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Exercise 5.4.19

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As Hint, we induct on $k$ . For $k = 1$ , the matrix is $(\begin{matrix} - a_{0} \end{matrix})$ and the characteristic polynomial is $- (a_{0} + t)$ . If the hypothesis holds for the case $k = m - 1$ , we may the expand the matrix by the first row and calculate the characteristic polynomial to be

\det (A - tI) = \det (\begin{matrix} - t & 0 & \dots & - a_{0} \\ 1 & - t & \dots & - a_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & - a_{m - 1} \end{matrix})

= - t \det (\begin{matrix} - t & 0 & \dots & - a_{0} \\ 1 & - t & \dots & - a_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & - a_{m - 2} \end{matrix}) + {(- 1)}^{m + 1} (- a_{0}) \det (\begin{matrix} 1 & - t & \dots & 0 \\ 0 & 1 & \dots & - a_{1} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & 1 \end{matrix})

= - t [{(- 1)}^{m - 1} (a_{1} + a_{2} t + \dots + a_{m - 1} t^{m - 2} + t^{m - 1})] + {(- 1)}^{m} a_{0}

= {(- 1)}^{m} (a_{0} + a_{1} t + \dots + a_{m - 1} t^{m - 1} + t^{m}) .

jephianlin

2011-06-27 00:00

Exercise 5.4.19

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