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{\bf Question}

Find the eigenvalues and normalised eigenvectors for each of the
following matrices.  In each case, write down an orthogonal matrix
$R$ such that $R^TAR$ is a diagonal matrix (you should verify this
by calculating $R^TAR$):

 $${\rm (i)}\  A = \left( \begin{array}{rr} 2 & {-4} \\ {-4} & 8 \\
\end{array} \right);\
{\rm (ii)}\  B = \left(\begin{array}{rr} 4 & 5 \\ 5 & 4
 \\ \end{array} \right);\
{\rm (iii)}\  C = \left( \begin{array}{rrr} 5 & 3 & 0 \\ 3 & 5 & 0
\\ 0 & 0 & 4 \end{array} \right). $$

{\bf Answer} \vspace{.2in}
\begin{description}
\item[(i)]
\begin{eqnarray*}
 \left| \begin{array}{cc} {2 - \lambda} & {-4} \\ {-4} & {8 - \lambda} \\
\end{array} \right| & = & (2 - \lambda)(8 - \lambda) - 16 \\ & = &
\lambda^2 - 10\lambda \\ & = & \lambda(\lambda - 10) = 0 \\ {\rm
so \ \ \ } \lambda & = & 0, 10
\end{eqnarray*}

$ \underline {\lambda = 0}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} 2 & {-4}\\ {-4} & 8\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $\left. \begin{array}{rcl} {2x - 4y} & = & 0 \\ {-4x + 8y} & =
& 0 \\ \end{array} \right\} \begin{array}{lrcl} {\rm let\ \ } & y
& = & \alpha \\ {\rm so\ \ } & x & = & 2\alpha \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} 2\alpha \\ \alpha
\\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{2}{\sqrt5} \\
\frac{1}{\sqrt5} \\ \end{array} \right)$

$ \underline {\lambda = 10}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} -8 & {-4}\\ {-4} & -2\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $\left. \begin{array}{rcl} {-8x - 4y} & = & 0 \\ {-4x - 2y} & =
& 0 \\ \end{array} \right\} \begin{array}{lrcl} {\rm let\ \ } & x
& = & \beta \\ {\rm so\ \ } & y & = & -2\beta \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \beta \\ -2\beta
\\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{1}{\sqrt5} \\
\frac{-2}{\sqrt5} \\ \end{array} \right)$

Take the orthogonal matrix $R = \left( \begin{array}{cc}
\frac{2}{\sqrt5} & \frac{1}{\sqrt5} \\ \frac{1}{\sqrt5} &
\frac{-2}{\sqrt5} \\ \end{array} \right)$ with $R^T = \left(
\begin{array}{cc} \frac{2}{\sqrt5} & \frac{1}{\sqrt5} \\
\frac{1}{\sqrt5} & \frac{-2}{\sqrt5} \\ \end{array} \right)$

[Note: check that the eigenvectors are orthogonal using the dot
product: $ \left( \begin{array}{c} \frac{2}{\sqrt5} \\
\frac{1}{\sqrt5}  \\ \end{array} \right)\cdot  \left(
\begin{array}{c}  \frac{1}{\sqrt5} \\  \frac{-2}{\sqrt5} \\ \end{array}
\right) = \frac{2}{5} - \frac{2}{5} = 0$]

Then \begin{eqnarray*} {\rm AR}  & = & \left( \begin{array}{rr} 2
& {-4}\\ {-4} & 8\\ \end{array} \right) \left( \begin{array}{cc}
\frac{2}{\sqrt5} & \frac{1}{\sqrt5} \\ \frac{1}{\sqrt5} &
\frac{-2}{\sqrt5} \\ \end{array} \right) = \left(
\begin{array}{cc} 0 & \frac{10}{\sqrt5} \\
0 & \frac{-20}{\sqrt5} \\ \end{array} \right)\\ {\rm R^TAR}  & = &
\left( \begin{array}{cc} \frac{2}{\sqrt5} & \frac{1}{\sqrt5} \\
\frac{1}{\sqrt5} & \frac{-2}{\sqrt5} \\ \end{array} \right) \left(
\begin{array}{cc} 0 & \frac{10}{\sqrt5} \\
0 & \frac{-20}{\sqrt5} \\ \end{array} \right) = \left(
\begin{array}{cc} 0 & 0 \\
0 & 10 \\ \end{array} \right)\\ \hspace{.2in}{\rm as\ required}
\end{eqnarray*}


