In the document Trapdoors for Lattices, section 5.4 Gaussian Sampling, they introduce the parameter $\sqrt{\Sigma_{\bf G}}$, which is related to the lattice $\Lambda^\perp(\bf G)$. They use it as a bound for the smoothing parameter of this lattice, therefore $\sqrt{\Sigma_{\bf G}}\in\mathbb{R}$. But later on, they do some calculation as if it were a matrx when they write $s_1(\sqrt{\Sigma_{\bf G}})$, where $s_1(\cdot)$ is the first singular value of the argument.

If this is not confusing enough, in 2.1 they use the $\Sigma$ notation extensively to refer to matrices. Even when defining the root of a matrix

I feel like there is a small detail I'm missing, so if anyone could help me understand what's going on here that'd be great.


1 Answer 1


While I haven't verified this, I think a more natural interpretation of this is that $\sqrt{\Sigma_{\mathbf{G}}}$ is simply the matrix square root (in the sense of the Cholesky decomposition) of the positive semidefinite $\Sigma_{\mathbf{G}}$. Your main concern seems to be the bound $$\sqrt{\Sigma_{\mathbf{G}}} \geq \eta_\epsilon(\Lambda^\perp(\mathbf{G})).$$ While it is natural to think this is an inequality between real numbers (and therefore $\Sigma_{\mathbf{G}}$ should be real, which would be confusing), there is another interpretation. Namely, one can write this (perhaps less ambiguously) as

$$\sqrt{\Sigma_{\mathbf{G}}} \succeq \eta_\epsilon(\Lambda^\perp(\mathbf{G}))\cdot I,$$ where $\succeq$ is the inequality in the the Loewner order on matrices, and $I$ is an appropriately-sized identity matrix.

This interpretation is consistent with the paper's (claimed) chosen notation. I quote from the first pagagraph of section 2

or convenience, we sometimes use a scalar $s$ to refer to the scaled identity matrix $sI$, where the dimension will be clear from context.

For the notation selected for the Loewner order, I quote the 4th paragraph of Section 2.1

A symmetric matrix $\Sigma\in\mathbb{R}^{n\times n}$ is positive definite (respectively, positive semidefinite), written $\Sigma > 0$ (resp., $\Sigma\geq 0$), if $x^t\Sigma x > 0$ (resp., $x^t\Sigma x \geq 0$) for all nonzero $x \in\mathbb{R}^n$. We have $\Sigma > 0$ if and only if $\Sigma$ is invertible and $\Sigma^{-1} > 0$, and $\Sigma \geq 0$ if and only if $\Sigma^+ \geq 0$. Positive (semi)definiteness defines a partial ordering on symmetric matrices: we say that $\Sigma_1 > \Sigma_2$ if $(\Sigma_1 − \Sigma_2) > 0$, and similarly for $\Sigma_1 \geq \Sigma_2$ ...

In general though, this is the natural interpretation solely because $\sqrt{\Sigma_{\mathbf{G}}}$ is mentioned as the parameter of a Gaussian. This should be a strong hint you are in the multi-dimensional setting (and this is the Cholesky decomposition of a covariance matrix, which is always positive-semidefinite, i.e. the Cholesky decomposition always exists). This is because in 1 dimensions it is horrible notation --- one would simplify $\sqrt{\sigma^2} = \sigma$ to get a much simpler expression in terms of more familiar parameters. While overloading $\geq$ and writing $s$ rather than $sI$ is somewhat ambiguous, it is much less of a notational crime than writing $\sqrt{\Sigma_{\mathbf{G}}}$ for $\sigma$, so is the natural interpretation.

  • $\begingroup$ Well that makes a lot of sense, thanks! I knew there was something fishy :) $\endgroup$ Commented Jun 27, 2023 at 9:20

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