Proof. As long as $c_2$ only depends on $n$, we may increase it whenever needed throughout the proof. We consider the normed space $(M_{p,q}, |\cdot|_{\infty})$ and we recall that for $M \in M_{p,q}$ and $N \in M_{q,r}$,

$$|MN|_{\infty} \le q |M|_{\infty} |N|_{\infty}. \qquad (165)$$

Let $Q \in M_k$ be such that $|Q|_{\infty} \le 1$ and let $B = \{M \in M_k : |I_k - M|_{\infty} \le 1\}$ be the closed unit ball centered at $I_k$. By compactness of $B$, we can introduce

$$c_{2,k} = \max \left\{ 1, \max_{M \in B} \|D \det(M)\|_{\infty}, \frac{1}{2} \max_{M \in B} \|D^2 \det(M)\|_{\infty} \right\},$$

where $\|\cdot\|_{\infty}$ and $\|\cdot\|_{\infty}$ denote, respectively, the linear and bilinear operator norms associated with $(M_{p,q}, |\cdot|_{\infty})$. Note that $c_{2,k}$ depends on $k$ (since $B$ depends on $k$) though this can be avoided by defining $c_2 = \max_{1 \le k \le n} c_{2,k}$. Recall that the differential of the determinant map $\det$ at $I_k$ is the trace map, i.e. $D \det(I_k) = \text{tr}$, therefore, the application of the Taylor-Lagrange inequality for $\det$ on the line segment $[I_k, I_k + Q] \subset B$ yields (161).

Let $L \in M_{d,n}$, $R \in M_n$ be such that $L \circ L^t = I_d$ and $|R|_{\infty} \le 1$ and let us use the notation $Q = L \circ (R^t + R) \circ L^t + L \circ R^t \circ R \circ L^t \in M_d$ hereafter, so that $((I_n+R) \circ L^t)^t \circ ((I_n+R) \circ L^t) = I_d+Q$. First note that $|L|_{\infty} \le 1$, indeed, the assumption $L \circ L^t = I_d$ can be reformulated as follows: the columns of $L^t$ (i.e. the rows of $L$) constitute an orthonormal family $(v_1, \dots, v_d)$ of $\mathbb{R}^n$ so that $|L|_{\infty} = \max_{ij} |L_{ij}| = \max_{ij} |v_i \cdot e_j| \le 1$. Using (165) and $|M^t|_{\infty} = |M|_{\infty}$, we have

$$|Q|_{\infty} = |L \circ (R^t + R) \circ L^t + L \circ R^t \circ R \circ L^t|_{\infty} \le n^2 |R + R^t|_{\infty} |L|_{\infty}^2 + n^3 |L|_{\infty}^2 |R|_{\infty}^2 \\ \le (2n^2 + n^3 |R|_{\infty}) |R|_{\infty} \le c_2 |R|_{\infty},$$

that is (162).

We now assume $c_2|R|_{\infty} \le 1$ and consequently $((I_n + R) \circ L^t)^t \circ ((I_n + R) \circ L^t) = I_d + Q$ with $|Q|_{\infty} \le 1$ so that the first part of (161) gives

$$|\det (I_d + Q) - 1| \le c_2 |Q|_{\infty} \le c_2^2 |R|_{\infty} \le 1 \quad \text{and in particular} \quad \det (I_d + Q) \ge 0. \qquad (166)$$

We infer (163) applying $|a - 1| \le |a^2 - 1|$ with $a = \det (I_d + Q) \ge 0$. We are left with the proof of (164). We now apply the second inequality in (161) to obtain

$$|\det (I_d + Q) - 1 - \operatorname{tr} Q| \le c_2 |Q|_{\infty}^2 \le c_2^3 |R|_{\infty}^2. \qquad (167)$$

Furthermore, using $\operatorname{tr}(A) = \operatorname{tr}(A^t)$ and $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ when both products make sense, we have

$$\operatorname{tr}(L \circ R^t \circ L^t) = \operatorname{tr}(L \circ R \circ L^t) = \operatorname{tr}(R \circ L^t \circ L)$$

and thus, by definition of $Q$ and (165),

$$\begin{aligned} |\operatorname{tr} Q - 2 \operatorname{tr}(R \circ L^t \circ L)| &= |\operatorname{tr}(L \circ R^t \circ R \circ L^t)| \le d |L \circ R^t \circ R \circ L^t|_{\infty} \le dn^3 |L|_{\infty}^2 |R|_{\infty}^2 \\ &\le n^4 |R|_{\infty}^2. \end{aligned} \qquad (168)$$

From (167) and (168) we obtain

$$|\det (I_d + Q) - 1 - 2 \operatorname{tr}(R \circ L^t \circ L)| \le (c_2^3 + n^4) |R|_{\infty}^2 \qquad (169)$$

