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Q21SE

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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)

It is proved that for \(n = k + 1\), \(\det \left( {{C_p} - \lambda I} \right) = {\left( { - 1} \right)^{k + 1}}p\left( \lambda \right)\).

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Step by Step Solution

Step 1: Find the companion matrix

Consider the polynomial \(p\left( t \right) = {a_0} + {a_1}t + ... + {a_{n - 1}}{t^{n - 1}} + {t^n}\).

The companion matrix of \(p\) is \({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1&0&{...}&0\\0&0&1&{}&0\\:&{}&{}&{}&:\\0&0&0&{}&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}&{...}&{ - {a_{n - 1}}}\end{aligned}} \right)\).

Apply mathematical induction to prove as shown below:

\(\begin{aligned}{c}\det \left( {{C_p} - \lambda I} \right) &= {\left( { - 1} \right)^n}\left( {{a_0} + {a_1}\lambda + ... + {a_{n - 1}}{\lambda ^{n - 1}} + {\lambda ^n}} \right)\\ &= {\left( { - 1} \right)^n}p\left( \lambda \right)\end{aligned}\)

Step 2: Apply mathematical induction to find whether it is true for \(n{\bf{ = 2}}\)

For \(n = 2\) the matrix is shown below:

\({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1\\{ - {a_0}}&{ - {a_1}}\end{aligned}} \right)\)

Now find the determinant.

\(\begin{aligned}{c}\det \left( {{C_p} - \lambda I} \right) &= \left( { - \lambda } \right)\left( { - {a_1} - \lambda } \right) + {a_0}\\ &= {a_0} + {a_1}\lambda + {\lambda ^2}\\ &= {\left( { - 1} \right)^2}\left( {{a_0} + {a_1}\lambda + {\lambda ^2}} \right)\end{aligned}\)

Thus, the result is true for \(n = 2\).

Step 3: Apply mathematical induction to find whether it is true for \(n = k\)

Assume the result is true for \(n = k\).

\(\det \left( {{C_p} - \lambda I} \right) = {\left( { - 1} \right)^k}q\left( \lambda \right)\)

Now check the result for \(n = k + 1\).

Now find the determinant.

\(\begin{aligned}{c}\det \left( {{C_p} - \lambda I} \right) &= \det \left( {\begin{aligned}{*{20}{c}}{ - \lambda }&1&0&{...}&0\\0&{ - \lambda }&1&{}&0\\:&{}&{}&{}&:\\0&0&0&{}&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}&{...}&{ - {a_k} - \lambda }\end{aligned}} \right)\\ &= \left( { - \lambda } \right)\det + \left( {\begin{aligned}{*{20}{c}}{ - \lambda }&1&{...}&0\\:&:&:&:\\0&{...}&{...}&1\\{ - {a_1}}&{ - {a_2}}&{...}&{ - {a_k} - \lambda }\end{aligned}} \right) + 0... + {\left( { - 1} \right)^{k + 1}}{a_0}et\left( {\begin{aligned}{*{20}{c}}1&0&{...}&0\\0&1&:&:\\:&{...}&{...}&1\\0&0&{...}&1\end{aligned}} \right)\\ &= \left( { - \lambda } \right){\left( { - 1} \right)^k}q\left( \lambda \right) + {\left( { - 1} \right)^{k + 1}}{a_0}\\ &= \left( \lambda \right)\left( { - 1} \right){\left( { - 1} \right)^k}\left( {{a_1} + ... + {a_k}{\lambda ^{k - 1}} + {\lambda ^k}} \right) + {\left( { - 1} \right)^{k + 1}}{a_0}\\ &= {\left( { - 1} \right)^{k + 1}}p\left( \lambda \right)\end{aligned}\)

Thus, the result is true for \(n \ge 2\).

Hence, it is proved that for \(n = k + 1\), \(\det \left( {{C_p} - \lambda I} \right) = {\left( { - 1} \right)^{k + 1}}p\left( \lambda \right)\).

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