Question

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).)

Short answer

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

Step-by-step solution

Step 1: Find the companion matrix

Considerthe 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 \(n = k + 1\), \(\det \left( {{C_p} - \lambda I} \right) = {\left( { - 1} \right)^{k + 1}}p\left( \lambda \right)\).