printbr(A:eq(eig.R * eig.Lambda * eig.L)) printbr('permuting to', A:eq(eig.R * eig.Lambda * eig.L))
local A = Matrix({-6, 3}, {4, 5}) local ch = A:charpoly(lambda) printbr(ch) local solns = table{ch:solve(lambda)} printbr(solns:mapi(tostring):concat',') simplifyAssertEq(ch, (lambda^2 + lambda - 42):eq(0)) local eig = A:eigen() if Constant.isValue(eig.Lambda[1][1], 6) then local P = Matrix({0,1},{1,0}) eig.R = (eig.R * P)() eig.Lambda = (P * eig.Lambda * P)() eig.L = (P * eig.L)() end simplifyAssertEq(eig.L, Matrix({-frac(4,13), frac(1,13)}, {frac(4, 13), frac(12,13)})) simplifyAssertEq(eig.R, Matrix({-3, frac(1,4)}, {1, 1})) simplifyAssertEq(eig.Lambda, Matrix.diagonal(-7, 6)) 		${-{\left({{12} + {{{\left({{5}{-{\lambda}}}\right)}} {{\left({{6} + {\lambda}}\right)}}}}\right)}} = {0}$
${\lambda} = {6}$,${\lambda} = {-{7}}$
${{-{\left({{12} + {{{\left({{5}{-{\lambda}}}\right)}} {{\left({{6} + {\lambda}}\right)}}}}\right)}} = {0}} = {{{{{\lambda}^{2}} + {\lambda}}{-{42}}} = {0}}$
${\left[\begin{array}{cc} -{\frac{4}{13}}& \frac{1}{13}\\ \frac{4}{13}& \frac{12}{13}\end{array}\right]} = {\left[\begin{array}{cc} -{\frac{4}{13}}& \frac{1}{13}\\ \frac{4}{13}& \frac{12}{13}\end{array}\right]}$
${\left[\begin{array}{cc} -{3}& \frac{1}{4}\\ 1& 1\end{array}\right]} = {\left[\begin{array}{cc} -3& \frac{1}{4}\\ 1& 1\end{array}\right]}$
${\left[\begin{array}{cc} -{7}& 0\\ 0& 6\end{array}\right]} = {\left[\begin{array}{cc} -7& 0\\ 0& 6\end{array}\right]}$
GOOD 		time: 63.789000ms
stack: size: 7
  • Init
  • Prune
  • Expand
  • Prune
  • Factor
  • Prune
  • Tidy
printbr('permuting to', A:eq(eig.R * eig.Lambda * eig.L))
local A = Matrix({2, 0, 0}, {0, 4, 5}, {0, 4, 3}) local ch = A:charpoly(lambda) printbr(ch) local solns = table{ch:solve(lambda)} printbr(solns:mapi(tostring):concat',') local eig = A:eigen() printbr(A:eq(eig.R * eig.Lambda * eig.L)) if Constant.isValue(eig.Lambda[1][1], 6) then local P = Matrix({0,1},{1,0}) eig.R = (eig.R * P)() eig.Lambda = (P * eig.Lambda * P)() eig.L = (P * eig.L)() printbr('permuting to', A:eq(eig.R * eig.Lambda * eig.L)) end simplifyAssertEq(eig.R, Matrix({1, 0, 0}, {0, -1, 1}, {0, frac(5,4), 1}):T()) simplifyAssertEq(eig.L, Matrix({1, 0, 0}, {0, -frac(4,9), frac(4,9)}, {0, frac(5,9), frac(4,9)}):T()) simplifyAssertEq(eig.Lambda, Matrix.diagonal(2, -1, 8)) 		${{{\left({{-{20}} + {{{\left({{3}{-{\lambda}}}\right)}} {{\left({{4}{-{\lambda}}}\right)}}}}\right)}} {{\left({{2}{-{\lambda}}}\right)}}} = {0}$
${\lambda} = {2}$,${\lambda} = {-{1}}$,${\lambda} = {8}$
${\left[\begin{array}{ccc} 2& 0& 0\\ 0& 4& 5\\ 0& 4& 3\end{array}\right]} = {{{\left[\begin{array}{ccc} 1& 0& 0\\ 0& -{1}& \frac{5}{4}\\ 0& 1& 1\end{array}\right]}} {{\left[\begin{array}{ccc} 2& 0& 0\\ 0& -{1}& 0\\ 0& 0& 8\end{array}\right]}} {{\left[\begin{array}{ccc} 1& 0& 0\\ 0& -{\frac{4}{9}}& \frac{5}{9}\\ 0& \frac{4}{9}& \frac{4}{9}\end{array}\right]}}}$
${\left[\begin{array}{ccc} 1& 0& 0\\ 0& -{1}& \frac{5}{4}\\ 0& 1& 1\end{array}\right]} = {\left[\begin{array}{ccc} 1& 0& 0\\ 0& -1& \frac{5}{4}\\ 0& 1& 1\end{array}\right]}$
${\left[\begin{array}{ccc} 1& 0& 0\\ 0& -{\frac{4}{9}}& \frac{5}{9}\\ 0& \frac{4}{9}& \frac{4}{9}\end{array}\right]} = {\left[\begin{array}{ccc} 1& 0& 0\\ 0& -{\frac{4}{9}}& \frac{5}{9}\\ 0& \frac{4}{9}& \frac{4}{9}\end{array}\right]}$
${\left[\begin{array}{ccc} 2& 0& 0\\ 0& -{1}& 0\\ 0& 0& 8\end{array}\right]} = {\left[\begin{array}{ccc} 2& 0& 0\\ 0& -1& 0\\ 0& 0& 8\end{array}\right]}$
GOOD 		time: 53.722000ms
stack: size: 7
  • Init
  • Prune
  • Expand
  • Prune
  • Factor
  • Prune
  • Tidy
printbr(A:eq(eig.R * eig.Lambda * eig.L)) OK HERE ... what about lambda, and their associations with R and L in the generalized system?
