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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: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

7. \(\left[ {\begin{array}{*{20}{c}}5&3\\- 4&4\end{array}} \right]\)

19鈥�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.

For the matrix A, find real closed formulas for the trajectory x鈫�(t+1)=Ax炉(t) where x鈫�=[01]. Draw a rough sketchA=[7-156-11]

Show that if \({\bf{x}}\) is an eigenvector of the matrix product \(AB\) and \(B{\rm{x}} \ne 0\), then \(B{\rm{x}}\) is an eigenvector of\(BA\).

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.
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