SlowGuess/NeurIPS_2023_OCR huggingface dataset:
An instance of  $\mathbf{X}^{(t)}$  is shown in Figure 4. The same conditions as in Figure 3 are used, except that the ground truth matrix is a rank-3 matrix with non-zero eigenvalues 1, 0.75, 0.5. The singular values of  $\mathbf{X}^{(t)}$  behave similarly to  $\left\| \mathbf{x}^{(t)} \right\|_2$  in the rank-1 case. In the early iterations, where the orthogonal components dominate, the singular values stay close to their initial scale  $\beta_0$ . After that, each singular value  $\sigma_i(\mathbf{X}^{(t)})$  increases at a rate of  $(1 + \eta \lambda_i^\star)$  and saturates at  $\sqrt{\lambda_i^\star}$ . We use  $\left\| \mathbf{X} - \mathbf{Y} \right\|_{\mathrm{R}}$  to denote the Frobenius norm between  $\mathbf{X}$  and  $\mathbf{Y}$  under best rotational alignment.  $\left\| \mathbf{X}^{(t)} - \mathbf{X}^\star \right\|_{\mathrm{R}}$  decreases exponentially and saturates at the level determined by the noise size  $\sigma$ , after all singular values have saturated, as local convergence begins.
To extend the results of the rank-1 case, we need to show that  $\left\| X^{(t)} - \widetilde{X}^{(t)}\right\|_{\mathrm{R}}$  remains much smaller than  $\left\| X^{(t)}\right\|_{\mathrm{F}}$  throughout the iterations, where  $\widetilde{X}^{(t)}$  is the trajectory of the fully observed case. Before  $\sigma_1(X^{(t)})$  saturates around  $\sqrt{\lambda_1^*}$ , it behaves similarly to  $\left\| x^{(t)} - \widetilde{x}^{(t)}\right\|_2$  of the rank-1 case, i.e., it expands at a rate of  $(1 + \eta \lambda_1^*)$  along with  $\left\| X^{(t)}\right\|_{\mathrm{F}}$  after the early iterations. However, because each singular value grows at a different rate, a different phenomenon is observed for the rank- $r$  case. During the iterations before  $\sigma_{i+1}(X^{(t)})$  saturates after  $\sigma_i(X^{(t)})$  does, both  $\left\| X^{(t)} - \widetilde{X}^{(t)}\right\|_{\mathrm{R}}$  and  $\left\| X^{(t)}\right\|_{\mathrm{F}}$  do not increase much. Our current theory can only show that  $\left\| X^{(t)} - \widetilde{X}^{(t)}\right\|_{\mathrm{R}}$  increases at a rate less than  $(1 + \eta \lambda_{i+1}^*)$ , and in order for  $\left\| X^{(t)} - \widetilde{X}^{(t)}\right\|_{\mathrm{R}}$  to remain much smaller than  $\left\| X^{(t)}\right\|_{\mathrm{F}}$ , additional sample complexity is required to compensate for the exponential increases. Therefore, we expect that the convergence of GD for the case of rank- $r$  can be proved with the techniques developed in this paper if  $n^{1+\Theta(\kappa-1)} \operatorname{poly}(\kappa, r, \log n)$  samples are provided, where  $\kappa = \frac{\lambda_1^*}{\lambda_r^*}$  is the condition number. Nevertheless, whether GD can converge with the optimal  $n \operatorname{poly}(\kappa, r, \log n)$  samples for the rank- $r$  matrix completion problem remains an open problem.
This research was supported by the National Research Foundation of Korea under grant 2021R1C1C11008539.
[19] S. Du, J. Lee, H. Li, L. Wang, and X. Zhai, "Gradient descent finds global minima of deep neural networks," in International Conference on Machine Learning, pp. 1675-1685, 2019.
[20] Y. Ma, A. Olshevsky, C. Szepesvari, and V. Saligram, "Gradient descent for sparse rank-one matrix completion for crowd-sourced aggregation of sparsely interacting workers," in International Conference on Machine Learning, pp. 3335-3344, 2018.
