I am reading through this paper to try and code the model myself. The specifics of the paper don't matter, however in the authors matlab code I noticed they use a Cholesky decomposition instead of computing the determinant of a covariance matrix directly.

Specifically, the author has

$$\log\det(\Sigma) = 2 \sum_i \log [ diag(L)_i ]$$

where $L$ is the lower triangular matrix produced by a Cholesky decomposition of the covariance $\Sigma$.

I tested this out myself using various covariance matrices and found the relation above always works to within 14 decimal places (it's probably just a machine precision issue). I believe the author uses a cholesky decomposition because it is slightly faster to compute than computing the determinant directly (at least when I timed it on my machine).

My question is, why does this relation hold true? I couldn't find any references in the paper / code, or any material online.

  • $\begingroup$ What does "computing the determinant directly" mean in this context? If you are using a library, the routine to compute determinants might well be using something like Gaussian elimination or Cholesky decomposition, or whatever, and not summing over all permutations or expanding by minors. $\endgroup$ Commented Mar 22, 2019 at 15:41
  • $\begingroup$ $\Sigma=LL^T$ so you take det, use eigenvalues, then log and it follows immediately. $\endgroup$ Commented Mar 22, 2019 at 15:52

2 Answers 2


Depending on the size of your matrix $\Sigma$ you probably shouldn't have been able to compute the determinant in any reasonable amount of the time. The time complexity is $\mathcal{O}(n!)$ where as when using the LU it is $\mathcal{O}(n^{3})$.

The reason it works is because of the following property of determinants.

$$ \Sigma = LL^{T} \implies \det(\Sigma) = \det(LL^{T}) = \det(L)\det(L^{T}) $$

For a triangular matrix, the determinant is the product of the diagonal.

$$ \det(L) = \prod_{i=1}^{n} L_{ii} $$

which means if you take the $\log$ then you get a sum.

$$ \log(\prod_{i=1}^{n} L_{ii}) = \sum_{i=1}^{n} \log(L_{ii}) $$

this follows from a basic property of logs. The $2$ there is because of the property of determinants. The determinant of the transpose of the matrix is the same as the determinant of the matrix.

$$ \det(A^{T}) = \det(A) $$

That means that when we take the log the power of $2$ comes down

$$ \det(\Sigma) = \det(LL^{T}) \\ \implies \det(L)\det(L^{T}) = \det(L)^{2} \\ \implies \log((\prod_{i=1}^{n} L_{ii})^{2}) = 2 \log(\prod_{i=1}^{n} L_{ii}) = 2 \sum_{i=1}^{n} \log(L_{ii}) $$


There is a very straightforward answer to your question. It all because the diagonal values of the Cholesky factor, L, are the square roots of the successive ratios of determinants. To learn more about the general structure of Cholesky decomposition (or factorization) please refer to https://www.sciencedirect.com/science/article/pii/S0167715215001273

That is, if you look at the diagonal values of the Cholseky factor L of the matrix, say Σ.

A simple example will be to look at Σ which is 3 by 3 covariance matrix.

    (σ11 σ12 σ13)
Σ = (σ12 σ22 σ23)
    (σ13 σ23 σ33)

The first value at the diagonal will be the covariance of the first variable, σ11, and you can see it as the square root of σ11/1. The second diagonal value will be the square root of the determinant of the upper left 2x2 covariance matrix:

  Σ12 =   (σ11 σ12)
          (σ12 σ22)

divided by σ11.

That is, sqrt(|Σ12|/σ11)

The third diagonal value will be sqrt(|Σ|/|Σ12|).


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