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Chapter 5: Q5.3-2E (page 267)

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

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

Short Answer

Expert verified

The required value is \({A^4} = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)\).

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

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

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

Compute \({P^{ - 1}}\).

\[\begin{array}{P^{ - 1}} = \frac{1}{{2 \times 5 - \left( { - 3} \right) \times \left( { - 3} \right)}}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \frac{1}{{10 - 9}}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \frac{1}{1}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\end{array}\]

03

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

\[\begin{array}{A^4} = P{D^4}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right){\left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{2}}\end{array}} \right)^4}\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{{\left( 1 \right)}^4}}&0\\0&{{{\left( {\frac{1}{2}} \right)}^4}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 3}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&{\frac{1}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{2 + 0}&{0 - \frac{3}{{16}}}\\{ - 3 + 0}&{0 + \frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}2&{ - \frac{3}{{16}}}\\{ - 3}&{\frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\end{array}\]

Further,

\[\begin{array}{c}{A^4} = \left( {\begin{array}{*{20}{c}}2&{ - \frac{3}{{16}}}\\{ - 3}&{\frac{5}{{16}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&3\\3&2\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{10 - \frac{9}{{16}}}&{6 - \frac{6}{{16}}}\\{ - 15 + \frac{{15}}{{16}}}&{ - 9 + \frac{{10}}{{16}}}\end{array}} \right)\\ = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)\end{array}\]

Thus, \({A^4} = \frac{1}{{16}}\left( {\begin{array}{*{20}{c}}{151}&{90}\\{ - 225}&{ - 134}\end{array}} \right)\).

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

11. \(\left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 2}\\{ - 3}&4&0\\{ - 3}&1&3\end{array}} \right)\)

Question 20: Use a property of determinants to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

Use mathematical induction to show that if \(\lambda \) is an eigenvalue of an \(n \times n\) matrix \(A\), with a corresponding eigenvector, then, for each positive integer \(m\), \({\lambda ^m}\)is an eigenvalue of \({A^m}\), with \({\rm{x}}\) a corresponding eigenvector.

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

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