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 \(p\): \({C_p} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right]\).

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

1. Write the companion matrix for \(p\).
2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

Short answer

1. The companion matrix for \(p\)is\({C_p} = \left[ {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right]\).
2. As \(\lambda \) is a zero of the matrix, this implies that \( - {a_0} - {a_1}\lambda - {a_2}{\lambda ^2} = {\lambda ^3}\). It is proved that \(\left( {1,\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix \(p\).

Step-by-step solution

Step 1: Write the companion matrix for \(p\)

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\({C_p} = \left[ {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right]\).

Thus, the companion matrix for \(p\)is\({C_p} = \left[ {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right]\).

Step 2: Show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\)

As \(\lambda \) is a zero of the given polynomial \(p\left( t \right)\) then we have,

\(\begin{aligned}{c}{a_0} + {a_1}\lambda + {a_2}{\lambda ^2} + {\lambda ^3} = 0\\ - {a_0} - {a_1}\lambda - {a_2}{\lambda ^2} = {\lambda ^3}\end{aligned}\)

Now check whether \(\left( {1,\lambda ,{\lambda ^2}} \right)\) is an eigenvector of \({C_p}\).

\(\begin{aligned}{c}{C_p} &= \left[ {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{*{20}{c}}\lambda \\{{\lambda ^2}}\\{ - {a_0} - {a_1}\lambda - {a_2}{\lambda ^2}}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{*{20}{c}}\lambda \\{{\lambda ^2}}\\{{\lambda ^3}}\end{aligned}} \right]\\ &= \lambda \left[ {\begin{aligned}{*{20}{c}}1\\\lambda \\{{\lambda ^2}}\end{aligned}} \right]\end{aligned}\)

Thus, \(\left( {1,\lambda ,{\lambda ^2}} \right)\) is an eigenvector of \({C_p}\).

Hence it is proved that \(\left( {1,\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix \(p\).