[21] Q. Ma and A. Olshevsky, "Adversarial crowdsourcing through robust rank-one matrix completion," in Advances in Neural Information Processing Systems, pp. 21841-21852, 2020.
[22] S. Gunasekar, B. E. Woodworth, S. Bhojanapalli, B. Neyshabur, and N. Srebro, "Implicit regularization in matrix factorization," in Advances in Neural Information Processing Systems, 2017.
[23] D. Stöger and M. Soltanolkotabi, "Small random initialization is akin to spectral learning: Optimization and generalization guarantees for overparameterized low-rank matrix reconstruction," in Advances in Neural Information Processing Systems, pp. 23831-23843, 2021.
[24] S. Arora, N. Cohen, W. Hu, and Y. Luo, "Implicit regularization in deep matrix factorization," in Advances in Neural Information Processing Systems, pp. 7413-7424, 2019.
[25] N. Razin and N. Cohen, "Implicit regularization in deep learning may not be explainable by norms," in Advances in Neural Information Processing Systems, pp. 21174-21187, 2020.
[26] Z. Li, Y. Luo, and K. L. Lyu, "Towards resolving the implicit bias of gradient descent for matrix factorization: Greedy low-rank learning," in In International Conference on Learning Representations, 2021.
[27] S. Bhojanapalli, B. Neyshabur, and N. Srebro, "Global optimality of local search for low rank matrix recovery," in Advances in Neural Information Processing Systems, 2016.
[28] R. Ge, J. D. Lee, and T. Ma, "Matrix completion has no spurious local minimum," in Advances in Neural Information Processing Systems, 2016.
Detailed proofs for the results explained in the main text are provided in this appendix. We say that an event happens with high probability if it happens with probability at least  $1 - \frac{1}{n^C}$  for a constant  $C > 0$  and  $C$  can be made arbitrary large by controlling constant factors. A union of  $\mathrm{poly}(n)$  number of events that happens with high probability still happens with high probability. For a matrix  $\mathbf{A}$ , we denote the spectral norm by  $\| \mathbf{A}\|$  and the maximum absolute value of entries by  $\| \mathbf{A}\|_{\infty}$ . Also, the largest  $\ell_2$ -norm of rows of  $\mathbf{A}$  is denoted as  $\| \mathbf{A}\|_{2,\infty}$ .
A Spectral Analysis
We introduce some spectral bounds related to random sampling and Gaussian noise.
Lemma A.1. If  $n^2 p \gtrsim n \log n$ , we have
$$
\left\| \frac {1}{p} \mathcal {P} _ {\Omega} (\boldsymbol {M} ^ {\star}) - \boldsymbol {M} ^ {\star} \right\| \lesssim \lambda^ {\star} \mu \sqrt {\frac {\log n}{n p}}
$$
with high probability.
Lemma A.2. If  $n^2 p \gtrsim \mu n \log n$ , for all  $l \in [n]$ , we have
$$
\left\| \frac {1}{p} \mathcal {P} _ {\Omega} (\boldsymbol {M} ^ {\star}) - \mathcal {P} _ {\Omega} ^ {(l)} (\boldsymbol {M} ^ {\star}) \right\| \lesssim \lambda^ {\star} \sqrt {\frac {\mu}{n p}}
$$
Lemma A.3. If  $n^2 p \gtrsim n\log^2 n$ , we have
$$
\left\| \frac {1}{p} \mathcal {P} _ {\Omega} (\boldsymbol {E}) \right\| \lesssim \sigma \sqrt {\frac {n}{p}}
$$
Note that Lemma A.3 also implies that  $\| \pmb{E}^{(l)}\| \lesssim \sigma \sqrt{\frac{n}{p}}$  for all  $l\in [n]$  with high probability. Combined with the condition  $\sigma \lesssim \frac{\lambda^{\star}\mu}{n}\sqrt{\log n}$ , we have
$$
\left\| \frac {1}{p} \mathcal {P} _ {\Omega} (\boldsymbol {E}) \right\| \lesssim \lambda^ {\star} \mu \sqrt {\frac {\log n}{n p}}, \quad \left\| \boldsymbol {E} ^ {(l)} \right\| \lesssim \lambda^ {\star} \mu \sqrt {\frac {\log n}{n p}}
$$
for all  $l \in [n]$ . Proofs for Lemmas A.1 and A.2 are provided in Appendix G. Check Lemma 11 of [12] for the proof of Lemma A.3.
