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Chapter 5: Q5.3-1E (page 267)

Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)

Short Answer

Expert verified

The required value is \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

Step by step solution

01

Write the Diagonalization Theorem

The Diagonalization Theorem: An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

02

Find the inverse of the invertible matrix

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\)and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

\[\begin{array}{P^{ - 1}} = \frac{1}{{5 \times 3 - 7 \times 2}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{{15 - 14}}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \frac{1}{1}\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\end{array}\]

03

Find \({{\bf{4}}^{{\bf{th}}}}\)the power of the diagonal matrix

\[\begin{array}{l}D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{D^2} = \left( {\begin{array}{*{20}{c}}{{2^2}}&0\\0&{{1^2}}\end{array}} \right)\\{D^3} = \left( {\begin{array}{*{20}{c}}{{2^3}}&0\\0&{{1^3}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{{2^4}}&0\\0&{{1^4}}\end{array}} \right)\\{D^4} = \left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\end{array}\]

04

Find \({A^{\bf{4}}}\)

Consider the matrix \(P = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\)and \(D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\).

As it is given that \(A = PD{P^{ - 1}}\)than by using the formula for \({n^{th}}\) power we get:

\[{A^n} = P{D^n}{P^{ - 1}}\].

Then we get:

\[\begin{array}{A^4} = P{D^4}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}5&7\\2&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}{16}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{80}&7\\{32}&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 7}\\{ - 2}&5\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\end{array}\]

Thus, \({A^4} = \left( {\begin{array}{*{20}{c}}{226}&{ - 525}\\{90}&{ - 209}\end{array}} \right)\).

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Most popular questions from this chapter

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

9. \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

Exercises 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 \(p\): \({C_p} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right]\).

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

  1. Write the companion matrix for \(p\).
  2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

For the matrices Afind real closed formulas for the trajectoryx→(t+1)=Ax→(t)wherex→(0)=[01]. Draw a rough sketchA=[1-31.2-2.6]

A particle moving in a planar force field has a position vector .\(x\). that satisfies \(x' = Ax\). The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and 2, with corresponding eigenvectors \({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right)\). Find the position of the particle at a time \(t\), assuming that \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).
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