using LinearAlgebra

A = [1 1
    -2 4]
eigvals(A)

X = [1 1
     1 2]

x = [1,0]
c = X \ x

A*x

Λ = Diagonal([2, 3])

X*Λ*c

X * Λ * inv(X) * x

X * Λ / X   # / X is equivalent to multiplying by inv(X), but is more efficient

A*X

X*Λ

X \ A * X

S = randn(2,2)

det(S) # a random S is almost certainly nonsingular

B = S \ A * S   # same as inv(S) * A * S, but more efficient since it avoids the explicit inverse

eigvals(B)

eigvals(A * S)

det(A), det(B)

det(A)

2 * 3

prod(eigvals(A))

Abig = randn(100,100)
det(Abig)

prod(eigvals(Abig))

logabsdet(Abig) # the log of the absolute value of the determinant, and the sign

s = sum(log.(abs.(eigvals(Abig)))) # the sum of the logs of |λ|

Abigger = randn(1000,1000)
det(Abigger)

floatmax(Float64) # the problem is that there is a maximum value the computer can represent, beyond which it gives "Inf"

logabsdet(Abigger) # but the log is fine

s = sum(log.(abs.(eigvals(Abigger)))) # the sum of the logs of |λ|

tr(A)

2 + 3

sum(eigvals(A))

tr(Abig)

sum(eigvals(Abig))

A = [1 1
    -2 4]

eigvals(A)

eigvals(A^4)

X = eigvecs(A)
X = X ./ X[1,:]'

X′ = eigvecs(A^4)
X′ = X′ ./ X′[1,:]'

eigvals(inv(A))

sqrt.(eigvals(A)) # takes the square root (elementwise) of each eigenvalue

Diagonal(sqrt.(eigvals(A))) # the diagonal matrix √Λ

Asqrt = X * Diagonal(sqrt.(eigvals(A))) / X   # the √A matrix

Asqrt * Asqrt

sqrt(A)

sqrt(A) ≈ Asqrt