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QUESTION


Consider the matrices

$$A=\left(\begin{array}{ccc}1&3&0\\3&1&4\\0&4&1\end{array}\right)\
\
B=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right).$$

\begin{description}

\item[(i)]
Compute the following matrices and their determinants:

$$A+B,AB,BA.$$

\item[(ii)]
Find the eigenvalues and the corresponding normalised eigenvectors
for the matrix $A$.

\item[(iii)]
Write down the quadratic form associated with $A$ and express it
in diagonal form. Say giving brief reasons for your answer,
whether or not this quadratic form represents an ellipsoid.

\end{description}



ANSWER

\begin{description}

\item[(i)]

$$A=\left(\begin{array}{ccc}1&3&0\\3&1&4\\0&4&1\end{array}\right)\
\
B=\left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right).$$

$$A+B=\left(\begin{array}{ccc}1&4&0\\4&1&5\\0&5&1\end{array}\right),\
AB=\left(\begin{array}{ccc}3&1&3\\1&7&1\\4&1&4\end{array}\right),\
BA=\left(\begin{array}{ccc}3&1&4\\1&7&1\\3&1&4\end{array}\right)$$

$\det(A+B)=-40,\ \det(AB)=\det(BA)=0$

\item[(ii)]
The eigenvalues of $A$ satisfy
$(1-\lambda)\left[(1-\lambda)(1-\lambda)-16\right]-3\left[3(1-\lambda)-0\right]+0=0$
which factorises as $(1-\lambda)\left[(1-\lambda)^2-25\right]=0$
so $\lambda=1$ or $1-\lambda=\pm5$ with roots $\lambda=-4,1,6$.

For the eigenvalues:

$\lambda=-4$ gives
$\left(\begin{array}{ccc}5&3&0\\3&5&4\\0&4&5\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$ so $5x=-3y$ and
$5z=-4y$.

This gives a normalised eigenvector
$\frac{1}{\sqrt{50}}\left(\begin{array}{c}-3\\5\\-4\end{array}\right)$

$\lambda=1$ gives
$\left(\begin{array}{ccc}0&3&0\\3&0&4\\0&4&0\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$ so $y=0$ and
$3x=-4z$.

This gives a normalised eigenvector
$\frac{1}{5}\left(\begin{array}{c}4\\0\\-3\end{array}\right)$

$\lambda=-4$ gives
$\left(\begin{array}{ccc}-5&3&0\\3&-5&4\\0&4&-5\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$ so $5x=3y$ and
$5z=4y$.

This gives a normalised eigenvector
$\frac{1}{\sqrt{50}}\left(\begin{array}{c}3\\5\\4\end{array}\right)$

\item[(iii)]
The quadratic form associated to $A$ is $x^2+y^2+z^2+6xy+8yz$
which does not represent an ellipsoid since it diagonalises to
$X^2-4Y^2+6Z^2$ which has both negative and positive coefficients.

\end{description}




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