[Solved] Determinant and random matrix generation

linearalgebra
#1

Hello,
I was playing with random matrices when I noticed this strange behaviour :

julia> using LinearAlgebra

julia> a=randn(10,11);

julia> isposdef(a*a')
true

julia> isposdef(a'*a)
false

I expected to have either way positive definite matrices since both matrices can be seen as covariance matrices of random data.

Am I doing something wrong here ?

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#2

Strictly speaking, this is not correct, since they can have zero eigenvalues. Eg

julia> isposdef(zeros(3,3) |> x -> x'*x)
false

What you are seeing is probably near-0, possibly negative, eigenvalues as a result of floating point error, which is normal.

Normally, reproducing the exact matrix would be needed to help investigate this issue, but in this case, one can easily generate examples with

using LinearAlgebra
global x
for _ in 1:1000
    a = randn(10, 11);
    if !isposdef(a'*a)
        @info "found one"
        global x = a
        break
    end
end
eigen(x'*x) # observe the eigenvalues
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#3

a’a has 0 as eigenvalue. (It is at most rank 10)

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#4

But it seems that the smallest dimension is favored

julia> using Statistics

julia> mean([ isposdef(randn(11,10) |> x->x*x') for i in 1:1000])
0.463

julia> mean([ isposdef(randn(10,11) |> x->x*x') for i in 1:1000])
1.0
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#5

Think of it this way: if a = [1;0;0], a*a’ = diag(1,0,0), which is not positive definite (but is semidefinite). However, a’a = 1, which is positive definite. In general, if a is N x M, then a a’ has rank at most M.

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#6

Thanks ! I forgot about that :zipper_mouth_face:

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#7

And numerical error is not how you get !isposdef. In fact, it’s the other way around. The reason you get positive definate a'a at all is due to numerical error shifting the 0 eigen value to the positive side…

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