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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}

Find the eigenvalues and eigenvectors of the following matrices
\begin{description}
\item[(i)]
$A=\left(\begin{array}{cc} 3 & 0\\ 0 & -4 \end{array}\right)$

\item[(ii)]
$B=\left(\begin{array}{cc} 1 & 2\\ 0 & 1 \end{array}\right)$

\item[(iii)]
$C=\left(\begin{array}{cc} 2 & 4\\ 3 & 6 \end{array}\right)$
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$A=\left(\begin{array}{cc} 3 & 0\\ 0 & -4 \end{array}\right)$

$A{\bf{x}}=\lambda{\bf{x}} \Rightarrow (A-\lambda)x=0 \Rightarrow
det(A-\lambda I_2)=0$

i.e., $\left|\begin{array}{cc} 3-\lambda & 0\\ 0 & -4-\lambda
\end{array}\right|=-(3-\lambda)(4+\lambda)=0$

Therefore \un{$\lambda=3\ \rm{or}\ -4$} Eigenvalues.

\un{Eigenvectors}:

\un{$\lambda=3$}

\begin{eqnarray*} & & \left(\begin{array}{cc} 3-3 & 0\\ 0 & -4-3
\end{array}\right)\left(\begin{array}{c}x\\ y \end{array}\right)=
\left(\begin{array}{c}0\\ 0\end{array}\right)\\ & \Rightarrow &
\left(\begin{array}{cc} 0 & 0\\ 0 & -7
\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right)=
\left(\begin{array}{c}0\\0\end{array}\right)\\ & \Rightarrow &
0x+0y=0\\ & & 0x-7y=0\\ & \Rightarrow & y=0 \end{eqnarray*}

Hence if $y=0$ and $0x=0$, $x$ can be anything, say, $k$, so
eigenvector is $\left(\begin{array} {c} k\\ 0 \end{array}\right)$
for eigenvalue =3.

\newpage
\un{$\lambda=-4$}

\begin{eqnarray*} & & \left(\begin{array}{cc} 3-(-4) & 0\\ 0 &
-4-4 \end{array}\right)\left(\begin{array}{c}x\\ y
\end{array}\right)= \left(\begin{array}{c}0\\
0\end{array}\right)\\ & \Rightarrow & \left(\begin{array}{cc} 7 &
0\\ 0 & -8 \end{array} \right)\left(\begin{array}{c}x\\ y
\end{array}\right)= \left(\begin{array}{c}0\\
0\end{array}\right) \end{eqnarray*}

$\Rightarrow \left.\begin{array}{r} 7x=0\\ -8y=0 \end{array}
\right\} \Rightarrow x=y=0$

so eigenvector is $\left(\begin{array}{c}0\\0 \end{array}\right)$
for eigenvalues=-4.

\item[(ii)]
$B=\left(\begin{array}{cc} 1 & 2\\0 & 1 \end{array} \right)$

Eigenvalues and eigenvectors determined from
$B{\bf{x}}=\lambda{\bf{x}}$

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

Eigenvalues:  \begin{eqnarray*} & & det\left(\begin{array}{cc}
1-\lambda & 2\\ 0 & 1-\lambda \end{array}\right)=0\\ & \Rightarrow
& (1-\lambda)(1-\lambda)=0\\ & \Rightarrow & \lambda=+1
\un\rm{twice} \end{eqnarray*}

Eigenvectors: $$\left(\begin{array}{cc} 0 & 2\\ 0 & 0 \end{array}
\right)\left(\begin{array}{c}x\\y \end{array}
\right)=\left(\begin{array}{c}0\\0 \end{array} \right)$$

$$ \Rightarrow \begin{array} {l} 2x=0 \Rightarrow x=0\\ 0y=0
\Rightarrow y \end{array}\ \rm{where}\ y\ \rm{can\ be\ anything}$$

Therefore \un{eigenvectors} are $\left(\begin{array}{c} 0\\k
\end{array} \right)$

\newpage
\item[(iii)]
$C=\left(\begin{array}{cc} 2 & 4\\3 & 6 \end{array} \right)$

Eigenvalues and eigenvectors determined from
$C{\bf{x}}=\lambda{\bf{x}}$

$$\left(\begin{array}{cc} 2-\lambda & 4\\ 3 & 6-\lambda
\end{array}\right)\left(\begin{array}{c}x\\ y
\end{array}\right)= \left(\begin{array}{c}0\\
0\end{array}\right)$$

Eigenvalues:  \begin{eqnarray*} & & det\left(\begin{array}{cc}
2-\lambda & 4\\ 3 & 6-\lambda \end{array}\right)=0\\ & \Rightarrow
& (2-\lambda)(6-\lambda)-12=0\\ & &
\lambda^2+12-6\lambda-2\lambda-12 =0\\ & \Rightarrow &
\lambda^2=8\lambda\\ \Rightarrow \un{\lambda=0\ \rm{or}\ 8}
\end{eqnarray*}

So eigenvectors are:

\un{$\lambda=0$}

$$ \left(\begin{array}{cc} 2 & 4\\ 3 & 6
\end{array}\right)\left(\begin{array}{c}x\\ y
\end{array}\right)= \left(\begin{array}{c}0\\
0\end{array}\right)$$

$$ \Rightarrow \begin{array}{l} 2x+4y=0 \Rightarrow x=-2y\\
3x+6y=0 \Rightarrow x=-2y \end{array} $$

so $\left(\begin{array}{c} -2\lambda\\ \lambda \end{array}
\right)$ is the eigenvector for $\lambda=0$.

\un{$\lambda=8$}

$$ \left(\begin{array}{cc} -6 & 4\\ 3 & -2
\end{array}\right)\left(\begin{array}{c}x\\ y
\end{array}\right)= \left(\begin{array}{c}0\\
0\end{array}\right)$$

$$ \Rightarrow \begin{array}{l} -6x+4y=0 \Rightarrow
y=\ds\frac{3x}{2}\\ 3x-2y=0 \Rightarrow y=\ds\frac{3x}{2}
\end{array} $$

so $\left(\begin{array}{c} \lambda\\ \frac{3\lambda}{2}
\end{array} \right)$ is the eigenvector for $\lambda=8$.

\end{description}
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