Suggested languages for you:

Americas

Europe

Q23SE

Expert-verified
Found in: Page 267

### 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

# 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 matrixV{\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.

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.

See the step by step solution

## Step 1: Write the companion matrix for $$p$$

Consider the 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.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.