Product of singular values = determinant proof

In summary, to prove that the determinant of an n x n square matrix is equal to the product of its singular values, one can use the singular value decomposition of the matrix A = U*E*V, where U and V are orthogonal matrices and E is a diagonal matrix with the singular values on its diagonal. By the definition of the singular value decomposition, the determinants of U and V are +/- 1. Therefore, the absolute value of the determinant of A is equal to the product of the singular values.
  • #1
Vai
5
0

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?
 
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  • #2
Hi Vai! :smile:

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?
 
  • #3
micromass said:
Hi Vai! :smile:

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)
 
Last edited:
  • #4
How can det(A) equal det(E) if det(E) is always positive, but if det(A) is not always positive??
 
  • #5
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
 
  • #6
Vai said:
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! :smile:
 
  • #7
Ok, thank you very much for your help; saved me a lot of time there.
 

Related to Product of singular values = determinant proof

1. What is the "product of singular values = determinant proof"?

The "product of singular values = determinant proof" is a mathematical theorem that states that the product of all the singular values of a square matrix is equal to the determinant of the matrix.

2. How is this proof used in science?

This proof is commonly used in various fields of science, such as physics, engineering, and computer science, to solve problems involving matrix operations and to understand the properties of matrices.

3. Can you provide an example of how this proof is applied?

Sure, let's say we have a 3x3 matrix A with singular values s1, s2, and s3. The product of these singular values would be s1 * s2 * s3. The determinant of A would be calculated as det(A) = s1^2 * s2^2 * s3^2. As you can see, the product of singular values is equal to the determinant of A.

4. What is the significance of this proof?

This proof helps us understand the relationship between singular values and determinants in matrices. It also allows us to simplify complex matrix operations by using the product of singular values instead of calculating the determinant.

5. Are there any limitations to this proof?

Yes, this proof only applies to square matrices. It also assumes that the matrix has a finite number of singular values. Additionally, it does not provide a proof for why this relationship holds, but rather shows that it is a true statement.

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