# Math Help - [SOLVED] Det A and singular values

1. ## [SOLVED] Det A and singular values

There is a rule that says:
$|det(A)| = \prod_{i=1}^m\sigma_i$ for an m x m matrix.
where $\sigma_i$
are singular values. It doesn't specify whether they are just singular values or non-zero singular values.

Does this mean just the non-zero singular values?

Because that means a determinant I'm working on is zero, not 12.

Thanks

2. Originally Posted by Bucephalus
There is a rule that says:
$|det(A)| = \prod_{i=1}^m$ for an mxm matrix.

What did you write here?? The product...of what?

Tonio

Does this mean just the non-zero singular values?

Because that means a determinant I'm working on is zero, not 12.

Thanks
.

3. ## I edited it up above...

Thanks.

4. I just thought about it some more though.
If for example, a 2 x 2 matrix had only one non-zero singular value it means that it's rank one, not full rank, which mean's it's not invertible, which means that it has a 0 determinant.
So I'm thinking that you have to go with how the equation reads and include all singular values non-zero or not.

Do you agree?

5. Originally Posted by Bucephalus
There is a rule that says:
$|det(A)| = \prod_{i=1}^m\sigma_i$ for an m x m matrix.
where $\sigma_i$
are singular values. It doesn't specify whether they are just singular values or non-zero singular values.

Does this mean just the non-zero singular values?

Because that means a determinant I'm working on is zero, not 12.

Thanks

Googling this stuff I found this SVD and LSI Tutorial 2: Computing Singular Values

Look at 1 just before figure 6 (almost at the bottom): it seems to be the determinant is the product of all the singular values.

Tonio

6. ## Thanks for that

Yes, I agree with you.

7. "Singular values" are just the eigenvalues of the matrix. Any matrix is similar to a matrix in either diagonal form or Jordan normal form where the determinant is just the product of the values on the diagonal. And the values on the diagonal are the eigenvalues.

And if a matrix has any 0 eigenvalue, then its determinant is 0.

(If the determinant were the product of the "non-zero" singular values, then NO matrix would have 0 determinant!)

8. ## thanks for your help.

Thanks.

9. Originally Posted by HallsofIvy
"Singular values" are just the eigenvalues of the matrix.

This isn't accurate: the singular values of a matrix $A$ (in general of an operator in a Hilbert space that blah-blah) are the eigenvalues of the normal matrix (operator) $A^{*}A$ which can be very different from the eigenvalues of $A$ itself.

That the determinant of any matrix is the product of its eigenvalues (in some field extension of the definition field) is basic stuff in linear algebra. Singular values, imo, aren't and, besides, belong a little more to functional analysis than to just linear algebra.

Tonio

Any matrix is similar to a matrix in either diagonal form or Jordan normal form where the determinant is just the product of the values on the diagonal. And the values on the diagonal are the eigenvalues.

And if a matrix has any 0 eigenvalue, then its determinant is 0.

(If the determinant were the product of the "non-zero" singular values, then NO matrix would have 0 determinant!)
.

10. Hey, yeah you're right.
The singular values are the square roots of the eigenvalues of $A^HA$

Good pick up Tonio.
Thanks.