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\begin{document}

\vspace*{-0.7true in}

\hbox to \hsize{Math 210 \hfill Jerry L. Kazdan}

\centerline{\Large \bf Quadratic Polynomials}

\vskip 20 pt
\noindent{\bf Polynomials in One Variable.}
\bigskip

After studying linear functions $y=ax + b$, the next step is to
study {\it quadratic polynomials}, $y=ax^2 + bx + c$, whose graphs
are parabolas.  Initially one studies the simpler special case
\begin{equation} \label{eq:quad1}
y = ax^2 + c
\end{equation}
If $a>0$ these parabolas have a minimum at $x=0$ and open upward,
while if $a<0$ they have a maximum at $x=0$ and open downward.

One can reduce the more general quadratic polynomial
 \begin{equation} \label{eq:quad2}
 y= ax^2 + bx + c
 \end{equation}
to the special case (\ref{eq:quad1}) by a change of variable,
$x = v + r$ translating $x$ by $r$, where $r$ is to be found.
Substituting this into (\ref{eq:quad2}) we find
 $$
 y = a(v+r)^2 + b(v+r) + c = av^2 +(b +2ar)v + ar^2 +br +c.
 $$
We now pick $r$ to remove the linear term in $v$, that is, $b+2ar=0$
so $r=-b/(2a)$.  Then
 \begin{equation} \label{eq:quad3}
 y= av^2 + k = a(x-r)^2 + k,
 \end{equation}
where $k = c - b^2/(4a)$.  Thus, $x=r$ is the {\it axis of
symmetry\/} of this parabola.

\vskip -12pt

\hbox{\vbox{\hangindent=-1.80in \hangafter=0
This procedure is equivalent to ``completing the square'', a
procedure that should be familiar from algebra.  Another way to find
$r$ is to observe that the only critical point of (\ref{eq:quad2})
is at $x=-b/(2a)$.  Thus translating by $b/(2a)$ places this
critical point on the vertical axis.
}
\hskip -0.8cm \vbox{\epsfig{figure=2parabolas.eps,scale=0.70} \vskip -2.5cm}
}
\smallskip

%\hbox{\begin{minipage}[b]{4.3in} 
\hbox{\vbox{\hangindent=-1.80in \hangafter=0
\noindent{\sc Example}.  On the right are the graphs of $y=x^2+1$ and
$y=(x-2)^2+1=x^2-2x+2$.   They clearly shows the graph on the right
is merely a translation of the graph on the left.  
}}
%\end{minipage}

\vskip20pt

\noindent{\bf Polynomials in Several Variables.}
\medskip

\noindent{\sc Maxima, minima, and saddle points.}
\medskip

There are more interesting possibilities for quadratic polynomials in two
variables. 
\vskip -2.3cm 
\hbox{\hskip-1.5cm\vbox{\epsfig{figure=paraboloidup.eps,scale=1.0}} 
\hskip -4.5in \raise 0.6cm \vbox{\epsfig{figure=paraboloiddown.eps,scale=1.0}}
\hskip -3.8in \raise 1.6cm\vbox{\epsfig{figure=hyperboloid1.eps,scale=0.70}}
}
\vskip -3cm
$\hskip 1.2cm w=2x^2 + y^2
\hskip 1.5cm  w= -(2x^2 + y^2)
\hskip 1.8cm w=-2x^2 + y^2$

\bigskip\noindent 
From the graphs it is clear that the first has a {\it minimum\/} at
the origin, the second a {\it maximum\/}, while the third has a {\it
saddle point\/} there.

\medbreak

It is less obvious how to treat polynomials such as
 \begin{equation} \label{eq:cross1}
   w=3x^2 - 2xy +y^2
 \end{equation}
or
 \begin{equation} \label{eq:cross2}
   w=3x^2 - 2xy +y^2 +  z^2
 \end{equation}
with $xy$ terms and possibly more variables.  These two examples can be
handled if one recognizes that $x^2-2xy+y^2 = (x-y)^2$ so, if one
makes the change of variables $r=x,\,s=x-y$ and $t=z$ then in the 
new coordinates these polynomials become
 $$
 w=2x^2 + (x-y)^2 = 2r^2 + s^2 \qquad \hbox{\rm and}\qquad
           w= 2r^2 + s^2 +t^2,
 $$
which clearly have minima at the origin in $rs$ and $rst$ space,
respectively.

\medskip

The primary task of this section is to give useful criteria for a
quadratic polynomial in several variables to have a maximum,
minimum, or saddle point.  This will then be used in the next
section to generalize the calculus of one variable second derivative
test for a local maximum to functions of several variables such as
our quadratic polynomials.

\medskip

The first step is to be a bit more systematic.  Rewrite
(\ref{eq:cross1}) as $w=3x^2 - xy - yx +y^2$ and observe that using
the inner (=dot) product it can be written in the more compact form
 \begin{equation} \label{eq:quad}
        w = {\bf X\cdot}A{\bf X},
 \end{equation}
where
$${\bf X}= \left(\!\! \begin{array}{c} x \\
                                   y \end{array}\!\!\right)
\qquad \hbox{\rm and $A$ is the {\it symmetric\/} matrix}\qquad
  A= \left(\!\! \begin{array}{cc}
               \phantom{-}3 &           -1\\
                         -1 & \phantom{-}1
 \end{array}\!\right).
$$
Similarly, we can also write
 \begin{eqnarray} \label{eq:xyz}
 w & = & 3x^2 + 2xy +y^2 -4yz + xz+ 5z^2 \\
   & = & 3x^2 + xy + yx +y^2 -2yz -2zy +
            \textstyle{1\over2}xz + \textstyle{1\over2}zx + 5z^2
 \end{eqnarray}
in the form (\ref{eq:quad}) where
$${\bf X}= \left(\!\! \begin{array}{c} x \\
                                   y \\
                                   z  \end{array}\!\!\right)
\qquad \hbox{\rm and $A$ is the {\it symmetric\/} matrix}\qquad
  A= \left(\!\! \begin{array}{ccc}
                  3  & \phantom{-}1 & \phantom{-}\textstyle{1\over 2}\\
                  1  & \phantom{-}1 &         -2 \\
\textstyle{1\over 2} &           -2 & \phantom{-}5
 \end{array}\!\right).
$$
\medskip

In the simplest cases when there are no cross-product terms,
 \begin{equation} \label{eq:diag}
 w = ax^2 + by^2 +cz^2
 \end{equation}
the matrix $A$ is a {\it diagonal\/} matrix
 $$
  A= \left(\!\! \begin{array}{ccc}
                  a & 0 & 0 \\
                  0 & b & 0 \\
                  0 & 0 & c
 \end{array}\!\right).
 $$
If $a, b$ and $c$ are all positive, then clearly $w$ has a minimum
at the origin.  Similarly if $a, b$ and $c$ are all negative then
clearly $w$ has a maximum at the origin.  However if at least one of
the coefficients $a, b, c$ is positive and at least one is negative,
then the origin is a {\it saddle point}.

\medbreak

For a given symmetric matrix $A$ we say that the matrix $A$ is
\smallskip

\begin{center}
 \begin{tabular}{lcrlr}
{\it positive definite\/} \quad& if & \quad${\bf X\cdot}A{\bf X}$ & $>0$
& for all ${\bf X}\ne 0$ \\
{\it negative definite\/} \quad& if & \quad${\bf X\cdot}A{\bf X}$ & $<0$
& for all ${\bf X}\ne 0$  \\
{\it indefinite\/} \quad& if & \quad${\bf X\cdot}A{\bf X}$ & changes sign &
 \end{tabular}
\end{center}
\smallskip

Thus, {\it positive definite\/} means precisely that the quadratic
polynomial $w= {\bf X\cdot}A{\bf X}$ has a minimum only at the origin,
{\it negative definite\/} means that $w$ has a maximum only at the
origin, and {\it indefinite\/} means that $w$ has a saddle point at
the origin.

