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-1 27 "Experiment of power method:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(LinearAlgebra):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "A := RandomMatrix(4,4,genera
tor=-5..5,outputoptions=[shape=symmetric])*1.0;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"AG-%'RTABLEG6%\"*g;&o9-%'MATRIXG6#7&7&$!\"&\"\"!$!
\"%F0F.$F0F07&F1$\"\"\"F0$!\"#F0F37&F.F7F5$!\"\"F07&F3F3F:F7%'MatrixG
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{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*%e*fY\"-%'MATRIXG6#7&7#
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"" {TEXT -1 74 "We expect to find eigenvalue -9.72... and the eigenvec
tor by power method " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 331 "x \+
:= RandomVector(4):\neps := 10.0^(-6):\nfor k from 1 to 20 do\n   y :=
 A.x:\n   x := y/Norm(y,2):\n   beta[k] := HermitianTranspose(x).A.x/N
orm(x,2):\n   if k > 1 then\n      print(beta[k], abs(beta[k]-beta[k-1
])):\n      if abs(beta[k]-beta[k-1]) < eps then\n         print( beta
[k], x):\n         break:\n      end if\n   end if\nend do:   " }}
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{PARA 11 "" 1 "" {XPPMATH 20 "6$$!+?'\\Fs*!\"*-%'RTABLEG6%\"*w'zz9-%'M
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{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "So we now have an approxi
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0 "" {TEXT -1 86 "Next question:  If we know there is an eigenvalue ne
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}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "(Q,R) := QRDecomposition( A
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x := RandomVector(4):\nfor k from 1 to 20 do\n   y := BackwardSubstitu
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x/Norm(x,2) ):\nend do:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+;vK*)R!
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