Demonstrate that the matrix
$$
\mathrm{A}=\left(\begin{array}{ccc}
2 & 0 & 0 \\
-6 & 4 & 4 \\
3 & -1 & 0
\end{array}\right)
$$
is defective, i.e. does not have three linearly independent eigenvectors, by showing the following:
(a) its eigenvalues are degenerate and, in fact, all equal;
(b) any eigenvector has the form $\left(\begin{array}{lll}\mu & (3 \mu-2 v) & v\end{array}\right)^{\mathrm{T}}$.
(c) if two pairs of values, $\mu_{1}, v_{1}$ and $\mu_{2}, v_{2}$, define two independent eigenvectors $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ then any third similarly defined eigenvector $\mathrm{v}_{3}$ can be written as a linear combination of $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$, i.e.
$$
\mathbf{v}_{3}=a \mathbf{v}_{1}+b \mathbf{v}_{2}
$$
where
$$
a=\frac{\mu_{3} v_{2}-\mu_{2} v_{3}}{\mu_{1} v_{2}-\mu_{2} v_{1}} \quad \text { and } \quad b=\frac{\mu_{1} v_{3}-\mu_{3} v_{1}}{\mu_{1} v_{2}-\mu_{2} v_{1}}
$$
Illustrate (c) using the example $\left(\mu_{1}, v_{1}\right)=(1,1),\left(\mu_{2}, v_{2}\right)=(1,2)$ and $\left(\mu_{3}, v_{3}\right)=$ $(0,1)$
Show further that any matrix of the form
$$
\left(\begin{array}{ccc}
2 & 0 & 0 \\
6 n-6 & 4-2 n & 4-4 n \\
3-3 n & n-1 & 2 n
\end{array}\right)
$$
is defective, with the same eigenvalues and eigenvectors as $A$.