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QUESTION


The matrices $A$ and $B$ are defined to be

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

\begin{description}

\item[(a)]
Compute the determinants of each of the matrices $A,B,a+B,AB$.

\item[(b)]
Find the eigenvalues for the matrix $A$ (you are NOT required to
find the eigenvectors). [Hint: Use row reduction to simplify the
computation.]

\item[(c)]
The matrix $B$ has three distinct eigenvalues. Compute them and
find, and normalise the corresponding eigenvectors. Verify that
these vectors form an orthonormal system.

\end{description}



ANSWER


\begin{description}

\item[(a)]
$$\det(A)=4\left|\begin{array}{cc}4&1\\1&4\end{array}\right|
-\left|\begin{array}{cc}1&1\\1&4\end{array}\right|+\left|
\begin{array}{cc}1&4\\1&1\end{array}\right|=4.15-4+1+1-4=54$$

det$(B)=-1\det(A)=-54$ since $B$ is obtained by switching rows 1
and 3.

det$(A+B)=\left|\begin{array}{ccc}5&2&5\\2&8&2\\5&2&5\end{array}\right|=0$
since row 1=row 3

det$(AB)=\det(A)\det(B)=-(54)^2=-2916$

\item[(b)]
The eigenvalues satisfy $|A-\lambda I|=0$

\begin{eqnarray*}
\left|\begin{array}{ccc}4-\lambda&1&1\\ 1&4-\lambda&1\\
1&1&4-\lambda\end{array}\right|&=&\left|
\begin{array}{ccc}6-\lambda&6-\lambda&6-\lambda\\1&4-\lambda&1\\
1&1&4-\lambda\end{array}\right|\\
&=&(6-\lambda)\left|\begin{array}{ccc}1&1&1\\1&4-\lambda&1\\1&1&4-
\lambda\end{array}\right|\\
&=&(6-\lambda)\left|\begin{array}{ccc}0&\lambda-3&0\\1&4-
\lambda&1\\1&1&4-\lambda\end{array}\right|\\
&=&(6-\lambda)(3-\lambda)(3-\lambda)
\end{eqnarray*}

$\lambda=3,3,6$.

\item[(c)]
\begin{eqnarray*}
|A-\lambda I|&=&\left|\begin{array}{ccc}1-\lambda&1&4\\1&4-
\lambda&1\\4&1&1-\lambda\end{array}\right|\\
&=&(6-\lambda)\left|\begin{array}{ccc}1&1&1\\1&4-\lambda&1\\4&1&1-
\lambda\end{array}\right|\\
&=&(6-\lambda)\left|\begin{array}{ccc}0&\lambda-3&0\\
1&4-\lambda&1\\4&1&1-\lambda\end{array}\right|\\
&=&(6-\lambda)(\lambda-3)(-3-\lambda)
\end{eqnarray*}

so $\lambda=6,3,-3$.

For the eigenvectors,

$\lambda=-3$

$$\left(\begin{array}{ccc}4&1&4\\1&7&1\\4&1&4\end{array}
\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$$

$$\left.\begin{array}{r}4x+y+4z\\
x+7y+z=0\end{array}\right\}\begin{array}{l}44+y+4z=0
\\4x+28y+4z=0\end{array}$$

$27y=0\ y=0,\ x=-z$ so putting $x=1$,
$\left(\begin{array}{c}1\\0\\-1\end{array}\right)$ which
normalises to $\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\0\\-1
\end{array}\right)=\mathbf{v}_1$

$\lambda=3$

$$\left(\begin{array}{ccc}-2&1&4\\1&1&1\\4&1&-2\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}0\\0\\0\end{array}\right)$$

$$\left.\begin{array}{r}-2x+y+4z=0\\x+y+z=0\end{array}
\right\}\left.\begin{array}{l}-2x+y+4z=0\\-2x-2y-2z=0
\end{array}\right\}3y+6z=0$$

so putting $z=1,y=-2\Rightarrow x=1,\
\left(\begin{array}{c}1\\-2\\1\end{array}\right)$ which normalises
to $\frac{1}{\sqrt{6}}\left(\begin{array}{c}1\\-2\\1
\end{array}\right)=\mathbf{v}_2$

$\lambda=6$

$$\left(\begin{array}{ccc}-5&1&4\\1&-2&1\\4&1&-5
\end{array}\right)\left(\begin{array}{c}x\\y\\z
\end{array}\right)=\left(\begin{array}{c}0\\0\\0 \end{array}\right)$$

$$\left.\begin{array}{r}-5x+y+4z=0\\x-2y+z\end{array}\right\}
\left.\begin{array}{r}-5x+y+4z\\-5x+10y-5z\end{array}\right\}0x-9y+9z$$

$y=z$ so putting $y=1,z=1\Rightarrow x=1,\
\left(\begin{array}{c}1\\1\\1\end{array}\right)$ which normalises
to
$\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\1\\1\end{array}\right)
=\mathbf{v}_3$

\begin{eqnarray*}
\mathbf{v}_1.\mathbf{v}_2&=&\frac{1}{\sqrt{12}}(1.1+0+1.-1)=0\\
\mathbf{v}_1\mathbf{v}_3&=&\frac{1}{\sqrt{6}}(1.1+0.1+-1.1)=0\\
\mathbf{v}_2.\mathbf{v}_3&=&\frac{1}{\sqrt{18}}(1.1-2.1+1.1)
\end{eqnarray*}

So the vectors do form an orthonormal system.



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