\vskip -1.2cm

\hbox{\vbox{\hangindent=-1.80in \hangafter=0
If $A$ is invertible, we will see that these three cases are the
only possibilities, while if $A$ is not invertible, there are a few
degenerate cases not included in the above.  One of the simplest
examples is the polynomial $w=x^2 + 2xy + y^2$.  Since $w=(x+y)^2
\ge 0$, it is almost positive definite.  The difficulty is that
$w=0$ at points other than just $x=y=0$ since it is zero along the
whole line $y=-x$.  This is clearer from its graph.
}
\hskip -1.0cm \vbox{\epsfig{figure=parabcyl.eps,scale=0.80} \vskip -1.5cm}
}

For situations such as these, we say that the quadratic polynomial
${\bf X\cdot}A{\bf X}$ is {\it positive semi-definite\/} if
${\bf X\cdot}A{\bf X} \ge 0$ for all ${\bf X}$ with the
understanding that it may be zero for values of ${\bf X}$ other than just
zero.  There is a similar definition of {\it negative semi-definite\/}.

\smallbreak

We can now rephrase the primary task for this section: give criteria
on the matrix $A$ for the quadratic polynomial $w={\bf X\cdot}A{\bf
X}$ to be positive definite, negative definite, or indefinite.  We
will give two different versions, one using eigenvalues and one
using determinants.  Both of these have their own virtues and are
applicable in different situations.

\medbreak

Before venturing further, we make four useful elementary observations.
\smallskip

\noindent {\sc Observation 1}. {\em A diagonal matrix is positive
definite if and only if all its diagonal elements are positive, it is
negative definite if and only if all its diagonal elements are
negative}.  We mentioned this above.  For a diagonal matrix, the
corresponding quadratic polynomial looks like (\ref{eq:diag}), so
this assertion should be obvious. \smallskip

\noindent {\sc Observation 2}. {\em If $A$ is positive definite,
then $-A$ is negative definite}.  The simplest example is $w=x^2 +
y^2$ where $A=I$ is the identity matrix.  $w$ has a minimum only at
the origin. The polynomial $w=-(x^2 + y^2)$, where $A=-I$, has a
maximum at the origin. \smallskip

\noindent {\sc Observation 3}. {\em If a symmetric matrix $A$, has
at least one positive diagonal element and at least one negative diagonal
element, then it is indefinite}. For instance, the matrix
$A$ associated with the quadratic polynomial
 $$
  w =  3x^2 + 2xy +y^2 -4yz + xz - 5z^2 \\
 $$
has two positive diagonal elements (the coefficients of $x^2$ and of
$y^2$), and one negative (the coefficient of $z^2$).  At points of the
form $(x,0,0)$ the polynomial $w=3x^2$ is clearly positive, while at
points of the form $(0,0,z)$, $w=-5z^2$ is clearly negative.  Thus
the origin  is neither a max nor a min, it is a saddle point.  We
conclude that $A$ is {\it indefinite}.

The same reasoning shows that {\em the diagonal elements of a
positive definite matrix are positive} while {\em the diagonal
elements of a negative definite matrix are negative}.
 \smallskip

\noindent {\sc Observation 4}. {\em If the symmetric matrix $A$ is
positive definite, then it is invertible}.  To see this, note that
if $A$ were not invertible, then there would be a vector ${\bf Z}\neq 0$
with $A{\bf Z}= 0$.  But then ${\bf Z\cdot}A{\bf Z}=0$ which
contradicts $A$ being positive definite.


\bigbreak
\noindent{\it Criterion for Positive Definite Using Eigenvalues}
\medskip

The key ingredients to this approach are {\sc Observation 1} just
above and the fundamental fact that one can always diagonalize a
symmetric matrix.  To be more specific, given a real symmetric
$n\!\times\!n$ matrix $A$ there is an orthogonal matrix  $R$ whose
columns are eigenvectors of $A$ so that the matrix $D:=R^{-1}AR$  is
a diagonal matrix.  The diagonal elements of $D$ are the eigenvalues
$\lambda_1,\, \lambda_2,\ldots,\lambda_n$ of $A$.  From this we can
write $A = RDR^{-1}$.  Thus, using the general rule ${\bf
X\cdot}R{\bf Y} = R^T{\bf X\cdot}{\bf Y}$ and since $R^{T}= R^{-1}$,
we find the quadratic polynomial is
 $$
 w = {\bf X\cdot}A{\bf X}= {\bf X\cdot}RDR^{-1}{\bf X}
                         = R^{-1}{\bf X\cdot}DR^{-1}{\bf X}.
 $$
This leads us to make the change of coordinates
${\bf V}=R^{-1}{\bf X}$.  In these ${\bf V}=(v_1,\ldots,v_n)$ coordinates
 \begin{equation} \label{eq:diag2}
 w = {\bf V\cdot}D{\bf V}=
     \lambda_1v_1^2 +\lambda_2v_2^2 +\cdots + \lambda_nv_n^2.
 \end{equation}
We now apply {\sc Observation 1} and conclude that $w$ has a minimum
only at the origin if and only if the eigenvalues
$\lambda_1,\, \lambda_2,\ldots,\lambda_n$ are positive.  To summarize:
\medbreak

\noindent{\sc Eigenvalue Test}. {\em A symmetric matrix $A$ is:

 \begin{tabular}{ll}
\quad {\em positive definite\/} \quad & if and only if all of its
eigenvalues are positive \\
\quad {\em negative definite\/} \quad & if and only if all of its
eigenvalues are negative \\
\quad {\em indefinite\/} \quad & if and only if some eigenvalues are
positive and some negative.
 \end{tabular}}

\bigskip

\noindent If $A$ is invertible, that is, if none of its eigenvalues
are zero, these three are the {\it only\/} possibilities.  However, some
degenerate cases, such as a matrix having some positive eigenvalues
and the remaining eigenvalues being zero, are not included in the
above.  This is treated further in Exercises 26-28.

\vskip 15 true pt

\noindent{\sc Example 1}.  The quadratic polynomial
 $$
       4x^2 -6xy + 4y^2 + 2z^2
 $$
is associated with the matrix
   $$
  A= \left(\!\! \begin{array}{ccc}
    \phantom{-}4 &           -3 & 0 \\
              -3 & \phantom{-}4 & 0 \\
    \phantom{-}0 & \phantom{-}0 & 2
 \end{array}\!\right)
 $$
whose characteristic polynomial is
$(2-\lambda)[(4-\lambda)^2-9]=(2-\lambda)(\lambda - 7)(\lambda - 1)$.
Thus its eigenvalues are $1,\ 2$ and $7$.  Since these eigenvalues
are all positive, the matrix $A$ is positive definite. Notice that
this matrix $A$ does have some negative elements, although by {\sc
Observation 3} they couldn't have been on the diagonal since then
$A$ could not possibly have been positive definite.
 \vskip 15 true pt

\bigbreak

\noindent{\it Criterion for Positive Definite Using Determinants}

\medskip

One can also use determinants to test if a matrix is positive
definite.  This has the advantage that one does not need to find the
eigenvalues.  Associated with an $n\!\times\!n$ matrix $A$ one has
the {\it principal minors\/}
 $$
  A_1= \left(a_{11}\right), \ %
  A_2= \left(\!\! \begin{array}{cc}
               a_{11} & a_{12}  \\
               a_{21} & a_{22}
 \end{array}\!\right), \ %
  A_3= \left(\!\! \begin{array}{ccc}
               a_{11} & a_{12} & a_{13} \\
               a_{21} & a_{22} & a_{23} \\
               a_{31} & a_{32} & a_{33}
 \end{array}\!\right), \ \ldots,\ %
  A_n= \left(\!\! \begin{array}{cccc}
               a_{11} & a_{12} & \cdots & a_{1n} \\
               a_{21} & a_{22} & \cdots & a_{2n} \\
                 :    &   :    & \cdots &   :    \\
               a_{n1} & a_{n2} & \cdots & a_{nn}
 \end{array}\!\right).
 $$
\medbreak

\noindent{\sc Determinant Test}. {\em The symmetric matrix $A$ is
positive definite if and only if the determinants of all the
principal minors are positive.  Similarly, by {\sc Observation 2},
$A$ is negative definite if and only if these determinants alternate 
in sign: $\det A_1 <0,\ \det A_2>0,\ \ldots$}. 

