# Determinant of a non-square matrix?

1. Apr 17, 2010

### GargleBlast42

"Determinant" of a non-square matrix?

Hi,

is there any numerical invariant that would characterize the rank of a non-square matrix, similar to the determinant for square matrices? I.e. having a matrix nxm, with n<m, I'm looking for a number that would be zero if the rank of the matrix is smaller than n and nonzero if the rank is n. By "similar to the determinant" I mean that it would be some number, which you could obtain by doing some arithmetic operations on the entries, but without the necessity to perform Gaussian Elimination.

2. Apr 18, 2010

### Lord Crc

Re: "Determinant" of a non-square matrix?

There's probably a better solution, but from what I can see, if the rank is n, then it should be http://en.wikipedia.org/wiki/Inverse_element#Matrices", and thus AA^T should exist and be invertible. So the determinant of AA^T could perhaps fit the bill?

Last edited by a moderator: Apr 25, 2017
3. Apr 18, 2010

### NaturePaper

Re: "Determinant" of a non-square matrix?

I hope you are right...by the way what the eigenvalues of AA^\daggers called? :)

Regards

4. Apr 18, 2010

### Landau

Re: "Determinant" of a non-square matrix?

Singular values.

5. Apr 19, 2010

### GargleBlast42

Re: "Determinant" of a non-square matrix?

Thanks Lord Crc, I have actually thought about this before, but it turns out that Mathematica can't handle this very well for large symbolic matrices.

Maybe a bit more about my problem: my matrix has one parameter and I want to find out for which values of this parameter this matrix doesn't have its full rank. Computing the determinant of the matrix isn't a problem for Mathematica, but the equation $$Det[A A^\dagger]=0$$ which I obtain contains complex conjugates and it seems that Mathematica is not able to deal with this kind of equation.

Finding singular values seems also uneffective. Any other ideas?

6. Apr 27, 2010

### trambolin

Re: "Determinant" of a non-square matrix?

QR decomposition might be fast. And solving linear equations maybe? Example:
$$\begin{pmatrix}a &b\\c&d\\e&f\\g&h \end{pmatrix}x = 0$$
gives you relations about the entries. You can check the roots of the resulting possible polynomials.