## Lower Bound for a Determinant

Hello,

I have the following determinant:

$$\text{det}\left(\mathbf{A}\mathbf{A}^H\right)$$

where H denoted complex conjugate transpose, and A is a circulant matrix. I am looking for a lower bound for the above determinant. Is there one?

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 Mentor I think det(AB)=det(A)=det(B) should help to express your determinant via det(A). If something is known about A, it might be possible to evaluate the expression or give some lower bound.

 Quote by mfb I think det(AB)=det(A)=det(B) should help to express your determinant via det(A). If something is known about A, it might be possible to evaluate the expression or give some lower bound.
I do not want to evaluate it, I need a lower bound to see what is the lowest value for the determinant.

Recognitions:

## Lower Bound for a Determinant

The the eigenvectors of a circulant matrix are independent of the matrix and related to the nth roots of unity, so there are some nice ways to write the value of the determinant. See http://en.wikipedia.org/wiki/Circulant_matrix.

For an arbitrary A the lower bound is zero, which isn't a very interesting result - but maybe you want a lower bound that depends on the elements of A in some way?

 Quote by AlephZero The the eigenvectors of a circulant matrix are independent of the matrix and related to the nth roots of unity, so there are some nice ways to write the value of the determinant. See http://en.wikipedia.org/wiki/Circulant_matrix. For an arbitrary A the lower bound is zero, which isn't a very interesting result - but maybe you want a lower bound that depends on the elements of A in some way?
Exactly, that is what I need, especially in term of A's columns.

Mentor
 Quote by S_David I do not want to evaluate it, I need a lower bound to see what is the lowest value for the determinant.
The evaluation gives a lower bound. In fact, it is the best possible lower bound.

The wikipedia article has an explicit formula for the determinant, you can just calculate it and use it as a lower bound. If that needs to much computing power (which I doubt, but I don't know what you are doing), you can produce a lot of lower bounds of variable quality. But in that case, it would be useful to know in which way you need this bound.

 Quote by mfb The evaluation gives a lower bound. In fact, it is the best possible lower bound. The wikipedia article has an explicit formula for the determinant, you can just calculate it and use it as a lower bound. If that needs to much computing power (which I doubt, but I don't know what you are doing), you can produce a lot of lower bounds of variable quality. But in that case, it would be useful to know in which way you need this bound.
I am working on a wireless communication system, and I need to know the worst performance of the system, which is equivalent to know the "lowest" bound of the above determinant. I do not need any lower bound, I need the lowest bound. A is circulant and toeplitz, so the determinant is strictly greater than zero, and hence , I need the lowest determinant greater than zero, and I need it, if possible, in terms of the the first (or any) row of A, since all contains the same elements. That would ease the analysis.

Thanks
 Blog Entries: 1 Recognitions: Homework Help Why is the value for the determinant in the wikipedia article not good enough?

 Quote by Office_Shredder Why is the value for the determinant in the wikipedia article not good enough?
Again, I do not need to find the determinant itself, I need a lower bound.

Mentor
As said before, the determinant IS a lower bound for the determinant of this matrix.
If you want a lower bound for all possible matrices at the same time, you need additional data about the entries of the matrix. Are they always bounded in some way?

 I do not need any lower bound, I need the lowest bound.
Do you mean "the highest"? The lowest lower bound is -infinity, which is not what you are looking for I think. The highest lower bound is your determinant.

 Quote by mfb As said before, the determinant IS a lower bound for the determinant of this matrix. If you want a lower bound for all possible matrices at the same time, you need additional data about the entries of the matrix. Are they always bounded in some way? Do you mean "the highest"? The lowest lower bound is -infinity, which is not what you are looking for I think. The highest lower bound is your determinant.
$$\text{det}\left(\mathbf{A}\mathbf{A}^H\right)\geq f(\mathbf{h})=w$$

Yeah, I need to find w>0 for all possible realizations of the random matrix A. Further information about A:

1- A is an N-by-N circulant matrix with first column:

$$\mathbf{h}^T=[h_0\,\,h_1,\cdots,\,h_L,\mathbf{0}_{N-L+1}]^T$$

2- The entries of h are i.i.d Gasussian random variables with zero-mean and unit variance.

Does that change any thing?

Mentor
I don't think your first equation is what you want:

f(h)=|det(A)|2 with

$$\mathrm{det}(A) = \prod_{j=0}^{n-1} (h_0 + h_{n-1} \omega_j + h_{n-2} \omega_j^2 + \dots + h_1\omega_j^{n-1}).$$
where $\omega_j=\exp \left(\frac{2\pi i j}{n}\right)$.

But as f(h) depends on h, this bound now depends on the matrix.

 2- The entries of h are i.i.d Gasussian random variables with zero-mean and unit variance.
In that case, 0 is the best lower bound you can give. The determinant of your expression can reach arbitrary small values, as it is possible that all entries are 0 (as an example) and therefore the determinant is 0.

 Quote by mfb I don't think your first equation is what you want: f(h)=|det(A)|2 with $$\mathrm{det}(A) = \prod_{j=0}^{n-1} (h_0 + h_{n-1} \omega_j + h_{n-2} \omega_j^2 + \dots + h_1\omega_j^{n-1}).$$ where $\omega_j=\exp \left(\frac{2\pi i j}{n}\right)$. But as f(h) depends on h, this bound now depends on the matrix. In that case, 0 is the best lower bound you can give. The determinant of your expression can reach arbitrary small values, as it is possible that all entries are 0 (as an example) and therefore the determinant is 0.
Just throw that exception. I need the determinant that is strictly greater than zero.
 Mentor But there is no lower bound larger than 0 with the distribution of hi you gave. In addition to 0, it can reach a value smaller than epsilon for all epsilon > 0. It should be possible to evaluate the distribution function and estimate how likely a value smaller epsilon is (for a given epsilon), but it will never be 0.

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