\medskip

\noindent The above assertion concerning negative definite matrices 
follows from the first part by using {\sc Observation 2}. 

This criterion is not obvious to most people.  However it is easy to
use, at least for matrices that are not too large.

\vskip 15 true pt
\noindent{\sc Example 1} (again). The principal minors are
  $$
  A_1 =\left( 4 \right),\quad
  A_2 = \left(\!\! \begin{array}{cc}
    \phantom{-}4 &           -3  \\
              -3 & \phantom{-}4
 \end{array}\!\right), \hbox{\rm \quad and \quad}
  A_3 = \left(\!\! \begin{array}{ccc}
    \phantom{-}4 &           -3 & 0 \\
              -3 & \phantom{-}4 & 0 \\
    \phantom{-}0 & \phantom{-}0 & 2
 \end{array}\!\right).
 $$
Since $\det A_1 = 4>0,\ \det A_2=7>0, \hbox{\rm \ and\ } \det A_3=14>0$,
by this determinant criterion the matrix $A$ is positive definite.
\medbreak

\noindent{\sc Example 2}. Although the {\it diagonal\/} elements of a
positive definite matrix must be positive ({\sc Observation 3}), the
sign of the off-diagonal elements is not significant.  For instance,
consider the matrices
 $$
   \left(\!\! \begin{array}{cc}
               1 &   2  \\
               2 &   1
 \end{array}\!\right), \hbox{\rm \quad and \quad}
   \left(\!\! \begin{array}{cc}
    \phantom{-}1 &           -2  \\
              -2 & \phantom{-}1
 \end{array}\!\right)
 $$
associated with the quadratic polynomials $w=x^2 \pm 4xy + y^2$.
By the determinant test both are indefinite so these quadratic
polynomials both have saddle points at the origin.  One can
transform from one of these polynomials to the other merely by
replacing $y$ with $-y$, that is, by making a reflection across the
$x-$axis; clearly this does not change the essential behavior of
the polynomial.

\medbreak

\noindent{\sc Example 3}. The symmetric matrix
 $$
   \left(\!\! \begin{array}{ccccc}
    \phantom{-}1 &   2  &           -2 & 5 & \phantom{-}3 \\
    \phantom{-}2 &   9  & \phantom{-}7 & 0 & \phantom{-}8 \\
              -2 &   7  & \phantom{-}7 & 0 & \phantom{-}3 \\
    \phantom{-}5 &   0  & \phantom{-}0 & 1 & \phantom{-}0 \\
    \phantom{-}3 &   8  & \phantom{-}3 & 0 &           -5
 \end{array}\!\right)
 $$
is indefinite.  To see this one could use the determinant test, but
it is simpler to use {\sc Observation 3} since at least one of the
diagonal elements is positive and a at least one is negative.

 \vskip 15 true pt

\noindent{\sc proof of the determinant test}.  The proof is instructive,
but complicated.  It is included so the more adventurous readers
can see how it works.  Upon first reading it is best just to skim it.

 \smallskip

{\small    % small type font

\noindent{\sc Part 1}. {\em If $A$ is positive definite then the
determinants of all its principal minors are positive}.
\smallskip

\noindent {\it Step 1}. We first show that if the symmetric matrix
$A$ is positive definite, then its determinant is positive.  In
fact, we give two different proofs of this.

The first proof begins from the basic fact that the determinant of
matrix is the product of its eigenvalues. If $A$ is positive
definite, above we showed that all of its eigenvalues must be
positive, hence the determinant is positive.

The second proof starts from the observation that the sum of two
positive definite symmetric matrices $A$ and $B$ is positive [proof:
 $$
{\bf X}\cdot(A+B){\bf V}={\bf X}\cdot A{\bf V} +{\bf X}\cdot B{\bf V} > 0
 $$
for all ${\bf X}\ne 0$.]  In particular, if $A$ is positive
definite, so is the matrix $C(t)=tA + (1-t)I$ for $0 \le t \le 1$.
Since $C(t)$ is positive definite, by {\sc Observation 4} above it
is invertible and hence $\det C(t) \neq 0$ for  $0 \le t \le 1$.
Since $\det C(0) = \det I >0$, then $\det C(1) = \det A >0$.
\smallskip

\noindent {\it Step 2}.  We claim that {\em if $A$ is positive 
definite, then so are its principal minors}.  For this we must show, 
for instance, that $A_3$ is positive definite. Let ${\bf 
V}=(v_1,\,v_2,\,v_3)$ and let ${\bf X}$ be a vector whose first part 
is ${\bf V}$ and the rest is $0$, so  ${\bf X}= 
(v_1,\,v_2,\,v_3,\,0.\ldots,0)$.  If ${\bf V}\neq 0$, then since $A$ 
is positive definite we know 
 $$
  0 < {\bf X\cdot}A{\bf X}= {\bf V\cdot}A_3{\bf V}.
 $$
Thus $A_3$ is positive definite.  One similarly shows that $A_k$ is
positive definite for any $1 \leq k \leq n$.  Since these $A_k$ are
positive definite, by Step 1 their determinants are all positive.
\medbreak

\noindent{\sc Part II}. Conversely, {\em if the determinants of all
its principal minors are positive, then $A$ is positive definite}.
\smallskip

\noindent {\it Step 1\/}. We need a preliminary computation relating $\det A$
and the matrix $A_{n-1}$.  For cleanliness, let $B =A_{n-1}$ and
write $A$ as
 $$
  A= \left(\!\! \begin{array}{crcc}
                      &       &              & a_{1n} \\
                      &   B   &              &  :       \\
                      &       &              & a_{n-1\,n}    \\
             a_{n1} &\cdots & a_{n\,n-1} & c
 \end{array}\!\right),
  $$
where $a_{nn}= c$. If we let ${\bf \alpha}$ be the column vector
${\bf \alpha} = (a_{1n},\,a_{2n},\,\ldots,a_{n-1\,n})$ this can be
further abbreviated as
 \begin{equation} \label{eq:block1}
  A= \left(\!\! \begin{array}{cc}
                                B       &  {\bf \alpha}  \\
                       {\bf \alpha}^T   & c
 \end{array}\!\right).
 \end{equation}
We want to compute $\det A$.  Perhaps the simplest approach is to
notice that matrix multiplication also works for {\it block
matrices\/} such as (\ref{eq:block1}).  In the formula below let $I$ 
be the $(n-1)\times (n-1)$ identity matrix.  We seek a column vector 
${\bf v}$ with the same shape as ${\bf \alpha}$ so that the product 

