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Chapter 2: Q2.5-26Q (page 93)

26. (Spectral Factorization) Suppose a \({\bf{3}} \times {\bf{3}}\) matrix Aadmits a factorization as \(A = PD{P^{ - 1}}\), where \(P\)is some invertible \({\bf{3}} \times {\bf{3}}\) matrix and D is the diagonal matrix \(D = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{{{\bf{1}} \mathord{\left/

{\vphantom {{\bf{1}} {\bf{2}}}} \right.

\kern-\nulldelimiterspace} {\bf{2}}}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{{{\bf{1}} \mathord{\left/

{\vphantom {{\bf{1}} {\bf{3}}}} \right.

\kern-\nulldelimiterspace} {\bf{3}}}}\end{aligned}} \right)\)

Show that this factorization is useful when computing high powers of A. Find fairly simple formulas for \({A^{\bf{2}}}\),\({A^{\bf{3}}}\), and \({A^k}\)(k a positive integer), using P and entries in D.

Short Answer

Expert verified

The formulas for \({A^2},{A^3},\) and \({A^k}\) are \({A^2} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 4}} \right.

\kern-\nulldelimiterspace} 4}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 9}} \right.

\kern-\nulldelimiterspace} 9}}\end{aligned}} \right){P^{ - 1}}\), \({A^3} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 8}} \right.

\kern-\nulldelimiterspace} 8}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 {27}}} \right.

\kern-\nulldelimiterspace} {27}}}\end{aligned}} \right){P^{ - 1}}\), and \({A^k} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^k}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^k}}\end{aligned}} \right){P^{ - 1}}\), respectively.

Step by step solution

01

Compute \({A^{\bf{2}}}\)

\(\begin{aligned}{c}{A^2} = AA\\ = \left( {PD{P^{ - 1}}} \right)\left( {PD{P^{ - 1}}} \right)\\ = PD\left( {{P^{ - 1}}P} \right)D{P^{ - 1}}\\ = PDID{P^{ - 1}}\\{A^2} = P{D^2}{P^{ - 1}}\end{aligned}\)

First, compute \({D^2}\):

\(\begin{aligned}{c}{D^2} = DD\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)}^2}}&0\\0&0&{{{\left( {{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}} \right)}^2}}\end{aligned}} \right)\\{D^2} = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^2}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^2}}\end{aligned}} \right)\end{aligned}\)

Then,

\(\begin{aligned}{c}{A^2} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^2}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^2}}\end{aligned}} \right){P^{ - 1}}\\ = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 4}} \right.

\kern-\nulldelimiterspace} 4}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 9}} \right.

\kern-\nulldelimiterspace} 9}}\end{aligned}} \right){P^{ - 1}}\end{aligned}\)

02

Compute \({A^{\bf{3}}}\)

\(\begin{aligned}{c}{A^3} = {A^2}A\\ = \left( {P{D^2}{P^{ - 1}}} \right)\left( {PD{P^{ - 1}}} \right)\\ = = P{D^2}\left( {{P^{ - 1}}P} \right)D{P^{ - 1}}\\ = P{D^2}ID{P^{ - 1}}\\{A^3} = P{D^3}{P^{ - 1}}\end{aligned}\)

First, Compute \({D^3}\):

\(\begin{aligned}{c}{D^3} = {D^2}D\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 {{2^2}}}} \right.

\kern-\nulldelimiterspace} {{2^2}}}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 {{3^2}}}} \right.

\kern-\nulldelimiterspace} {{3^2}}}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{\left( {{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}} \right)}^3}}&0\\0&0&{{{\left( {{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}} \right)}^3}}\end{aligned}} \right)\\{D^3} = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^3}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^3}}\end{aligned}} \right)\end{aligned}\)

Then,

\(\begin{aligned}{c}{A^3} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^3}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^3}}\end{aligned}} \right){P^{ - 1}}\\ = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{1 \mathord{\left/

{\vphantom {1 8}} \right.

\kern-\nulldelimiterspace} 8}}&0\\0&0&{{1 \mathord{\left/

{\vphantom {1 {27}}} \right.

\kern-\nulldelimiterspace} {27}}}\end{aligned}} \right){P^{ - 1}}\end{aligned}\)

03

Compute \({A^k}\)

In general, \({A^k} = P{D^k}{P^{ - 1}}\) with \({D^k} = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^k}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^k}}\end{aligned}} \right)\). That is

\({A^k} = P\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{{{{1 \mathord{\left/

{\vphantom {1 2}} \right.

\kern-\nulldelimiterspace} 2}}^k}}&0\\0&0&{{{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}^k}}\end{aligned}} \right){P^{ - 1}}\)

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Most popular questions from this chapter

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?
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