We now apply the following inequality, valid for any $z \ge -1$,

$$\left| \sqrt{1+z} - 1 - \frac{1}{2}z \right| \le \frac{1}{2}z^2$$

with $z = \det (I_d + Q) - 1$, from (166) we know that $-1 \le z \le c_2^2 |R|_{\infty}$, we hence obtain

$$\left| \sqrt{\det(I_d + Q)} - 1 - \operatorname{tr}(R \circ L^t \circ L) \right| \le \left| \sqrt{1+z} - 1 - \frac{1}{2}z \right| + \left| \frac{1}{2}z - \operatorname{tr}(R \circ L^t \circ L) \right|$$
$$ \begin{aligned} & \le \frac{1}{2} z^2 + \frac{1}{2} (c_2^3 + n^4) |R|_{\infty}^2 \text{ thanks to (169)}, \\ & \le \frac{1}{2} (c_2^4 + c_2^3 + n^4) |R|_{\infty}^2 \end{aligned} $$

which concludes the proof of (164). □

The following lemma stems directly from Lemma A.2.

**Lemma A.3.** Let $P, N \in \mathcal{M}_d$ and assume that $P$ is invertible, then

$$ \|P^{-1}\| \|P - N\| \le 1 \Rightarrow |\det(P) - \det(N)| \le c_2 |\det(P)| \|P^{-1}\| \|P - N\|. \quad (170) $$

*Proof.* Indeed, first note that

$$ |\det(P) - \det(N)| = |\det(P)| |1 - \det(P^{-1}N)| \quad \text{and} \quad P^{-1}N = I_d + P^{-1}(N - P). $$

Furthermore $|P^{-1}(N - P)|_{\infty} \le \|P^{-1}(N - P)\| \le \|P^{-1}\| \|P - N\| \le 1$ so that applying (161) with $k = d$ and $Q = P^{-1}(N - P)$ we can assert that

$$ |1 - \det(P^{-1}N)| \le c_2 |P^{-1}(N - P)|_{\infty} \le c_2 \|P^{-1}\| \|N - P\| $$

which concludes the proof of (170). □

The following is a crucial step to prove Proposition 3.10.

**Lemma A.4.** Let $S, T \in G_{d,n}$, there exist $\tilde{S} = (s_1|\dots|s_d)^t$, $\tilde{T} = (t_1|\dots|t_d)^t \in \mathcal{M}_{d,n}$ where $\{s_i\}_{i=1}^d$ and $\{t_i\}_{i=1}^d$ are two orthonormal bases of $S$ and $T$ such that

$$ \|\tilde{S} - \tilde{T}\| \le 2\|S - T\|. $$

*Proof.* Let $\theta$ be the largest principal angle between the subspaces $S$ and $T$, which can be characterized by:

$$ \sin(\theta) = \max_s \min_t \sqrt{1 - \langle s, t \rangle^2}, \quad s \in S, t \in T \quad \text{and} \quad |s| = |t| = 1. $$

We infer from [1, Proposition III.29] that $\|S - T\| = \sin(\theta)$, furthermore, there exists a rotation $r$ of $\mathbb{R}^n$ such that $r(S) = T$, with

$$ \|r - I_n\| \le 2 \sin(\theta/2). $$

Let $\tilde{S} = (s_1|\dots|s_d)^t \in \mathcal{M}_{d,n}$, with $\{s_i\}_{i=1}^d$ being an orthonormal basis of $S$, the matrix $\tilde{T} = r \circ \tilde{S} \in \mathcal{M}_{d,n}$ can be written as $(t_1|\dots|t_d)^t$ where $\{t_i\}_{i=1}^d$ is an orthonormal basis of and $T$. We have then, using that $\|\tilde{S}\| = 1$

$$ \|\tilde{S} - \tilde{T}\| = \|\tilde{S} - r \circ \tilde{S}\| \le \|I_n - r\| \|\tilde{S}\| \le 2 \sin(\theta/2); $$

the result follows from noting that $2 \sin(\theta/2) \le 2 \sin(\theta)$ as $\theta \in [0, \pi/2]$. □

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UNIVERSITÉ PARIS-SACLAY, INRIA, CNRS, LMO UMR8628, 91400 ORSAY, FRANCE

Email address: blanche.buet@universite-paris-saclay.fr

DIPARTIMENTO DI MATEMATICA, VIA SOMMARIVE 14, IT-38123 POVO - TRENTO, ITALY

Email address: gianpaolo.leonardi@unitn.it

UNIVERSITÉ CLAUDE BERNARD LYON 1, CNRS, ECOLE CENTRALE DE LYON, INSA LYON, UNIVERSITÉ JEAN MONNET, ICJ UMR5208, 69622 VILLEURBANNE, FRANCE

Email address: masnou@math.univ-lyon1.fr

UNIVERSITÉ CLAUDE BERNARD LYON 1, CNRS, ECOLE CENTRALE DE LYON, INSA LYON, UNIVERSITÉ JEAN MONNET, ICJ UMR5208, 69622 VILLEURBANNE, FRANCE

Email address: sagueni@math.univ-lyon1.fr