Question

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

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

Short answer

All the columns of \(V\) are linearly independent and \(V\) is invertible. Therefore, the matrix \({C_p}\) is diagonalizable and \({V^{ - 1}}{C_p}V\) is a diagonal matrix.

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 general companion matrix is\({C_p} = \left( {\begin{aligned}{*{20}{c}}0&1&0\\0&0&1\\{ - {a_0}}&{ - {a_1}}&{ - {a_2}}\end{aligned}} \right)\).

As \(p\left( t \right) = {a_0} + {a_1}t + {a_2}{t^2} + {t^3}\) and \({\lambda _1},{\lambda _2},{\lambda _3}\) are the distinct roots of the equation \(p\left( t \right) = 0\).

Consider \(V = \left( {\begin{aligned}{*{20}{c}}1&1&1\\{{\lambda _1}}&{{\lambda _2}}&{{\lambda _3}}\\{\lambda _1^2}&{\lambda _2^2}&{\lambda _3^2}\end{aligned}} \right)\).

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

As \(\lambda \) is a zero of \(p\left( t \right)\), then we get,

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

Step 2: Check whether \(\left( {{\bf{1}},\lambda ,{\lambda ^{\bf{2}}}} \right)\) 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 \({C_p}\).

Since all eigenvalues \({\lambda _1},{\lambda _2},{\lambda _3}\) are distinct the eigenvectors of \({C_p}\) are linearly independent.

Thus, all the columns of\(V\)are linearly independent and\(V\)is invertible. Therefore, the matrix\({C_p}\)is diagonalizable and\({V^{ - 1}}{C_p}V\) is a diagonal matrix.