Next, we state bounds on the eigenvalues of  $M^{\circ}$  and  $M^{(l)}$ . The first eigenvalues of  $M^{\circ}$  and  $M^{(l)}$  are denoted as  $\lambda^{\circ}$  and  $\lambda^{(l)}$ , respectively. The following lemma is derived from Lemmas A.1 and A.2 with Weyl's Theorem.
Lemma A.4. If  $n^2 p \gtrsim \mu n \log n$ , we have
$$
\left| \lambda^ {\circ} - \lambda^ {\star} \right| \lesssim \lambda^ {\star} \mu \sqrt {\frac {\log n}{n p}}, \tag {A.1}
$$
$$
\left| \lambda^ {(l)} - \lambda^ {\star} \right| \lesssim \lambda^ {\star} \mu \sqrt {\frac {\log n}{n p}} \tag {A.2}
$$
for all  $l \in [n]$  with high probability.
Lastly, Lemmas A.1 and A.2 with Davis-Kahan Theorem give the following lemma.
Lemma A.5. If  $n^2 p \gtrsim \mu n \log n$ , we have
$$
\left\| \boldsymbol {u} ^ {(l)} - \boldsymbol {u} ^ {\star} \right\| _ {2} \lesssim \mu \sqrt {\frac {\log n}{n p}}
$$
In this section, we introduce some properties that the initialization vector  $\pmb{x}^{(0)}$  satisfies. Recall that each entry of  $\pmb{x}^{(0)}$  is sampled from  $\mathcal{N}(0,\frac{1}{n}\beta_0^2)$  independently. We use  $\pmb{H}$  to denote the perturbation  $M^{\circ} - M^{\star}$ .
Lemma B.1. The initialization vector  $\pmb{x}^{(0)}$  satisfies
$$
\frac {1}{2} \beta_ {0} \leq \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} \leq \frac {3}{2} \beta_ {0} \tag {B.1}
$$
with probability at least  $1 - e^{-n / 32}$ , and
$$
\left\| \boldsymbol {x} ^ {(0)} \right\| _ {\infty} \leq 2 \sqrt {\log n} \frac {\beta_ {0}}{\sqrt {n}}, \tag {B.2}
$$
$$
\left| \boldsymbol {u} ^ {\star \top} \boldsymbol {H} ^ {s} \boldsymbol {x} ^ {(0)} \right| \leq 2 \sqrt {\log n} \frac {\beta_ {0}}{\sqrt {n}} \| \boldsymbol {H} \| ^ {s}, \quad \forall s \leq 3 0 \log n, \tag {B.3}
$$
with probability at least  $1 - \frac{1}{n} - \frac{30\log n}{n^2}$ . It also satisfies
$$
\frac {1}{\sqrt {\log n}} \frac {\beta_ {0}}{\sqrt {n}} \leq \left| \boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)} \right| \tag {B.4}
$$
with probability at least  $1 - \frac{1}{2\sqrt{\log n}}$ .
Proof. To bound  $\left\| \boldsymbol{x}^{(0)}\right\|_2$ , we use the following basic concentration inequality that holds for i.i.d. standard normal variables  $\{X_i\}_{i\in [n]}$ .
$$
\mathbb {P} \left[ \left| \frac {1}{n} \sum_ {i = 1} ^ {n} X _ {i} ^ {2} - 1 \right| \geq t \right] \leq 2 e ^ {- n t ^ {2} / 8}
$$
If we put  $t = \frac{1}{2}$ , with probability at least  $1 - e^{-n / 32}$ , we have
$$
\frac {1}{2} \beta_ {0} ^ {2} \leq \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} ^ {2} \leq \frac {3}{2} \beta_ {0} ^ {2},
$$
and this implies (B.1).
