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Exercises 19–23 concern the polynomial and an
Chapter , Problem 23E(choose chapter or problem)
Exercises 19–23 concern the polynomial
\(p(t)=a_{0}+a_{1} t+\cdots+a_{n-1} t^{n-1}+t^{n}\)
and an \(n \times n\) matrix \(C_p\) called the companion matrix of p:
\(C_{p}=\left[\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & & 0 \\ \vdots & & & & \vdots \\ 0 & 0 & 0 & & 1 \\ -a_{0} & -a_{1} & -a_{2} & \cdots & -a_{n-1} \end{array}\right]\)
Let p be the polynomial in Exercise 22, and suppose the equation p(t) = 0 has distinct roots \(\lambda_{1}, \lambda_{2}, \lambda_{3}\). Let V be the Vandermonde matrix
\(V=\left[\begin{array}{ccc} 1 & 1 & 1 \\ \lambda_{1} & \lambda_{2} & \lambda_{3} \\ \lambda_{1} & \lambda_{2} & \lambda_{3} \end{array}\right]\)
(The transpose of V was considered in Supplementary Exercise 11 in Chapter 2.) Use Exercise 22 and a theorem from this chapter to deduce that V is invertible (but do not compute \(V ^{-1}\)). Then explain why \(V^{-1}C_pV\) is a diagonal matrix.
Questions & Answers
QUESTION:
Exercises 19–23 concern the polynomial
\(p(t)=a_{0}+a_{1} t+\cdots+a_{n-1} t^{n-1}+t^{n}\)
and an \(n \times n\) matrix \(C_p\) called the companion matrix of p:
\(C_{p}=\left[\begin{array}{ccccc} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & & 0 \\ \vdots & & & & \vdots \\ 0 & 0 & 0 & & 1 \\ -a_{0} & -a_{1} & -a_{2} & \cdots & -a_{n-1} \end{array}\right]\)
Let p be the polynomial in Exercise 22, and suppose the equation p(t) = 0 has distinct roots \(\lambda_{1}, \lambda_{2}, \lambda_{3}\). Let V be the Vandermonde matrix
\(V=\left[\begin{array}{ccc} 1 & 1 & 1 \\ \lambda_{1} & \lambda_{2} & \lambda_{3} \\ \lambda_{1} & \lambda_{2} & \lambda_{3} \end{array}\right]\)
(The transpose of V was considered in Supplementary Exercise 11 in Chapter 2.) Use Exercise 22 and a theorem from this chapter to deduce that V is invertible (but do not compute \(V ^{-1}\)). Then explain why \(V^{-1}C_pV\) is a diagonal matrix.
ANSWER:Solution 23E
So, by the definition of an eigenvector, the vector (1,,
) is an eigenvector of
corresponding to the eigenvalue
Thus, all columns of the matrix