<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{amsmath}
\begin{document}

{\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 $R^TAR$):

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

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

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

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

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 $\left(
\begin{array}{rr} -8 &amp; {-4}\\ {-4} &amp; -2\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

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

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 $R = \left( \begin{array}{cc}
\frac{2}{\sqrt5} &amp; \frac{1}{\sqrt5} \\ \frac{1}{\sqrt5} &amp;
\frac{-2}{\sqrt5} \\ \end{array} \right)$ with $R^T = \left(
\begin{array}{cc} \frac{2}{\sqrt5} &amp; \frac{1}{\sqrt5} \\
\frac{1}{\sqrt5} &amp; \frac{-2}{\sqrt5} \\ \end{array} \right)$

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


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

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

or $ \begin{array}{rcl} {5x + 5y} &amp; = &amp; 0 \\ \end{array}
\begin{array}{lrcl} {\rm let\ \ } &amp; x &amp; = &amp; \alpha \\ {\rm so\ \ }
&amp; y &amp; = &amp; -\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 $\left(
\begin{array}{rr} -5 &amp; {5}\\ {5} &amp; -5\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\ \end{array} \right)$

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

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

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

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

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

or $ \left. \begin{array}{rcl} {3x + 3y} &amp; = &amp; 0 \\ 2z &amp; = &amp; 0
\\ \end{array} \right\}
\begin{array}{lrcl} {} &amp; z &amp; = &amp; 0 \\ {\rm let\ \ } &amp; x &amp; = &amp; \alpha \\ {\rm so\ \ } &amp; y &amp;
= &amp; -\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 $\left(
\begin{array}{rrr} 1 &amp; 3 &amp; 0 \\ 3 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rclr} {x + 3y} &amp; = &amp; 0 &amp;{(1)} \\ {3x +y
} &amp; = &amp; 0 &amp; {(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 $\left(
\begin{array}{rrr} -3 &amp; 3 &amp; 0 \\ 3 &amp; -3 &amp; 0 \\ 0 &amp; 0 &amp; -4\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {-3x + 3y} &amp; = &amp; 0 \\ {3x - 3y} &amp; =
&amp; 0 \\ 2z &amp; = &amp; 0 \\ \end{array} \right\}
\begin{array}{lrcl}{} &amp; z &amp; = &amp; 0 \\ {\rm let\ \ } &amp; x &amp; = &amp; \gamma \\ {\rm so\ \ }
&amp; y &amp; = &amp; \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 $R = \left( \begin{array}{ccc}
\frac{1}{\sqrt2} &amp; 0 &amp; \frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} &amp; 0 &amp;
\frac{1}{\sqrt2} \\ 0 &amp; 1 &amp; 0 \\
\end{array} \right)$

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

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

\end{document}
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