\item[(ii)]
\begin{eqnarray*}
 \left| \begin{array}{cc} {4 - \lambda} & {5} \\ {5} & {4 - \lambda} \\
\end{array} \right| & = & (4 - \lambda)^2 - 25 \\ & = &
\lambda^2 - 8\lambda - 9 \\ & = & (\lambda - 9)(\lambda + 1) = 0
\\ {\rm so \ \ \ } \lambda & = & -1, 9
\end{eqnarray*}

$ \underline {\lambda = -1}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} 5 & {5}\\ {5} & 5\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $ \begin{array}{rcl} {5x + 5y} & = & 0 \\ \end{array}
\begin{array}{lrcl} {\rm let\ \ } & x & = & \alpha \\ {\rm so\ \ }
& y & = & -\alpha \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \alpha \\ -\alpha
\\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{1}{\sqrt2} \\
\frac{-1}{\sqrt2} \\ \end{array} \right)$

$ \underline {\lambda = 9}$ \hspace{.2in} Solve $\left(
\begin{array}{rr} -5 & {5}\\ {5} & -5\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

or $\left. \begin{array}{rcl} {-5x + 5y} & = & 0 \\ {5x - 5y} & =
& 0 \\ \end{array} \right\} \begin{array}{lrcl} {\rm let\ \ } & x
& = & \beta \\ {\rm so\ \ } & y & = & \beta \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \beta \\ \beta
\\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{1}{\sqrt2} \\
\frac{1}{\sqrt2} \\ \end{array} \right)$

Take the orthogonal matrix $R = \left( \begin{array}{cc}
\frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} &
\frac{1}{\sqrt2} \\ \end{array} \right)$ with $R^T = \left(
\begin{array}{cc} \frac{1}{\sqrt2} & \frac{-1}{\sqrt2} \\
\frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\ \end{array} \right)$

Then \begin{eqnarray*} {\rm AR}  & = & \left( \begin{array}{rr} 4
& 5\\ 5 & 4\\ \end{array} \right) \left( \begin{array}{cc}
\frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} &
\frac{1}{\sqrt2} \\ \end{array} \right) = \left(
\begin{array}{cc} \frac{-1}{\sqrt2} & \frac{9}{\sqrt2} \\
\frac{1}{\sqrt2} & \frac{9}{\sqrt2} \\ \end{array} \right)\\ {\rm
R^TAR} & = & \left( \begin{array}{cc} \frac{1}{\sqrt2} &
\frac{-1}{\sqrt2} \\ \frac{1}{\sqrt2} & \frac{1}{\sqrt2} \\
\end{array} \right) \left( \begin{array}{cc} \frac{-1}{\sqrt2} & \frac{9}{\sqrt2} \\
\frac{1}{\sqrt2} & \frac{9}{\sqrt2} \\ \end{array} \right) =
\left(
\begin{array}{cc} -1 & 0 \\
0 & 9 \\ \end{array} \right)\\ \hspace{.2in}{\rm as\ required}
\end{eqnarray*}

\item[(iii)]
\begin{eqnarray*}
 \left| \begin{array}{ccc} {5 - \lambda} & {3} & 0 \\ {3} & {5 -
 \lambda}& 0  \\0 & 0 & {4 - \lambda} \\ \end{array} \right|& = & (5 - \lambda) \left|
 \begin{array}{cc} {5 - \lambda} & 0 \\ 0 & {4 - \lambda} \\
\end{array} \right| - 3 \left| \begin{array}{cc} 3 & 0 \\ 0 & {4 - \lambda} \\
\end{array} \right|  +0 \\ & = & (5 - \lambda)^2(4 - \lambda) - 9(4 - \lambda) \\ & = &
\ (\lambda - 4)(\lambda^2 - 10\lambda +16) \\ & = & (\lambda -
4)(\lambda - 2)(\lambda - 8) = 0
\\ {\rm so \ \ \ } \lambda & = & 2, 4, 8
\end{eqnarray*}