For a centered Gaussian random variable with standard deviation  $\sigma$ , we have
$$
\mathbb {P} [ | X | \geq t ] \leq e ^ {- \frac {t ^ {2}}{2 \sigma^ {2}}}.
$$
Hence, an entry of  $\pmb{x}^{(0)}$  is less than  $2\sqrt{\log n}\frac{\beta_0}{\sqrt{n}}$  with probability at least  $1 - \frac{1}{n^2}$ , and all entries of  $\pmb{x}^{(0)}$  are less than  $2\sqrt{\log n}\frac{\beta_0}{\sqrt{n}}$  with probability at least  $1 - \frac{1}{n}$ .  $\pmb{u}^{\star \top}\pmb{H}^{s}\pmb{x}^{(0)}$  follows a centered Gaussian distribution with standard deviation  $\| H^{s}\pmb{u}^{\star}\|_{2} \leq \| H\|^{s}\| \pmb{u}^{\star}\|_{2}$  for all  $s$ , and (B.3) holds with probability at least  $1 - \frac{10\log n}{n^2}$ .
For a random variable  $X$  that is sampled from  $\mathcal{N}(0,\sigma^2)$ , we have
$$
\mathbb {P} [ | X | \leq t ] \leq \frac {t}{\sqrt {2 \pi \sigma^ {2}}}.
$$
$$
\frac {1}{\sqrt {\log n}} \frac {\beta_ {0}}{\sqrt {n}} \leq \left| \boldsymbol {u} ^ {\star^ {\top}} \boldsymbol {x} ^ {(0)} \right|
$$
Lemma B.1 implies that the  $\pmb{u}^{\star}$  component of  $\pmb{x}^{(0)}$  is in the range
$$
\frac {1}{\sqrt {\log n}} \frac {\beta_ {0}}{\sqrt {n}} \leq \left| \boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)} \right| \leq 2 \sqrt {\log n} \frac {\beta_ {0}}{\sqrt {n}}. \tag {B.5}
$$
$$
\left\| a \boldsymbol {x} ^ {(0)} + b \boldsymbol {u} ^ {\star} \right\| _ {\infty} \geq \left(| a | \beta_ {0} + | b |\right) \frac {1}{\sqrt {n}} \tag {B.6}
$$
for all  $a, b$ , with probability at least  $1 - \exp \left(-\frac{n}{2\mu}\right)$ .
Proof. The probability that an entry of  $\pmb{x}^{(0)}$  is less than  $\frac{\beta_0}{\sqrt{n}}$  is bounded by  $\frac{1}{\sqrt{2\pi}}$ . Without loss of generality, let us assume that all entries of  $\pmb{u}^{\star}$  are not negative and  $a, b \geq 0$ . There are at least  $\frac{n}{\mu}$  entries of  $\pmb{u}^{\star}$  that are larger than  $\frac{1}{\sqrt{n}}$ . For such entries, the probability that all entries of  $\pmb{x}^{(0)}$  is less than  $\frac{\beta_0}{\sqrt{n}}$  is bounded by  $\left(\frac{1}{\sqrt{2\pi}}\right)^{\frac{n}{\mu}} \leq \exp \left(-\frac{n}{2\mu}\right)$ . Hence, for at least one position, both entries of  $\pmb{x}^{(0)}$  and  $\pmb{u}^{\star}$  are larger than  $\frac{\beta_0}{\sqrt{n}}$  and  $\frac{1}{\sqrt{n}}$ , respectively, with probability at least  $1 - \exp \left(-\frac{n}{2\mu}\right)$ .
