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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

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 \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).

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 \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

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

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: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)
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