Question

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

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

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}\)

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}\)

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}}\)