In the following sections, we assume that we are given an initialization vector  $\pmb{x}^{(0)}$  that satisfies (B.1) to (B.6).
C Fully Observed Case
We provide some lemmas related to  $\widetilde{\pmb{x}}^{(t)}$  in this section. We first note that  $\widetilde{\pmb{x}}^{(t)}$  is explicitly written as
$$
\widetilde {\boldsymbol {x}} ^ {(t)} = \prod_ {s = 1} ^ {t} \left(1 - \eta \widetilde {\beta} _ {s} ^ {2}\right) \boldsymbol {x} ^ {(0)} + \prod_ {s = 1} ^ {t} \left(1 - \eta \widetilde {\beta} _ {s} ^ {2} + \eta \lambda^ {\star}\right) \left(\boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)}\right) \boldsymbol {u} ^ {\star} := A ^ {(t)} \boldsymbol {x} ^ {(0)} + B ^ {(t)} \boldsymbol {u} ^ {\star}. \tag {C.1}
$$
Let us define  $T_2'$  as the last  $t$  such that  $\widetilde{\beta}_t^2 \leq \frac{\lambda^\star}{64\log n}$ . We claim that  $T_2' \leq \frac{64\log n}{\eta\lambda^\star}$  and prove this later. Then, for all  $t \leq T_2'$ , we have
$$
\frac {1}{4} (1 + \eta \lambda^ {\star}) ^ {t} \leq \prod_ {s = 1} ^ {t} \left(1 - \eta \widetilde {\beta} _ {s} ^ {2} + \eta \lambda^ {\star}\right) \leq \left(1 + \eta \lambda^ {\star}\right) ^ {t} \tag {C.2}
$$
$$
\frac {1}{4} \leq \prod_ {s = 1} ^ {\tau} (1 - \eta \widetilde {\beta} _ {s} ^ {2}) \leq 1
$$
$$
\prod_ {s = 1} ^ {T _ {2} ^ {\prime}} \left(\frac {1 + \eta \lambda^ {\star} - \eta \widetilde {\beta} _ {s} ^ {2}}{1 + \eta \lambda^ {\star}}\right) \geq \prod_ {s = 1} ^ {T _ {2} ^ {\prime}} (1 - \eta \widetilde {\beta} _ {s} ^ {2}) \geq \left(1 - \frac {\eta \lambda^ {\star}}{6 4 \log n}\right) ^ {\frac {6 4 \log n}{\eta \lambda^ {\star}}} \geq \frac {1}{4}
$$
if  $\frac{\eta\lambda^{\star}}{64\log n} \leq \frac{1}{2}$ . Note that the upper bounds in (C.2) hold even if  $t > T_{2}$ .
From (C.2), we have the approximation  $\widetilde{\pmb{x}}^{(t)}\approx \pmb{x}^{(0)} + (1 + \eta \lambda^{\star})^{t}(\pmb{u}^{\star \top}\pmb{x}^{(0)})\pmb{u}^{\star}$  for all  $t\leq T_2'$  and the  $\ell_{2}$ -norm of  $\widetilde{\pmb{x}}^{(t)}$  is also approximately given by  $\left(1 + \frac{(1 + \eta\lambda^{\star})^t}{\sqrt{n}}\right)\beta_0$ . The  $\ell_{\infty}$ -norm is about  $\frac{1}{\sqrt{n}}$  times smaller than the  $\ell_{2}$ -norm. We make this observation rigorous with the following lemma.