\begin{equation} \label{eq:block2}
   \left(\!\! \begin{array}{cc}
                             B          &  {\bf \alpha}  \\
                       {\bf \alpha}^T   & c
 \end{array}\!\right)
   \left(\!\! \begin{array}{cc}
                             I   &  {\bf v}  \\
                             0   &    1
 \end{array}\!\right)
  = \left(\!\! \begin{array}{cc}
            B          &  B{\bf v + \alpha}  \\
            {\bf \alpha}^T        &  {\bf \alpha}^T{\bf v}+c
 \end{array}\!\right)
 \end{equation}
is as simple as possible.  Notice that the second matrix on the left 
is upper triangular, so its determinant is the product of the 
diagonal elements and hence is $1$ no matter how we pick ${\bf v}$. 
We'll use $B{\bf v + \alpha}=0$; that is, assuming $B$ is 
invertible, then ${\bf v}= - B^{-1}{\bf \alpha}$. With this choice 
of ${\bf v}$ we can evaluate the determinant on the right easily, 
expanding by minors using the last column.  Thus, we take the 
determinant of both sides of (\ref{eq:block2}) to conclude that 
 \begin{equation}
\label{eq:induction} \det A =(c-{\bf \alpha}^TB^{-1}{\bf
\alpha})\det B
       =(c-{\bf \alpha\cdot}B^{-1}{\bf \alpha})\det B.
\end{equation}
\smallskip

\noindent {\it Step 2\/}. We will use this to show that {\em if $B$ is a
positive definite symmetric matrix and if $A$ in {\rm(\ref{eq:block1})}
has $\det A > 0$, then $A$ is also positive definite}.

We must show that ${\bf X\cdot}A{\bf X}>0$ for all ${\bf X}\ne 0$.
To utilize the block structure of (\ref{eq:block1}), write
${\bf X}$ as the column vector ${\bf X}= ({\bf x}, z)$ where ${\bf
x}=(x_1,\ldots,x_{n-1})$ and $z$ is a scalar.  Then from
(\ref{eq:block1})
 \begin{equation}
  A{\bf X}= \left(\!\! \begin{array}{cc}
               B       &  {\bf \alpha}  \\
      {\bf \alpha}^T   & c
                     \end{array}\!\right)
           \left(\!\! \begin{array}{c}
               {\bf x} \\
                  z
                \end{array}\!\right)
         =  \left(\!\! \begin{array}{c}
               B{\bf x} +{\bf \alpha}z \\
                {\bf \alpha}^T{\bf x} +cz
                \end{array}\!\right).
 \end{equation}
Consequently
 \begin{equation} \label{eq:main1}
   {\bf X\cdot}A{\bf X} =
  {\bf x\cdot}B{\bf x} + 2{\bf \alpha\cdot x}z + cz^2.
 \end{equation}
Since $B$ is positive definite, in Part 1 we showed that $\det B >0$
so $B$ is invertible.  Because $\det A>0$, equation
(\ref{eq:induction}) tells us that $c > {\bf \alpha\cdot}B^{-1}{\bf
\alpha}$.  We use this in (\ref{eq:main1}) to find
 \begin{equation} \label{eq:main2}
   {\bf X\cdot}A{\bf X} \geq
  {\bf x\cdot}B{\bf x} + 2{\bf \alpha\cdot x}z +
     {\bf \alpha\cdot}B^{-1}{\bf \alpha}z^2
          \qquad (\ >0 \hbox{\rm \ unless\ } z=0),
 \end{equation}
To make this equation clearer, as above let ${\bf v}=-B^{-1}{\bf
\alpha}$, so ${\bf \alpha} = -B{\bf v}$.  Then we can rewrite
(\ref{eq:main2}) as \begin{eqnarray} \label{eq:main3}
   {\bf X\cdot}A{\bf X}  & \geq &
  {\bf x\cdot}B{\bf x} - 2B{\bf v\cdot x}z + B{\bf v\cdot v}z^2 \\
            & = & ({\bf x} - {\bf v}z){\bf \cdot}B({\bf x} -{\bf v}z).
 \end{eqnarray}
Since the right side if this is of the form ${\bf w\cdot}B{\bf w}$
and because $B$ is positive definite, we conclude that
(\ref{eq:main3}) is positive unless ${\bf x} - {\bf v}z = 0$ and,
from (\ref{eq:main2}), $z=0$.  Thus (\ref{eq:main3}) is positive
unless ${\bf x}=0$ and $z=0$, so it is positive for all ${\bf
X}\neq 0$.  This proves that $A$ is positive definite.
\smallbreak

\noindent {\it Step 3\/}.  We finally show that if the principal
minors of $A$ all have positive determinant, then $A$ is positive
definite.  If $a_{11}>0$ and $\det A_2>0$, then by Step 2, $A_2$ is
positive definite.  But since $\det A_3>0$ Step 2 again implies that
$A_3$ is positive definite.  Repeating this we conclude eventually 
that $A_n = A$ is positive definite.  This completes the proof.
}     % end \small

\vskip20pt

\noindent{\sc Some Pictures.}
\medskip

One way to understand the quadratic polynomial $w = {\bf X\cdot}A{\bf X}$
is to look at the level sets; thus so we are looking at the points 
{\bf X} where ${\bf X\cdot}A{\bf X} = k$, for the particular values 
$k=1,\,0,\,\hbox{\rm\ and\ } -1$.  As we saw in equation 
(\ref{eq:diag2}), by  using a rotation, that is, an orthogonal 
transformation, we need only consider the special case when $A$ is 
diagonal.  For pictures, we will look at the special cases of two 
and three dimensions. 

\medskip

\noindent{\bf Two dimensions\/}
 \begin{equation} \label{eq:pictures2}
  \lambda_1x_1^2 + \lambda_2x_2^2 =c.
 \end{equation}

There are two cases, depending if $\lambda_1$ and $\lambda_2$ have
the same or have opposite sign.

\noindent If $\lambda_1$ and $\lambda_2$ have the {\it same\/} sign, then
since we can always change the sign of $c$, we make assume that both
$\lambda$'s are positive and write $\lambda_1=1/a^2,\
\lambda_2=1/b^2$. Then our equation is
 $$
   {x_1^2\over a^2} + {x_2^2 \over b^2} =k.
 $$
If $k=1$, this defines an ellipsoid whose semi-axes have lengths $a$
and $b$, respectively.  If $k=0$, only the origin satisfies the
equation, while if $k=-1$ no points satisfy the equation.  Of course
all of this is obvious from the three dimensional graph of
$w=x_1^2/a^2 + x_2^2/b^2$.  The alternate approach here will make it
easier to compare with the other cases below.

\noindent If $\lambda_1$ and $\lambda_2$ have the {\it opposite\/}
sign, then say $0<\lambda_1 = 1/a^2$ and $0>\lambda_2 = -1/b^2$.
The equation becomes
 $$
   {x_1^2\over a^2} - {x_2^2 \over b^2} =k.
 $$
If $k=1$ this is a hyperbola that opens on the horizontal axis, if
$k=0$ the equation is $(x_1 - x_2)(x_1+x_2)=0$ which is the pair of
straight lines $x_1=\pm x_2$.  Finally if $k=-1$ this is a hyperbola
that opens on the vertical axis.

