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{\bf Question}

Given that

$$\left( \begin{array}{ccccc}
     a & b & 0 & 0 & 0\\
     b & a & b & 0 & 0\\
     0 & b & a & b & 0\\
     0 & 0 & b & a & b\\
     0 & 0 & 0 & b & a
 \end{array}  \right)$$
and

$$ {\bf c_1} = \left( \begin{array}{rrrrr}
     1\\
     \sqrt 3\\
     2\\
     \sqrt 3\\
     1
 \end{array}  \right),\ \
{\bf c_2} =\left( \begin{array}{rrrrr}
     1\\
     -\sqrt 3\\
     2\\
     -\sqrt 3\\
     1
 \end{array}  \right),\ \
{\bf c_3} =\left( \begin{array}{rrrrr}
     1\\
     1\\
     0\\
     -1\\
     -1
 \end{array}  \right),\ \
{\bf c_4} =\left( \begin{array}{rrrrr}
     1\\
     -1\\
     0\\
     1\\
     -1
 \end{array}  \right),\ \
{\bf c_4} =\left( \begin{array}{rrrrr}
     1\\
     0\\
     -1\\
     0\\
     1
 \end{array}  \right),$$
verify that ${\bf c_1}, {\bf c_2}, {\bf c_3}, {\bf c_4}, {\bf
c_5}$ are all eigenvectors of $A$. Determine the eigenvalues
corresponding to these eigenvectors.


{\bf Answer} $$\left(\begin{array}{ccccc}a & b & 0 & 0 & 0\\b & a
& b & 0 & 0\\0 & b & a & b & 0\\0 & 0 & b & a & b\\0 & 0 & 0 & b &
a\end{array}\right)
\left(\begin{array}{r}1\\\sqrt3\\2\\\sqrt3\\1\end{array}\right)=
\left(\begin{array}{c}a+\sqrt{3}b\\3b+\sqrt{3}a\\2a+2\sqrt{3}b\\3b+\sqrt{3}a\\
\sqrt{3}b+a\end{array}\right)=a+\sqrt{3}b\left(\begin{array}{r}1\\\sqrt{3}\\2\\
\sqrt{3}\\1\end{array}\right)$$

eigenvalue = $a+\sqrt{3}b$
\bigskip
$$\left(\begin{array}{ccccc}a & b & 0 & 0 & 0\\b & a & b & 0 &
0\\0 & b & a & b & 0\\0 & 0 & b & a & b\\0 & 0 & 0 & b &
a\end{array}\right)
\left(\begin{array}{r}1\\-\sqrt3\\2\\-\sqrt3\\1\end{array}\right)=
\left(\begin{array}{c}a-\sqrt{3}b\\3b-\sqrt{3}a\\2a-2\sqrt{3}b\\3b-\sqrt{3}a\\
a-\sqrt{3}b\end{array}\right)=a-\sqrt{3}b\left(\begin{array}{r}1\\-\sqrt{3}\\2\\
-\sqrt{3}\\1\end{array}\right)$$

eigenvalue = $a-\sqrt{3}b$
\bigskip
$$\left(\begin{array}{ccccc}a & b & 0 & 0 & 0\\b & a & b & 0 &
0\\0 & b & a & b & 0\\0 & 0 & b & a & b\\0 & 0 & 0 & b &
a\end{array}\right)
\left(\begin{array}{r}1\\1\\0\\-1\\-1\end{array}\right)=
\left(\begin{array}{c}a+b\\a+b\\0\\-(a+b)\\
-(a+b)\end{array}\right)=(a+b)\left(\begin{array}{r}1\\1\\0\\
-1\\-1\end{array}\right)$$

eigenvalue = $a+b$
\bigskip
$$\left(\begin{array}{ccccc}a & b & 0 & 0 & 0\\b & a & b & 0 &
0\\0 & b & a & b & 0\\0 & 0 & b & a & b\\0 & 0 & 0 & b &
a\end{array}\right)
\left(\begin{array}{r}1\\-1\\0\\1\\-1\end{array}\right)=
\left(\begin{array}{c}a-b\\b-a\\0\\a-b\\
b-a\end{array}\right)=(a+b)\left(\begin{array}{r}1\\-1\\0\\
1\\-1\end{array}\right)$$

eigenvalue = $a-b$
\bigskip
$$\left(\begin{array}{ccccc}a & b & 0 & 0 & 0\\b & a & b & 0 &
0\\0 & b & a & b & 0\\0 & 0 & b & a & b\\0 & 0 & 0 & b &
a\end{array}\right)
\left(\begin{array}{r}1\\0\\-1\\0\\1\end{array}\right)=
\left(\begin{array}{c}a\\0\\-a\\0\\
a\end{array}\right)=a\left(\begin{array}{r}1\\0\\-1\\
0\\1\end{array}\right)$$

eigenvalue = $a$



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