Lemma C.1. For all  $t \leq T_2'$ , we have
$$
\frac {1}{8} \frac {1}{\sqrt {\log n}} \left(1 + \frac {(1 + \eta \lambda^ {\star}) ^ {t}}{\sqrt {n}}\right) \beta_ {0} \leq \left\| \widetilde {\boldsymbol {x}} ^ {(t)} \right\| _ {2} \leq 2 \sqrt {\log n} \left(1 + \frac {(1 + \eta \lambda^ {\star}) ^ {t}}{\sqrt {n}}\right) \beta_ {0},
$$
$$
\frac {1}{4} \frac {1}{\sqrt {\log n}} \left(1 + \frac {(1 + \eta \lambda^ {\star}) ^ {t}}{\sqrt {n}}\right) \frac {\beta_ {0}}{\sqrt {n}} \leq \left\| \widetilde {\boldsymbol {x}} ^ {(t)} \right\| _ {\infty} \leq 2 \sqrt {\log n} \left(1 + (1 + \eta \lambda^ {\star}) ^ {t} \sqrt {\frac {\mu}{n}}\right) \frac {\beta_ {0}}{\sqrt {n}}.
$$
Proof. For brevity, let us drop the superscript  $(t)$  and write  $\widetilde{\pmb{x}} = A\pmb{x}^{(0)} + B\pmb{u}^{\star}$ . For the upper bounds, we may use the triangle inequality
$$
\left\| A \boldsymbol {x} ^ {(0)} + B \boldsymbol {u} ^ {\star} \right\| \leq A \left\| \boldsymbol {x} ^ {(0)} \right\| + | B | \| \boldsymbol {u} ^ {\star} \|.
$$
If we use (B.5) and (C.2), we get the upper bounds for  $A$  and  $B$ . We have  $\left\| \boldsymbol{x}^{(0)}\right\|_2 \leq 2\beta_0$  by (B.1), and the  $\ell_{\infty}$ -norm of  $\boldsymbol{x}^{(0)}$  is controlled through (B.2). These finish the proof for the upper bounds.
From the definition of  $B$ , we have  $B(\pmb{u}^{\star \top}\pmb{x}^{(0)})\geq 0$ , and
$$
\begin{array}{l} \left\| A \boldsymbol {x} ^ {(0)} + B \boldsymbol {u} ^ {\star} \right\| _ {2} ^ {2} = A ^ {2} \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} ^ {2} + B ^ {2} + 2 A B (\boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)}) \\ \geq A ^ {2} \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} ^ {2} + B ^ {2} \\ \geq \frac {1}{4} \left(A \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} + | B |\right) ^ {2}. \\ \end{array}
$$
(B.1) and (B.4) together with the lower bound in (C.2) give the desired lower bound for  $\| \widetilde{x}\| _2$ . The lower bound for  $\ell_{\infty}$ -norm is directly implied from Lemma B.2 together with (B.4) and (C.2).
With Lemma C.1, we have
$$
\frac {1}{8} \frac {1}{\sqrt {\log n}} \frac {(1 + \eta \lambda^ {\star}) ^ {T _ {2} ^ {\prime}}}{\sqrt {n}} \beta_ {0} \leq \frac {1}{8} \frac {1}{\sqrt {\log n}} \left(1 + \frac {(1 + \eta \lambda^ {\star}) ^ {T _ {2} ^ {\prime}}}{\sqrt {n}}\right) \beta_ {0} \leq \left\| \widetilde {\boldsymbol {x}} ^ {(T _ {2} ^ {\prime})} \right\| _ {2} \leq \frac {\sqrt {\lambda^ {\star}}}{8 \sqrt {\log n}},
$$
$$
T _ {2} ^ {\prime} \leq \frac {1}{\log (1 + \eta \lambda^ {\star})} \log \frac {\sqrt {\lambda^ {\star} n}}{\beta_ {0}} \leq \frac {1 1 \log n}{\log (1 + \eta \lambda^ {\star})} \leq \frac {6 4 \log n}{\eta \lambda^ {\star}}.