\hbox{\hskip 0.0cm\vbox{\epsfig{figure=hyperbx.eps,scale=0.6}} 
\hskip -4.4in \raise 0.0cm \vbox{\epsfig{figure=stlines.eps,scale=0.6}}
\hskip -4.5in \raise 0.0cm\vbox{\epsfig{figure=hyperby.eps,scale=0.6}}
}
\vskip -1cm
$\hskip 1.0cm x^2 - \frac{1}{4}y^2 = 1
\hskip 2.1cm  x^2 - \frac{1}{4}y^2 = 0
\hskip 1.7cm x^2 - \frac{1}{4}y^2 = -1$

\vskip15pt

\noindent{\bf Three dimensions\/}
 \begin{equation} \label{eq:pictures3}
  \lambda_1x_1^2 + \lambda_2x_2^2 + \lambda_3x_3^2 =1.
 \end{equation}
The three cases depend on how many of the $\lambda$'s are positive or
negative.
\smallskip

\noindent{\em All $\lambda$'s positive}.  For reasons that will be
clear shortly, it is convenient to write $\lambda_1=1/a^2$,
$\lambda_2=1/b^2$, and $\lambda_3=1/c^2$, so the equation is
 $$
  {x_1^2\over a^2} + {x_2^2\over b^2} + {x_3^2\over c^2} =k.
 $$
\vskip -2.0cm
\hbox{\begin{minipage}[b]{4.5in} 
This is an {\it ellipsoid\/}. If $k > 0$ we can always divide by it
to reduce to the case where $k=1$ (the cases $k\le 0$ are not
interesting, Why?). Thus, we assume $k=1$.  Since $x_1$
is largest when $x_2=x_3 = 0$, in which case $x_1=\pm a$, we see that
the length of this semi-axis is $a$.  Similarly the lengths of the
other semi-axes are $b$ and $c$, respectively. If $k=0$ the only point
that satisfies this is the origin, while if $k=-1$, no real points
satisfy this.
\end{minipage}
\hskip -0.55in \raise -1.0cm 
\vbox{\epsfig{figure=ellipsoid.eps,scale=0.8}}
}
\vskip -1.7cm\hskip 4.4in $x^2 + \frac{1}{25}y^2 +\frac{1}{9}z^2=1$.

 \bigskip

\noindent{\em Two $\lambda$'s positive, one negative}.  In this case
we write, say, $ \lambda_1=a^2$, $ \lambda_2=b^2$, while
$ \lambda_1=-c^2$ and the equation is
 $$
 {x_1^2\over a^2} + {x_2^2\over b^2} - {x_3^2\over c^2} =k.
 $$
If $k=1$ this is a {\it hyperboloid of one sheet\/}.  If $k=0$ this is a
{\it cone\/}, while if $k = -1$ this is a {\it hyperboloid of two sheets\/}.

 \smallskip

\noindent{\em One $\lambda$ positive, two negative}.   Multiplying
the equation by $-1$ reduces this case to the previous case (just
change the sign of $k$).
\vskip-1cm

\hbox{\hskip-1.0cm\vbox{\epsfig{figure=hyperboloid2.eps,scale=0.80}} 
\hskip -4.0in \raise 0.0cm \vbox{\epsfig{figure=cone.eps,scale=0.8}}
\hskip -4.2in \raise 0.0cm\vbox{\epsfig{figure=hyperboloid3.eps,scale=0.8}}
}
\vskip -1.3cm
\noindent$\hspace*{0.8cm}  x^2 + y^2/25 -z^2/9=1 
	\hskip 1.5cm  x^2 + y^2/25 -z^2/9=0
        \hskip1.0cm  x^2 + y^2/25 -z^2/9=-1
$

\noindent \hspace*{0.50cm}  hyperboloid of one sheet \hskip2.5cm  cone
\hskip2.4cm  hyperboloid of two sheets

\newpage

\noindent{\sc Quadratic polynomials with lower order terms}
\medskip

We close this section by showing how to treat quadratic polynomials
such as
 $$
     w=3x^2 - 2xy +y^2 +  z^2 -2x +y -5z +11
 $$
with lower order terms.  Write this in the form
 \begin{equation} \label{eq:quad10}
      w= {\bf X\cdot}A{\bf X} + {\bf b\cdot X} + c,
 \end{equation}
where the matrix $A$ is as before, ${\bf b}= (-2,\,1,\,-5)$ and
$c=11$.  We will use the approach mentioned at the beginning of this
section for a polynomial in one variable and make a change of
variable ${\bf X}= {\bf V} + {\bf r}$ where {\bf r} is to be found.
Substituting and computing gives
 \begin{eqnarray}
   w & = & ({\bf V}+{\bf r})\cdot A({\bf V}+{\bf r})
           + {\bf b\cdot}({\bf V}+{\bf r}) +c \\
     & = & {\bf V\cdot}A{\bf V} +({\bf b} + 2A{\bf r})\cdot {\bf V}
              + {\bf r\cdot}A{\bf r} + {\bf b\cdot r}+ c.
 \end{eqnarray}
Now we choose ${\bf r}$ to eliminate the linear term in ${\bf v}$;
thus $2A{\bf r}=-{\bf b}$, so ${\bf r}=-{1\over 2}A^{-1}{\bf b}$.  Then $w$
has the desired simpler form
 $$
 w = {\bf V\cdot}A{\bf V} + k =({\bf X}-{\bf r})\cdot A({\bf X}-{\bf r}) + k ,
 $$
where $k$ is the constant
 $$
   k = {\bf r\cdot}A{\bf r} + {\bf b\cdot r}+ c
     = c - \textstyle{1\over 4}{\bf b\cdot}A^{-1}{\bf b}.
 $$

\noindent This procedure is the generalization of ``completing the 
square'' to quadratic polynomials in several variables.  

Just as for a quadratic polynomial in one variable, another way to 
find ${\bf r}$ is to observe from (\ref{eq:quad10}) that since 
$\hbox{\rm grad}\,w = 2A{\bf X} + {\bf b}$, the only critical point 
of (\ref{eq:quad10}) is at ${\bf X} =-(1/2)A^{-1}{\bf b}$.  Thus 
translating by $(1/2)A^{-1}{\bf b}$ places this critical point on 
the vertical axis of symmetry.  Observe that these formulas are 
essentially identical with (\ref{eq:quad3}) for a polynomial in one 
variable. 

We leave it as an exercise for you to find the particular value of
$r$ for the special example we have.  Note that this procedure
presumes that $A$ is invertible, or at least that one can solve
$2A{\bf r}+{\bf b}=0$.

\vskip -1.7cm

\hbox{\begin{minipage}[b]{4.5in} 
\noindent{\sc Example}.  To the right are plots of $z = x^2 +4y^2$
and its translate, $z=x^2 + 4(y-2)^2$ so the axis of symmetry is the
vertical line through $(0,2,0)$.  Note that for $z=c$, the level curve on
the left is the ellipse $x^2 + 4y^2=c$, while on the right is the same
ellipse, just translated: $x^2 + 4(y-2)^2=c$.
\end{minipage}
\hskip -0.55in \lower 2.0cm 
\vbox{\epsfig{figure=2paraboloids.eps,scale=0.8}}
}
\vskip -1cm

\centerline{\large \bf Problems}
\parindent=0pt
\bigskip
{\small 

1. Write the following quadratic polynomials in the form
$a(x-r)^2+k$, that is, find the axis of symmetry $r$, and the
height $k$.

\hbox{\begin{minipage}[t]{3truein}
\begin{parts}
\item $x^2 +4x -3$
\end{parts}
\end{minipage}
\begin{minipage}[t]{3truein}
\begin{parts}\addtocounter{bean}{1}
\item $-2x^2 -6x +5$
\end{parts}
\end{minipage}
}
\bigbreak

2.  Write each of the following quadratic polynomials in the form
${\bf X\cdot}A{\bf X}$, that is, find the symmetric matrix $A$.

