using LinearAlgebra

A = randn(5,5)
b = randn(5)
x = A \ b # solve Ax = b
b - A*x # this "residual" should be zero

A = [1    2
     2.01 3.99]
b = [1,2]
x = A \ b # "exact" answer

Δb = [-0.01, 0.004]
x′ = A \ (b + Δb)

x′ - x # the error Δx

Δx = A \ Δb

norm(Δb) # the size of the error in the input

norm(Δx) # the size of the error in the output

opnorm(A)        # ‖A‖

opnorm(inv(A))   # ‖A⁻¹‖

using PyPlot
θ = range(0,2π,length=100)
plot(θ/(2π), [norm(A \ [cos(θ), sin(θ)]) for θ in θ], "r-")
plot(θ/(2π), ones(length(θ))*norm(inv(A)), "k--")
xlabel(L"\theta / 2\pi")
ylabel(L"\Vert A^{-1} y \Vert")
text(0.16,opnorm(inv(A))*0.95, L"\Vert A^{-1} \Vert")
title(L"maximizing $\Vert A^{-1} y \Vert$ over angle $\theta$")

eigvals(inv(A))

abs.(eigvals(inv(A)))

abs.(eigvals(inv(A)))[1] - opnorm(inv(A))

B = randn(2,2)
opnorm(B)

abs.(eigvals(B))

svdvals(B)

svdvals(B)[1] - opnorm(B)

svdvals(inv(A))[1] - opnorm(inv(A))

opnorm(B)

sqrt.(eigvals(B'B))

svdvals(B)

opnorm(10*B) / opnorm(B)

B₂ = randn(size(B))
opnorm(B + B₂) / (opnorm(B) + opnorm(B₂))

opnorm(B * B₂) / (opnorm(B) * opnorm(B₂))

opnorm(inv(A))

1 ./ svdvals(A)

opnorm(inv(A)) - 1/svdvals(A)[2]

cond(A)

opnorm(A) * opnorm(inv(A))

σ = svdvals(A)
maximum(σ) / minimum(σ)

cond(A)

A

norm(Δx)/norm(x) / (norm(Δb)/norm(b))

X = [1  1
     0  1e-14]
M = X * Diagonal([1,1+1e-14]) / X

exp(M)

X * exp(Diagonal([1,1+1e-14])) / X