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\bf{Question}

\quad For each of the following quadratic forms $Q$ on $\br^n$
find the nature of the critical point at the origin. Describe also
the geometry of the level set $Q^{-1}(1)$:
\medskip

\begin{tabular}{lll}
$n=2;$& $\quad Q(x,y)=$\\ &(i)\ $x^2+y^2-xy$ \hspace{1cm}&(ii)\
$9x^2-12xy+4y^2$\\ $n=3;$&$\quad Q(x,y,z)=$\\ &(iii)\
$x^2+y^2+z^2+xy+yz+zx$\hspace{.5cm} &(iv)\ $x^2+y^2+z^2-xy$ \\
&(v)\ $2x^2+y^2-4xy-4yz$ &(vi)\ $3x^2+y^2+z^2-2xy+2xz-2yz$
\end{tabular}

\medskip

In each case (i) -- (vi) say whether the addition of higher order
terms to $Q$ could change the nature of the level sets close to
the origin.



\bf{Answer}


\begin{description}
\item{(i)}
$x^TAx$ with $A= \left ( \begin{array}{cc} 1 & -frac{1}{2} \\
-\frac{1}{2} & 1 \end{array} \right )$.

Eigenvalues
\begin{eqnarray*}
(1-\lambda)^2 - \frac{1}{4} & = & 0\\ \Rightarrow
\lambda^2-2\lambda+ \frac{3}{4} & = & 0\\ \Rightarrow \lambda_1
\lambda_2 & = & \textrm{det}A = \frac{3}{4}>0\\ \lambda_1 +
\lambda_2 & = & 2 >0
\end{eqnarray*}

So as $\lambda_1$, $\lambda_2$ both $>0$: minimum ($Q^{-1}(1)$:
ellipse).

[* $Q=(x_1-\frac{1}{2}x_2)^2+\frac{3}{4}x_2^2$, hence minimum.]

\item{(ii)}

$A = \left ( \begin{array}{cc} 9 & -6\\ -6 & 4 \end{array} \right
)$.

Eigenvalues $0$, $13$: ``trough'' ($Q^{-1}(1)$: pair of lines).

[* $Q=(3x-2y)^2$.]

\item{(iii)}

$A = \left ( \begin{array}{ccc} 1 & \frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & 1 & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & 1
\end{array} \right )$.

Eigenvalues
\begin{eqnarray*}
& & (1- \lambda)((1-\lambda)^2 -\frac{1}{4}) -
\frac{1}{2}(\frac{1}{2}( 1-\lambda) - \frac{1}{4} ) +
\frac{1}{2}(\frac{1}{4} - \frac{1}{2} (1-\lambda))\\ & = &
(1-\lambda)(\lambda^2-2\lambda+\frac{3}{4}) - \frac{1}{2}
(\frac{1}{4} - \frac{1}{2}\lambda) + \frac{1}{2}(-\frac{1}{4} +
\frac{1}{2} \lambda)\\ & = & -\lambda^3 +
3\lambda^2-\frac{9}{4}\lambda+\frac{1}{2}\\ & = & -\frac{1}{4}
(4\lambda^3 - 12\lambda^2 + 9\lambda -2)\\ & = & 0
\end{eqnarray*}
Clearly $\lambda=\frac{1}{2}$ is an eigenvalue, so
\begin{eqnarray*}
(2\lambda-1)(2\lambda^2-5\lambda+2) & = & 0\\
(2\lambda-1)^2(\lambda-2) & = & 0
\end{eqnarray*}

Thus eigenvalues are $\frac{1}{2}$, $\frac{1}{2}$, $2$: all
positive.

$\Rightarrow$ minimum ($Q^{-1}(1)$: ellipsoid).

\item{(iv)}

$A= \left ( \begin{array}{ccc} 1 7 -\frac{1}{2} & 0\\ -\frac{1}{2}
& 1 & 0 \\ 0 & 0 & 1 \end{array} \right )$

Eigenvalues $\lambda=1$ and two other positives (see (i) above)
$\Rightarrow$ minimum

[* $Q = (x_1-\frac{1}{2}x_2)^2 + \frac{3}{4} x_2^2 + x_3^2]
\leftarrow$ ($Q^{-1}(1)$: ellipsoid.]

\item{(v)}

$A = \left ( \begin{array}{ccc} 2 & -2 & 0\\ -2 & 1 & -2\\ 0 & -2
& 0
\end{array} \right )$.

Eigenvalues
\begin{eqnarray*}
(2-\lambda)((1-\lambda)(-\lambda)-4) + 2(2\lambda) & = & 0\\
(2-\lambda) (\lambda^2-\lambda-4) + 4\lambda & = & 0\\ -\lambda^3
+ 3\lambda^2 + 6\lambda -8 & = & 0 \ : \ \lambda =1\\ -(\lambda
-1)(\lambda^2-2\lambda -8) & = & 0
\end{eqnarray*}
The quadratic equation $\lambda^2 -2\lambda -8=0$ has roots
$\lambda =-2,4$, $\Rightarrow$ saddle ($Q^{-1}(1)$: hyperboloid (1
sheet)).

\item{(vi)}

$A = \left ( \begin{array}{ccc} 3 & -1 & 1\\ -1 & 1 & -1\\ 1 & -1
& 1
\end{array} \right )$

$\lambda =0,1,4$. Degenerate critical point (line of zeros).
($Q^{-1}(1)$: elliptical tube).

\end{description}
(* indicates short cuts)

[Addition of higher order terms can affect conclusion only when
$\det A=0$ ($\exists$ zero eigenvalue), namely cases (ii), (vi).]


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