AB = A[j$th  column of $B]$
\vskip -7pt
In other words $AB = A[b_1 ... b_n] = [Ab_1 ... Ab_n]$
\vskip 5pt
\hrule
%%\vskip -6pt
$i$th row of $AB = [i$th row of $A]B$
\vskip -7pt

In other words $AB =~ \left[\matrix{a(1)  \cr . \cr . \cr . \cr
a(n)}\right]B
~~=~~ \left[\matrix{a(1)B \cr  . \cr . \cr . \cr a(n)B}\right]$

\eject
Thm (Properties of matrix arithmetic)
Let $A, B, C$ be matrices.  Let $a, b$ be scalars. 
 Assuming   that the
following operations are defined, then
\vskip -5pt
a.) $A + B = B + A$
\v

b.) $A + (B + C) = (A + B) + C$
\v

c.) $A + 0 = A$
\v

d.) $A + (-A) = 0$
\v

e.) $A(BC) = (AB)C$
\v

f.) $AI = A, IB = B$
\v

g.) $A(B + C) = AB + AC, \hfill \break
.\hskip 13.5pt (B + C)A = BA + CA$
\v

h.) $a(B + C) = aB + aC$
\v

i.) $(a + b)C = aC + bC$
\v

j.) $(ab)C = a(bC)$
\v
k.) $a(AB) = (aA)B = A(aB)$
\v

l.) $1A = A$
\hfill Defn.) $-A = -1A$
\v

Cor.) $A0 = 0, 0B = 0$
\hfill
Cor.) $a0 = 0~~~~~~~$


a.)$(A^T)^T = A$
\hfill
~~~~b.) $(A + B)^T = A^T + B^T$
\v
c.) $(kA)^T = kA^T$
\hfill
d.) $(AB)^T = B^TA^T~~~~~$

\eject

Note

\centerline{$AB \not= BA$}

It is also possible that $AB = AC$, but $B \not = C$.

In particular it is possible for $AB = 0$, but $A \not= 0$ AND $B \not= 0$


\vfill
\vfill

Defn:  $A$ is invertible if there exists a matrix $B$ such that $AB = BA = I$, and 
$B$ is called the inverse of $A$.  If the inverse
of
$A$ does not exist, then $A$ is said to be singular.

Ex:  $\left[\matrix{1 & 3  \cr 2 & 7}\right]$
	$\left[\matrix{7 & -3  \cr -2 & 1}\right] =$


Ex:  $\left[\matrix{a & b  \cr c & d}\right] \left[\matrix{~~~ & ~~~  \cr ~~~ & ~~~}\right] =$ 

Thus  $\left[\matrix{a & b  \cr c & d}\right]^{-1} =$




\vfill
\vfill
\vfill

Note that if $A$ is invertible, then $A$ is a square matrix.


Thm:  If $A$ is invertible, then its inverse is unique. 

Proof:  Suppose $AB = I$ and $CA = I$.  
\vskip -5pt
\hskip 30pt Then, $B = IB = CAB = CI = C$.
\vfill





\eject
Defn: A {\bf linear transformation} 
 is a function $T: \R^n \rightarrow \R^m$ that satisfies the following 
two 
conditions.
For each {\bf u} and {\bf v} in $\R^n$ and scalar $a$,

i.)  $T(a{\bf u}) = aT({\bf u})$

ii.)  $T({\bf u + v}) = T({\bf u}) + T({\bf v})$

%%In this case, we say, $T \in {\cal L}(R^n;R^m)$.
\vskip 5pt
\hrule


Thm:  Let $T: V \rightarrow W$ be a linear transformation.  Then
$T({\bf 0}) = {\bf 0}$

Pf:  $T({\bf 0}) = T({\bf 0} +  {\bf 0}) = T({\bf 0}) +  T({\bf 0})$

\vskip 5pt
\hrule

Thm:  Let $A$ be an $m \times n$ matrix.  Then the function
\vskip 5pt
\centerline{$\matrix{ T:\R^n \rightarrow \R^m \cr 
T({\bf x}) = A{\bf x}}$}
\vskip -5pt
 is a linear transformation.

\vfill\vfill


\vskip 5pt
\hrule

Prove that $T:R^3 \rightarrow R^2$, $T(x, y, z) = (4x - 5y, 8)$ is NOT a
linear transformation.

\vfill\vfill

\eject

Examples of linear transformations:

\vfill
\vfill\vfill\vfill


\vskip 45pt
Ex:  If $T: \R^2 \rightarrow \R$ is a linear transformation and \hfil \break
$T(1, 0) = 3$, $T(0, 1) = 4$, then 

$T(x, y) = $


\eject

Thm 4.3.3:  If $T: R^n \rightarrow R^m$ is a linear transformation,
then $T({\bf x}) = A{\bf x}$ where $A = [T({\bf e_1}) ... T({\bf e_n})]$.
\vskip -5pt

Proof:
Let ${\bf x} = \left[\matrix{x_1 \cr . \cr . \cr
.
\cr
x_n}\right] = x_1{\bf e_1} + ... + x_n{\bf e_n}$.

Thus $T({\bf x}) = T(x_1{\bf e_1} + ... + x_n{\bf e_n})$

\hskip 0.67in $= x_1T({\bf e_1}) + ... + x_nT({\bf e_n})$

\hskip 0.67in $=[T({\bf e_1})  ...  T({\bf e_n})]\left[\matrix{x_1 \cr . \cr . \cr
.
\cr
x_n}\right]$

\end
Thm:  Let $A$ be a square matrix.  If there exists a square matrix $B$ such that 
$AB = I$, then $BA = I$ and thus $B = A^{-1}$



Defn: $A^0 = I$, 
and if $n$ is a positive integer \break
 $A^n = AA \cdot
\cdot
\cdot A$ and $A^{-n} = A^{-1}A^{-1} \cdot \cdot
\cdot A^{-1}$.  

Thm: 
 If r, s are integers, $A^rA^s = A^{r+s}$, \break
 $(A^r)^s = A^{rs}$

Thm: If $A^{-1}$ and $B^{-1}$exist, then 
\v

~\hskip 4pt i.) $AB$ is invertible and
$(AB)^{-1} =
B^{-1}A^{-1}$
\v
~ii.) $A^{-1}$ is invertible and $(A^{-1})^{-1} =
A$
\v
iii.) $A^{r}$ is invertible and $(A^{r})^{-1} =
(A^{-1})^r$ 

\rightline{where $r$ is any integer}
\v
iv.) For any nonzero scalar $k$, 

\rightline{$kA$ is invertible and
$(kA)^{-1} =
{1 \over k}A^{-1}$}
\v
~v.) $A^T$ is invertible  and $(A^T)^{-1} = (A^{-1})^T$

Thm: If $A = \left[\matrix{a & b \cr c & d}\right]$, then $A$ is
invertible if
and only if $ad - bc \not= 0$, in which case
\vskip 5pt
\centerline{$A^{-1} = {1 \over ad - bc}\left[\matrix{~d & -b \cr -c &
~a}\right]$}

\end