local A = Matrix({1, 1}, {0, 1}) printbr'without generalization:' local eig = A:eigen() for k,v in pairs(eig) do printbr(k, '=', v) end assert(eig.defective) assertEq(eig.R, Matrix{1, 0}:T()) simplifyAssertAllEq(eig.allLambdas, {1}) printbr'with generalization:' eig = A:eigen{generalize=true, verbose=true} for k,v in pairs(eig) do printbr(k, '=', v) end printbr('#allLambdas', #eig.allLambdas) printbr('#lambdas', #eig.lambdas) printbr('lambdas', eig.lambdas:mapi(function(l) return '{mult='..l.mult..', expr='..l.expr..'}' end):concat',') simplifyAssertEq(eig.L, Matrix.identity(2)) simplifyAssertEq(eig.R, Matrix.identity(2)) 		without generalization:
allLambdas = table: 0x70b33a63cf50
lambdas = table: 0x70b33a63ce68
R = $\left[\begin{array}{c} 1\\ 0\end{array}\right]$
defective = true
charPoly = ${{\left({{1}{-{λ}}}\right)}^{2}} = {0}$
Lambda = $\left[\begin{array}{c} 1\end{array}\right]$
${\left[\begin{array}{c} 1\\ 0\end{array}\right]} = {\left[\begin{array}{c} 1\\ 0\end{array}\right]}$
${1} = {1}$
${1} = {1}$
${1} = {1}$
with generalization:
${A} = {\left[\begin{array}{cc} 1& 1\\ 0& 1\end{array}\right]}$
${I} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
${{A}{-{{{λ}} \cdot {{I}}}}} = {\left[\begin{array}{cc} {1}{-{λ}}& 1\\ 0& {1}{-{λ}}\end{array}\right]}$
charPoly ${{{\left({{1}{-{λ}}}\right)}} {{\left({{1}{-{λ}}}\right)}}} = {0}$
after simplify(), charPoly ${{\left({{1}{-{λ}}}\right)}^{2}} = {0}$
${λ} = {1}$, ${λ} = {1}$
$λ$ $= \{$ $1$, $1$ $\}$
mult 2 expr $1$
finding nullspace for $1$
$\left[\begin{array}{c} 1\\ 0\end{array}\right]$
...is defective, gathering more generalized eigenvectors
trying again for generalized eigensystem...
finding nullspace of (A - I λ)^2 = $\left[\begin{array}{cc} 0& 0\\ 0& 0\end{array}\right]$
which is $\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]$
nullspace dim of (A - I λ)^2 = 2
...times (A - λ I) is $\left[\begin{array}{cc} 0& 1\\ 0& 0\end{array}\right]$
checking new potential vector product $\left[\begin{matrix} 0 \\ 0\end{matrix}\right]$
...and it's all zeroes so we already considered this vector
checking new potential vector product $\left[\begin{matrix} 1 \\ 0\end{matrix}\right]$
adding col 2
right eigenvector of λ= $1$ is $\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]$
${R} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
${L} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
${Λ} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
verify:
${{{\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}} {{\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}} {{\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}}} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
allLambdas = table: 0x70b33a51bd18
lambdas = table: 0x70b33a51ba28
L = $\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]$
R = $\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]$
defective = true
charPoly = ${{\left({{1}{-{λ}}}\right)}^{2}} = {0}$
Lambda = $\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]$
#allLambdas 2
#lambdas 1
lambdas {mult=2, expr=$1$}
${\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
${\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]} = {\left[\begin{array}{cc} 1& 0\\ 0& 1\end{array}\right]}$
GOOD 		time: 13.345000ms
stack: size: 7
  • Init
  • Prune
  • Expand
  • Prune
  • Factor
  • Prune
  • Tidy