\hbox{\begin{minipage}[t]{3truein}
\begin{parts}
\item $3x^2 +3y^2$
\item $3x^2 -4xy -2y^2$
\item $-2x^2 +xy + y^2$
\item $4xy$
\item $x^2 + (x+2y)^2$
\item $x^2 + y^2 +2xz +5z^2$
\end{parts}
\end{minipage}
\begin{minipage}[t]{3truein}
\begin{parts}\addtocounter{bean}{6}
\item $-2x^2 +xy + y^2 +4xz -2yz +6z^2$
\item $2xy -6yz +z^2$
\item $(x +y -z+2u)^2$
\item $4xy +y^2 -7xz +2z^2$
\item $x^2+2y^2+3z^2 -4w^2 +4wz$
\item $x^2+2y^2+3z^2 +4w^2 +6yw$
\end{parts}
\end{minipage}
}
\bigbreak


3. Determine which of the quadratic forms in Problem 2 are positive
definite (or semi-definite), negative definite (or semi-definite), or
indefinite.

\bigbreak

4. For the polynomial \ $3x_1^2 -4x_1x_2 +3x_2^2$:
\begin{parts}
\item Determine if it is positive-definite, negative-definite,
or indefinite.
\item Use Maple to graph it.
\item Find an orthogonal matrix $R$ that diagonalizes the symmetric
matrix $A$ associated with this quadratic polynomial, so
$R^{-1}AR=D$ is a diagonal matrix.  Then make the change of variable
{\bf X}=R{\bf V}, where ${\bf V}=(v_1,\,v_2)$, and write the
polynomial in these new coordinates.
\end{parts}
 \bigbreak


5. Repeat Problem 4. for the polynomial \ $3x_1^2 +8x_1x_2 - 3x_2^2$.

 \bigbreak


6. Repeat Problem 4. for the polynomial \ $3x_1^2 +8x_1x_2 - 3x_2^2 -z^2$.

 \bigbreak


7.  Use Maple to graph \ $3x^2 -4xy +3y^2=1$.  Use all of the three
following approaches:
\begin{parts}
\item {\tt plot3d}: \ $z = x_1^2 -4x_1x_2 +3x_2^2$ \ for
$0 \le z \le 1$ (Maple uses {\tt view=0..1}), or else show the 
contour line $z=1$.
 \item Plot this curve implicitly.
 \item Use the procedure of Problem 4c to write this curve as \ 
$a^2v_1^2 + b^2v_2^2=1$, and then use the parameterization $v_1 
=(\cos t)/a,\ v_2=(\sin t)/b$ for $0\le t \le 2\pi$.
 \end{parts} 

\bigbreak

8. Repeat Problem 7 for the quadratic polynomial
\ $3x_1^2 +8x_1x_2 - 3x_2^2$, only for part c) use hyperbolic
functions $\cosh t = {1\over 2}(e^t+e^{-t})$ and $\sinh t={1\over
2}(e^t - e^{-t})$.  The only property one needs of these hyperbolic
functions is the easily verified identity $\cosh^2t - \sinh^2t = 1$,
which serves as a replacement for $\sin^2 t + \cos^2t=1$.

\bigbreak


9. Let \ $q(x,y,r,s)= 5x^2  + 2xy - 6xs + 5y^2 - 6ys - 9s^2 - 18xr - 18yr +
3r^2 + 36rs$.
Determine if $q$ is positive definite, negative definite, or indefinite.

\bigbreak


10. Let \ $h(x,y,r,s)= 5x^2  + 2xy - 6xs + 5y^2 - 6ys + 9s^2 - 18xr - 18yr +
3r^2 + 36rs$.
\begin{parts}
\item Determine if $h$ is positive definite, negative definite,
or indefinite.
\item Find the eigenvalues of the symmetric matrix associated
with $h$. [Remark: Maple does this quite quickly.]
\end{parts}


\bigbreak

11. Let \ $p(x,y)=3x^2-4xy+3y^2-14x+16y +25$.
\begin{parts}
\item Write this in the form
\ $p(x,y)=({\bf X}-{\bf X}_0)\cdot A({\bf X}-{\bf X}_0) +c$, that is,
find the axis of symmetry ${\bf X}_0$ and the height $c$.  Here, as
usual, ${\bf X}$ is the {\it column\/} vector ${\bf X}=(x,\,y)$.
\item Graph this polynomial.
\end{parts}

\bigbreak


12. Repeat Problem 11  for \ $q(x,y)=3x^2 +8xy -3y^2 +24x -18y +1$.

 \bigbreak


13. Let $A$ be a positive definite symmetric matrix.
\begin{parts}
\item Show that $A^2$ and $A^{-1}$ are also positive
definite.
\item Better yet, show that $A^{k}$ is positive definite
for {\it any\/} integer $k$.
\end{parts}

\bigbreak


14. If $A$ is a symmetric matrix, let \ $f({\bf X})={\bf X\cdot}A{\bf
X}$.  Use Lagrange multipliers to show that the maximum value of $f$
for all points on the unit ball, $\|{\bf X}\|=1$, is the largest
eigenvalue of $A$.

 \bigbreak


15. If $A$ is any symmetric matrix, show that there is some constant $c$
so that the matrix $A+cI$ is positive definite.  Can you find the
optimal value of $c$?

 \bigbreak


16. Let $A$ be a symmetric $n\times n$ matrix with eigenvalues
$\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$.  Show that for
any vector {\bf X}
 $$
 \lambda_1\|{\bf X}\|^2\le {\bf X\cdot}A{\bf X} \le \lambda_n\|{\bf X}\|^2.
 $$
[Suggestion: Use equation (\ref{eq:diag2}).]

\bigbreak


17. Let $B$ be any invertible matrix.  Show that $B^TB$ and $BB^T$ are both
positive definite.   If $B$ is not necessarily invertible -- or even 
a square matrix -- show that these are at least positive semi-definite. 

 \bigbreak


18.  If $A$ is a symmetric positive definite matrix and $C$ is any 
invertible matrix, show that $C^TAC$ is also positive definite.

\bigbreak


19. A matrix $C$ is called {\it skew-symmetric\/} if $C^T=-C$.
\begin{parts}
\item Give an example of a skew-symmetric $2\times 2$ matrix (other
than the $0$ matrix).
\item Show that $C$ is skew-symmetric if and only if
    $C{\bf X}\cdot {\bf Y}= -{\bf X}\cdot C{\bf Y}$ for all vectors
${\bf X}$ and ${\bf Y}$.
\item If $C$ is skew-symmetric, show that ${\bf X\cdot}C{\bf X}=0$
for all vectors {\bf X}.
\end{parts}

\bigbreak


20. Let $M$ be any square matrix and write $M = S + T$, where
$S={1\over 2}(M+M^T)$ and $T={1\over 2}(M-M^T)$.
\begin{parts}
\item Show that $S$ is symmetric and $T$ is skew-symmetric.
\item Show that if $M = A + B$, where $A$ is symmetric and $B$ is
skew symmetric,
then in fact $A=S$ and $B=T$.
\item Prove the identity ${\bf X\cdot}M{\bf X} = {\bf X\cdot}S{\bf X}$  
for all vectors {\bf X}.  Thus, the quadratic polynomial depends 
only on the symmetric part of $M$.  (This is why we only use 
symmetric matrices in quadratic polynomials).
 \end{parts} 