$ \underline {\lambda = 2}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} 3 & 3 & 0 \\ 3 & 3 & 0 \\ 0 & 0 & 2\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {3x + 3y} & = & 0 \\ 2z & = & 0
\\ \end{array} \right\}
\begin{array}{lrcl} {} & z & = & 0 \\ {\rm let\ \ } & x & = & \alpha \\ {\rm so\ \ } & y &
= & -\alpha \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \alpha \\ -\alpha
\\0 \\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{1}{\sqrt2} \\
\frac{-1}{\sqrt2} \\ 0 \\ \end{array} \right)$

$ \underline {\lambda = 4}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} 1 & 3 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 0\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rclr} {x + 3y} & = & 0 &{(1)} \\ {3x +y
} & = & 0 & {(2)} \\ \end{array} \right\}$ (2) - 3 (1) gives $-8y
= 0 \Rightarrow y = 0 \Rightarrow x = 0$. Let $z = \beta$

Suitable eigenvector $\left( \begin{array}{c} 0 \\ 0 \\ \beta
\\ \end{array} \right)$ which normalises to $\left(
\begin{array}{c} 0 \\ 0 \\ 1 \\
\end{array} \right)$

$ \underline {\lambda = 8}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} -3 & 3 & 0 \\ 3 & -3 & 0 \\ 0 & 0 & -4\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {-3x + 3y} & = & 0 \\ {3x - 3y} & =
& 0 \\ 2z & = & 0 \\ \end{array} \right\}
\begin{array}{lrcl}{} & z & = & 0 \\ {\rm let\ \ } & x & = & \gamma \\ {\rm so\ \ }
& y & = & \gamma \\ \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \gamma \\ \gamma
\\0 \\ \end{array} \right)$ which normalises to $\left( \begin{array}{c} \frac{1}{\sqrt2} \\
\frac{1}{\sqrt2} \\ 0 \\ \end{array} \right)$

Take the orthogonal matrix $R = \left( \begin{array}{ccc}
\frac{1}{\sqrt2} & 0 & \frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} & 0 &
\frac{1}{\sqrt2} \\ 0 & 1 & 0 \\
\end{array} \right)$

with $R^T = \left( \begin{array}{ccc} \frac{1}{\sqrt2} &
\frac{-1}{\sqrt2} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt2} &
\frac{1}{\sqrt2} & 0 \\ \end{array} \right)$

Then \begin{eqnarray*} {\rm AR}  & = & \left( \begin{array}{rrr}
{5 } & {3} & 0 \\ {3} & {5} & 0 \\0 & 0 & {4} \\ \end{array}
\right) \left( \begin{array}{ccc} \frac{1}{\sqrt2} & 0 &
\frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} & 0 & \frac{1}{\sqrt2} \\ 0
& 1 & 0 \\ \end{array} \right) = \left(
\begin{array}{ccc} \frac{2}{\sqrt2} & 0 & \frac{8}{\sqrt2} \\
\frac{-2}{\sqrt2} & 0 & \frac{8}{\sqrt2} \\ 0 & 4 & 0 \\
\end{array} \right)\\ {\rm R^TAR} & = & \left(
\begin{array}{ccc} \frac{1}{\sqrt2} & \frac{-1}{\sqrt2} & 0 \\ 0 & 0 & 1 \\
\frac{1}{\sqrt2} & \frac{1}{\sqrt2} & 0 \\ \end{array} \right)
\left( \begin{array}{ccc} \frac{2}{\sqrt2} & 0 & \frac{8}{\sqrt2}
\\ \frac{-2}{\sqrt2} & 0 & \frac{8}{\sqrt2} \\ 0 & 4 & 0 \\
\end{array}\right) = \left( \begin{array}{ccc} 2 & 0 & 0 \\
0 & 4 & 0 \\ 0 & 0 & 8 \\ \end{array} \right)\\ \hspace{.2in}{\rm
as\ required}
\end{eqnarray*}
\end{description}

\end{document}
