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Chapter 5: Q5.5-14E (page 267)

In Exercises 13-20, find an invertible matrix \(P\) and a matrix \(C\) of the form \(\left( {\begin{aligned}{}a&{}&{ - b}\\b&{}&a\end{aligned}} \right)\) such that the given matrix has the form\(A = PC{P^{ - 1}}\). For Exercises 13-16, use information from Exercises 1-4.

14. \(\left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&1\end{aligned}} \right)\)

Short Answer

Expert verified

The invertible matrix \(P\)and matrix \(C\) are \(P = \left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right),C = \left( {\begin{aligned}{}3&{}&{ - 1}\\1&{}&3\end{aligned}} \right)\).

Step by step solution

01

 Finding the matrix \(P\)  and the matrix \(C\) such that \(A = PC{P^{ - 1}}\)  

Let \(A\) be a \(2 \times 2\) with eigenvalues of the form \(a \pm bi\) , and let \(v\) be the eigenvector corresponding to the eigenvalue \(a - bi\), then the matrix \(P\) will be \(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right)\).

The matrix \(C\) can be obtained by \(C = {P^{ - 1}}AP\).

02

Find the Invertible matrix \(P\) and \(C\) 

\(A = \left( {\begin{aligned}{}5&{}&{ - 5}\\1&{}&1\end{aligned}} \right)\).

From Exercise 2, the eigenvalues of A is\(\lambda = 3 \pm i\), and the eigenvector corresponds to\(\lambda = 3 - i\)is \(v = \left( {\begin{aligned}{}{2 - i}\\1\end{aligned}} \right)\).

By Theorem 9,\(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} v}&{{\mathop{\rm Im}\nolimits} v}\end{aligned}} \right) \Rightarrow P = \left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\).

This implies that,\({P^{ - 1}} = \left( {\begin{aligned}{}0&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\).

Substitute the value of\({P^{ - 1}}\), \(P\) and \(A\) into \(C = {P^{ - 1}}AP\) to obtain matrix \(C\).

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\C &= \left( {\begin{aligned}{}0&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\left( {\begin{aligned}{}5&{}&{ - 5}\\1&1\end{aligned}} \right)\left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\\C &= \left( {\begin{aligned}{}3&{}&{ - 1}\\1&{}&3\end{aligned}} \right)\end{aligned}\)

Hence, the\(P\)and\(C\)are \(P = \left( {\begin{aligned}{}2&{}&{ - 1}\\1&{}&0\end{aligned}} \right)\)and \(C = \left( {\begin{aligned}{}3&{}&{ - 1}\\1&{}&3\end{aligned}} \right)\) respectively.

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Most popular questions from this chapter

Question: A is a \({\bf{4}} \times {\bf{4}}\) matrix with three eigenvalues. One eigenspace is one-dimensional and one of the other eigenspaces is two-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

In Exercises 9鈥�16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

The trace of a square matrix \(A\) is the sum of the diagonal entries in A and is denoted by \({\mathop{\rm tr}\nolimits} A\). It can be verified that \({\mathop{\rm tr}\nolimits} \left( {FG} \right) = {\mathop{\rm tr}\nolimits} \left( {GF} \right)\) for any \(n \times n\) matrices Fand G. Show that if A and B are similar, then \({\mathop{\rm tr}\nolimits} A = {\mathop{\rm tr}\nolimits} B\).

Let \(A\) be a real \(2 \times 2\) matrix with a complex eigenvalue \(\lambda = a - bi\)\(\left( {b \ne 0} \right)\) and an associated eigenvector \({\bf{v}}\) in \({\mathbb{}^2}\).

  1. Show that \(A({\mathop{\rm Re}\nolimits} {\bf{v}}) = a{\mathop{\rm Re}\nolimits} {\bf{v}} + b{\mathop{\rm Im}\nolimits} {\bf{v}}\) and \(A({\mathop{\rm Im}\nolimits} {\bf{v}}) = - b{\mathop{\rm Re}\nolimits} {\bf{v}} + a{\mathop{\rm Im}\nolimits} {\bf{v}}\). (Hint: Write \({\bf{v}} = {\mathop{\rm Re}\nolimits} {\bf{v}} + i{\mathop{\rm Im}\nolimits} {\bf{v}}\), and compute \(A{\bf{v}}\).)
  2. Verify that if \(P\) and \(C\) are given as in Theorem 9, then \(AP = PC\)

Question: Consider an \(n \times n\) matrix A with the property that the row sums all equal the same number s. Show that s is an eigenvalue of A. (Hint: Find an eigenvector.)
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