 \bigbreak


21. Let $A,\, B$, and $C$ be $n \times n$ symmetric matrices.
\begin{parts}
\item Expand $({\bf X}+{\bf Y})\cdot C ({\bf X}+{\bf Y})$ to obtain
the formula
 $$
         {\bf X}\cdot C{\bf Y}= \textstyle{1\over 2}
            [({\bf X}+{\bf Y})\cdot C ({\bf X}+{\bf Y})
             - {\bf X}\cdot C{\bf X}  - {\bf Y}\cdot C{\bf Y.}]
 $$
The point is that the expressions on the right are all of the form
${\bf V}\cdot C{\bf V}$ for various vectors {\bf V}, while the left 
side has a more general form. \item  Use this to show that if ${\bf 
X}\cdot C{\bf X}=0$ for all vectors ${\bf X}$, then the only 
possibility is that $C=0$. [There is an alternate proof by 
diagonalizing $C$]. \item Show that if the two quadratic polynomials 
${\bf X\cdot}A{\bf X}$ and ${\bf X\cdot}B{\bf X}$ are equal for all 
values of {\bf X}, then $A=B$.   Thus a quadratic polynomial 
uniquely determines its associated symmetric matrix. \end{parts} 

\bigbreak


22.  Let $A$ be a positive definite matrix.
\begin{parts}
\item Show $A$ has a {\it square root\/},
that is, there is a positive definite symmetric matrix $P$ so that
$A=P^2$.  [Suggestion: First do the special case when $A$ is a
diagonal matrix.  For the general case, begin by diagonalizing $A$].
\item As an example, find the square root of
$
   \left(\!\! \begin{array}{cc}
               \phantom{-}4 & -1  \\
                         -1 & \phantom{-}4
 \end{array}\!\right)
$
\item Let $g({\bf X})={\bf X\cdot}A {\bf X}$.  Show that
if one makes the change of variable ${\bf Y}=P{\bf X}$, then $g$
takes the much simpler form
$g({\bf X})= \|{\bf Y}\|^2 = y_1^2 + y_2^2 + \cdots +y_n^2$.  This
change of variable is often valuable.  Problems 24d, 35, 36 and 37
all benefit from this observation, or variants of it.
\item  If $P$ is a positive definite square root of $A$
and $B$ is any other symmetric matrix, show that $A-B=P(I-C)P$,
where $C=P^{-1}BP^{-1}$  is a symmetric matrix.  This is usually
more convenient than writing $A-B=A(I-A^{-1}B)$ since $C$ is a
symmetric matrix while $A^{-1}B$ is probably not.
\end{parts}

\bigbreak

23. If $M$ is any invertible matrix, the point of this problem is to
show that there is a positive definite matrix $P$ and an orthogonal
matrix $R$ so that $M=PR$. This is analogous to writing a complex
number $z$ in the polar form $z=re^{i\theta}$. In the complex case,
one finds $r$ by the observation that $z\overline{z}= r^2$ and then
finds $e^{i\theta}$ using  $|z/r|=1$.  The same procedure is used
here.
\begin{parts}
\item Observe that if $M=RP$, then $MM^T=P^2$.  Since $MM^T$ is
positive definite (Problem 17), it has a square root $P$ (Problem 22).
This determines $P$.
\item Now that one knows $P$, define the matrix $R$ by
$R=MP^{-1}$.  Show that this $R$ is an orthogonal matrix.
\item Use an analogous procedure to find a positive
definite matrix $Q$ and an orthogonal matrix $S$ so that $M=SQ$.
\end{parts}


 \bigbreak

24. For parts a)-b), consider the ellipse $x^2 + y^2/4=1$ and the
lines $2x+y = c$.
\begin{parts}
 \item Use Maple to plot the graphs of the ellipse and the line (on 
the same plot) for various values of $c$, both positive and 
negative. 
\item  For which value(s) of $c$ does this line intersect
the ellipse in exactly one point?
\item Repeat parts a)-b). for the ellipsoid $x^2 + y^2/4
+ z^2=1$ and, on the same plot, the planes $2x+y +z = c$.
\item More generally, let $A$ be a positive definite
symmetric matrix and {\bf b} a given non-zero vector.  For which
value(s) of the constant $c$ does the ``plane'' ${\bf b\cdot X}=c$
intersect the ellipsoid ${\bf X\cdot}A{\bf X}=1$ in exactly one
point? [Suggestion: First do the case when $A=I$, then do the case
when $A$ is a diagonal matrix.  The answer is $c=\pm\sqrt{{\bf
b\cdot}A^{-1}{\bf b}}$.]
  \end{parts}

 \bigbreak


25. Let ${\bf V}= (v_1,\ldots,v_n)$ be a non-zero column vector  and
let $C$ be the matrix $C={\bf V}{\bf V}^T=(v_iv_j)$, so the $j^{th}$
column of $C$ is $v_j{\bf V}$.
\begin{parts} \item Show that $C$ is positive semi-definite.
\item Compute the inverse of $I + C$.
\item Find the eigenvalues and eigenvectors of $C$.
\item If $A$ is a symmetric matrix with ${\bf V}$ as an eigenvector,
find the eigenvalues and eigenvectors of $A+C$.  When is this matrix
positive definite?
\end{parts}

 \bigbreak


26. Show that both of the following  are positive semi-definite, but
is not positive definite.  Also find the eigenvalues of the
associated symmetric matrix, and, for part a), graph $w=q(x,y)$.
\begin{parts}
\item $q(x,y)=x^2 -6xy +  9y^2$.
\item $Q(x,y,z)=x^2 -6xy + 9y^2 + z^2$
\end{parts}

 \bigbreak


27. Show that a symmetric matrix is positive semi-definite if and only
if all of its eigenvalues satisfy $\lambda_k \ge 0$.  State and
prove the related assertion for a negative semi-definite matrix.

\bigbreak


28. Let $A$ be a symmetric matrix.
\begin{parts}
\item If $A$ is positive
semi-definite, show that its principle minors all have non-negative
determinant: $\det A_1\ge 0$, $\det A_2\ge 0$, etc.
\item Use the example  $ A =
   \left(\!\! \begin{array}{ccc}
               1 & 1 & 1  \\
               1 & 1 & 1  \\
               1 & 1 & 0
 \end{array}\!\right)
 $
to show that the converse of part a) is {\it false\/}.  Do this by
showing that $\det A_1>0$, $\det A_2 = \det A_3 = 0$ but that $A$ is 
not positive semi-definite.
 \end{parts} 

\bigbreak


29. Let $A=(a_{ij})$ be a symmetric matrix.
\begin{parts}
 \item If $a_{11}=0$ but $a_{12} \ne 0$,
show that $A$ is indefinite.
\item More generally, if some diagonal element of $A$ is
zero but some other element in the {\it same\/} column is not zero,
show that $A$ is indefinite.  This can be thought of as an extension
of {\sc Observation 3}.
 \end{parts}

\bigbreak


30. Some people define an {\it orthogonal matrix\/} $R$ by the
property $R^T = R^{-1}$ (transpose is the inverse).  The following
presents an different approach that is more fundamental.
\begin{parts}
 \item  If  $R$ is an orthogonal matrix,
show that for {\it any\/} vectors {\bf X} and {\bf Y}  one has
 $$
   R{\bf X}\cdot R{\bf Y}= {\bf X}\cdot {\bf Y}.
 $$
Thus, {\it orthogonal matrices preserve inner products\/} -- and
hence the angles between vectors.  As a special case, show that  for
all vectors ${\bf X}$ one has $\|R{\bf X}\|=\|{\bf X}\|$, that is, {\it
orthogonal matrices preserve the distance from the origin\/}.
\smallskip
\item Conversely, if for {\it any\/} vectors ${\bf X}$
and ${\bf Y}$ one has
 \begin{equation} \label{eq:orthog}
 R{\bf X}\cdot R{\bf Y}= {\bf X}\cdot {\bf Y},
 \end{equation}
show that $R^T = R^{-1}$ (transpose is the inverse).  Thus,
preserving inner products is equivalent to the other definition of
an orthogonal matrix.
 \smallskip
\item Use the property (\ref{eq:orthog}) directly to show
that of $R$ and $S$ are both $n\times n$ orthogonal matrices, then
so their product, $RS$.
 \end{parts}

