Product of singular values = determinant proof

[Vai]

Homework Statement

So I'm working on this proof. Given an n x n (square) matrix, prove that it's determinant is equal to the product of it's singular values.

Homework Equations

We are given A = U*E*V as a singular value decomposition of A.

The Attempt at a Solution

I was thinking that det(A) = det(U) * det(E) * det(V)

and since E is the diagonal matrix with singular values on it's diagonal, it's determinant is the product of those singular values.

But then what to do about det(U) and det(V)? I guess it's logical that the product of their determinants is 1, but how do I show that?

 

[micromass]

Hi Vai!

Do your singular values always need to be positive?? Does det(E) always need to be positive and det(A)??

U and V are unitary, what do you know about the determinant of unitary matrices?

 

[Vai]

Hi Vai!

Do your singular values always need to be positive?? Does det(E) always need to be positive and det(A)??

U and V are unitary, what do you know about the determinant of unitary matrices?

Thanks for the quick response.

I'm not sure about the sign on the singular values. Since they are the square roots of the eigenvalues of A' * A, then I assume that they are all positive. So then that means det(E) is also positive.

I wasn't aware that U and V are unitary matrices. But your comment made me think, and according to the definition of the singular value decomposition, they are orthogonal matrices. So then I can say their determinants are +/- 1.

I think that is enough then to show the det(A) is the product of the singular values since:

det(A) = det(U) * det(E) * det(V)
= (+/- 1) * (product of singular values) * (+/- 1)

 

[micromass]

How can det(A) equal det(E) if det(E) is always positive, but if det(A) is not always positive??

 

[Vai]

Whoops, I'm really sorry. I had to prove that the absolute value of the determinant of A is equal to the product of the singular values.

then:

|det(A)| = |det(U) * det(E) * det(V)|
= | (+/- 1) * (product of singular values) * (+/- 1) |
= product of singular values

 

[micromass]

Whoops, I'm really sorry. I had to prove that the absolute value of the determinant of A is equal to the product of the singular values.

then:

|det(A)| = |det(U) * det(E) * det(V)|
= | (+/- 1) * (product of singular values) * (+/- 1) |
= product of singular values

That is correct!

 

[Vai]

Ok, thank you very much for your help; saved me a lot of time there.