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Chapter 5: Q5.5-20E (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.

20. \(\left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\)

Short Answer

Expert verified

The invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right),\;{\rm{and}}\,P = \left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\)

Step by step solution

01

Finding the matrix \(P\) and the matrix  \(C\)

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

Given that \(A = \left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\).

The characteristic polynomial is \({\lambda ^2} - .56\lambda + 1\). Hence the eigenvalues of \(A\) are \(\lambda = - .28 \pm .96i\).

To find an eigenvector corresponding to \( - .28 - .96i\) we compute,

\(A - \left( { - .28 - .96i} \right)I = \left( {\begin{aligned}{}{1.92 + .96i}&{}&{ - 2.4}\\{1.92}&{}&{ - 1.92 + .96i}\end{aligned}} \right)\)

The equation \(\left( {A - \left( { - .28 - .96i} \right)I} \right)x = 0\) has a system of equations,

\(\begin{aligned}{}\left( {1.92 + .96i} \right){x_1} - 2.4{x_2} &= 0\\1.92{x_1} + ( - 1.92 + .96i){x_2} &= 0\end{aligned}\)

Hence \({x_1} = \frac{{ - 2 - i}}{2}{x_2}\) , \({x_2}\)is a free variable.

03

Find the matrix further

Thus, the eigen vector corresponds to the eigenvalue\( - .28 - .96i\).

\(v = \left( {\begin{aligned}{}{ - 2 - i}\\2\end{aligned}} \right)\)

By Theorem 9,

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

And

\(\begin{aligned}{}C &= {P^{ - 1}}AP\\ &= \frac{1}{2}\left( {\begin{aligned}{}0&{}&1\\{ - 2}&{}&{ - 2}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 1.64}&{}&{ - 2.4}\\{1.92}&{}&{2.2}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right)\end{aligned}\)

Thus, the invertible matrix and matrix are \(C = \left( {\begin{aligned}{}{.28}&{}&{ - .96}\\{.96}&{}&{.28}\end{aligned}} \right),\;{\rm{and}}\;P = \left( {\begin{aligned}{}{ - 2}&{}&{ - 1}\\2&{}&0\end{aligned}} \right)\).

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Most popular questions from this chapter

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)

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

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)

Let\(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula\(\left( {\bf{1}} \right)\)for the determinant in section\({\bf{5}}{\bf{.2}}\)to explain why\(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of\(G\)is the product of the characteristic polynomials of\(A\)and\(B\).
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