\bigbreak


31. If a matrix $R$ has the property that for {\it any\/} vector
${\bf X}$ one has $\|R{\bf X}\|=\|{\bf X}\|$, show that $R^T = R^{-1}$
(transpose is the inverse).  Thus, preserving distance to the origin
is also equivalent to other definitions of an orthogonal matrix.
[Suggestion: use Problem 21a].
\bigbreak


32.  Let $F({\bf X})$ be a (not necessarily linear) map from $R^3$ 
(or $R^n$) to itself.  
 \begin{parts}
 \item If $F$ has the property that it preserves inner products (see
Problem 30. above):
 \begin{equation} \label{eq:FR}
    F({\bf X})\cdot F({\bf Y})= {\bf X}\cdot {\bf Y}.
 \end{equation}
show that in fact $F$ is linear:         
 $$
 F(c{\bf X})=cF({\bf X}) \hbox{\rm \ \ for any scalar\ } c, \quad
\hbox{\rm and}\quad 
F({\bf X}+{\bf Y})=F({\bf X})+F({\bf Y}) \hbox{\rm \ \ for any\ }{\bf X}
\hbox{\rm \ and any\ }{\bf Y}.
$$
Thus $F$ is an orthogonal mapping, $F({\bf X})=R{\bf X}$): 
[Suggestion: Show that $\|F({\bf X}+{\bf Y})-[F({\bf X})+F({\bf 
Y})]\|^2 = 0$, and $\|F(c{\bf X})-cF({\bf X})\|^2=0$]. 
 \item Show that if a $F$ both fixes the origin and preserves 
distance between points:
 $$
 F(0)=0 \quad\hbox{\rm and} \quad
        \|F({\bf X}) - F({\bf Y})\|= \|{\bf X}-{\bf Y}\|,
 $$
then $F$ has the property (\ref{eq:FR}). Hence it is an orthogonal mapping.
 \item Let $T({\bf X})$ be a {\it rigid mapping\/}, that is, it 
preserves the distance between any two points: 
 $$
 \|T({\bf X}) - T({\bf Y})\|= \|{\bf X}-{\bf Y}\|.
 $$
Show that $T$ is the sum of a translation by some vector ${\bf X}_0$ 
plus an orthogonal transformation, $R$:
 $$
 T(X) = {\bf X}_0  + R{\bf X}.
 $$
[Suggestion: Apply the preceding part to $F({\bf X}):=T({\bf X})-T(0)$.]
 \end{parts}

\bigbreak 


33.  Use both the eigenvalue and determinant tests to decide
if the following matrix is positive definite.
 $$
   \left(\!\! \begin{array}{ccccc}
    \phantom{-}3 &           -2 &  \phantom{-}3 &           -1 & 0 \\
              -2 & \phantom{-}6 &            -6 & \phantom{-}2 & 0 \\
    \phantom{-}3 &           -6 & \phantom{-}11 &           -3 & 0 \\
              -1 & \phantom{-}2 &            -3 & \phantom{-}3 & 0 \\
    \phantom{-}0 & \phantom{-}0 &  \phantom{-}0 & \phantom{-}0 & 2
 \end{array}\!\right)
 $$

\bigbreak


34.  One can think of the vector function ${\bf X}(t)$, depending
on a parameter $t$ as describing the position of a particle at time
$t$.  If
 $$
      {d{\bf X}(t)\over dt}= S(t){\bf X},
 $$
where $S(t)$ is a skew-symmetric matrix, show that the distance from
${\bf X}(t)$ to the origin is a constant, that is, ${\bf X}(t)$
moves on a sphere of fixed distance from the origin.  [Suggestion:
Problem 19 is useful here].

\bigbreak

35. This problem assumes you know how to make a change of variable
in a multiple integral by using the Jacobian determinant.  Let
${\bf X}= (x_1,\ldots,x_n)$.
\begin{parts}
\item Using the result from the special case $n=1$, Evaluate
 $$
  \int_{R^n} e^{-\|\hbox{\scrb X}\|^2}\,dx_1\ldots dx_n.
 $$
\item Let $A$ be a symmetric positive definite $n
\times n$ matrix and let
 $$
  Q = \int_{R^n} 
 e^{-\hbox{\scrb X}\cdot A\hbox{\scrb X}}\,dx_1\ldots dx_n.     
 $$
Use the change of variable in Problem 22c to show that
$Q=(\pi/\det A)^{n/2}$.
\item As an example, apply this to evaluate
 $$
   \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2 -2xy +3y^2)}\,dx\,dy.
 $$
\item Generalize part a) and evaluate
 $$
 \int_{R^n} e^{-(\hbox{\scrb X}\cdot A\hbox{\scrb X}+
 2\hbox{\scrb b}\cdot\hbox{\scrb X} +c)}\,dx_1\ldots dx_n,
 $$
where ${\bf b}$ is a specified vector and $c$ a scalar.
\end{parts}

 \bigbreak


36.  If $A=a_{ij}$ is a positive definite symmetric matrix and $B=(b_{ij})$ is
positive semi-definite symmetric (but not identically zero), show that
 $$
 \hbox{\rm trace}\,(AB) = \sum_{1,j=1}^n a_{ij}b_{ij} >0.
 $$
Here $\hbox{\rm trace}\,(A)$ is the sum of the diagonal elements of the square
matrix $A$. [Suggestion: First try the case where $A=I$ and $B$ is a
diagonal matrix. For the general case it will help to use the fact
that if $A$ and $C$ are similar matrices, then
$\hbox{\rm trace}\,(A)=\hbox{\rm trace}\,(C)$, and also that for any
matrices $C,\,D$ we have $\hbox{\rm trace}\,(CD)=
\hbox{\rm trace}\,(DC)$.]

 \bigbreak


37.  The following problem shows some of the ideas in this section 
at work in a related setting.
 \begin{parts}
 \item Let $u(x_1,\ldots,x_n)$ be a given function and say one makes 
the change of variable ${\bf Y}= S{\bf X}$, where $S=(s_{ij})$ is an 
invertible matrix.  Show that 
 $$
{\partial^2 u \over \partial x_i \partial x_j } = \sum_{k,\ell=1}^n 
s_{ki}s_{\ell j}{\partial^2 u\over \partial y_k \partial y_\ell}
 $$
\item If $A=(a_{ij})$ is a positive definite $2 \times 2$ matrix and
$u(x_1,\,x_2)$ is a given function, let $L$ be the partial
differential operator
 $$
 Lu :=  a_{11}{\partial^2 u\over \partial x_1^2}
 + 2a_{12}{\partial^2 u \over \partial x_1 \partial x_2}
 + a_{22}{\partial^2 u \over \partial x_2^2}.
 $$
Show that one can introduce new coordinates $(y_1,y_2)$ so that in
these new coordinates $L$ is the standard {\it Laplace operator\/}
 $$
 Lu = {\partial^2 u\over \partial y_1^2} + {\partial^2 u\over \partial y_2^2}.
 $$
[Suggestion: If $A=(a)$ is a $1\times 1$ matrix and $L$ is the
corresponding ordinary differential operator $Lu= au_{xx}$, then
under the change of variable $x=\sqrt{a}y$ we get $Lu= u_{yy}$.  This
leads one to try the change of variable ${\bf X}=\sqrt{A}{\bf Y}$, so
Problem 22 may help here.  From another view it may be useful to
think of $Lu:=\hbox{\rm trace}\,(Au'')$]
\end{parts}

 } % end \small
\end{document}