$$
For  $t \leq T_1$  where  $(1 + \eta \lambda^{\star})^{t}$  is not big, the bounds in Lemma C.1 are simplified to
$$
\frac {1}{\sqrt {\log n}} \beta_ {0} \lesssim \left\| \widetilde {\boldsymbol {x}} ^ {(t)} \right\| _ {2} \lesssim \sqrt {\log n} \beta_ {0}, \tag {C.3}
$$
$$
\frac {1}{\sqrt {\log n}} \frac {\beta_ {0}}{\sqrt {n}} \lesssim \left\| \widetilde {\boldsymbol {x}} ^ {(t)} \right\| _ {\infty} \lesssim \sqrt {\log n} \frac {\beta_ {0}}{\sqrt {n}}. \tag {C.4}
$$
After  $\widetilde{\pmb{x}}^{(t)}$  becomes almost parallel to  $\pmb{u}^{\star}$  and before  $T_{2}^{\prime}$ , we could approximate  $\widetilde{\beta}_{t}$  as increasing with the rate  $(1 + \eta \lambda^{\star})$ . However, after  $T_{2}^{\prime}$ , this approximation is invalid, and  $\widetilde{\beta}_{t}$  grows at a slower rate as it increases and it eventually converges to  $\sqrt{\lambda^{\star}}$ . How much iterations will be required for it to reach  $\sqrt{\lambda^{\star}}\sqrt{1 - \frac{1}{\log n}}$  after  $T_{2}^{\prime}$ ? With Lemma C.2, we will prove that  $O(\log \log n)$  iterations are required after  $T_{2}^{\prime}$ .
Lemma C.2. At  $t = T_2' + \frac{6\log\log n}{\log(1 + \eta\lambda^\star)}$ , we have  $\widetilde{\beta}_t^2 \geq \lambda^\star \left(1 - \frac{1}{\log n}\right)$ .
Proof. From the decomposition (C.1), we have
$$
\left| \left| \left| \tilde {\boldsymbol {x}} ^ {(t)} \right| \right| _ {2} - \left| B ^ {(t)} \right| \right| \leq \left| \left| \boldsymbol {x} ^ {(0)} \right| \right| _ {2}
$$
$$
\left| \left| \boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)} \right| - \left| B ^ {(t)} \right| \right| \leq \left| A ^ {(t)} \right| \left| \boldsymbol {u} ^ {\star \top} \boldsymbol {x} ^ {(0)} \right| \leq \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2},
$$
$$
\left| \widetilde {\alpha} _ {t} - \widetilde {\beta} _ {t} \right| \leq 2 \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} \leq \frac {\sqrt {\lambda^ {\star}}}{3 \log^ {2} n}. \tag {C.5}
$$
holds for all  $t$ . Because  $\widetilde{\beta}_t \gtrsim \frac{1}{\log n}$  for all  $t \geq T_2'$ , (C.5) implies that  $\widetilde{\beta}_t$  is well approximated by  $\widetilde{\alpha}_t$ . Hence, we will focus on  $\widetilde{\alpha}_t$ , which is an increasing sequence that evolves with
$$
\widetilde {\alpha} _ {t + 1} = (1 - \eta \widetilde {\beta} _ {t} ^ {2} + \eta \lambda^ {\star}) \widetilde {\alpha} _ {t}.
$$
For all  $i \geq 1$ , let  $N_i$  be the last  $t$  such that  $\lambda^{\star} - \widetilde{\alpha}_t^2 \geq \frac{\lambda^{\star}}{e^i}$ . Then, we have
$$
\lambda^ {\star} - \widetilde {\alpha} _ {N _ {i}} ^ {2} \geq \frac {\lambda^ {\star}}{e ^ {i}} > \lambda^ {\star} - \widetilde {\alpha} _ {N _ {i} + 1} ^ {2}. \tag {C.6}
$$
Let  $i \geq 2$ . For all  $N_{i-1} < t \leq N_i$ ,
$$
\frac {\widetilde {\alpha} _ {t + 1}}{\widetilde {\alpha} _ {t}} = 1 - \eta \widetilde {\beta} _ {t} ^ {2} + \eta \lambda^ {\star} = 1 + \eta (\lambda^ {\star} - \widetilde {\alpha} _ {t} ^ {2}) + \eta (\widetilde {\alpha} _ {t} ^ {2} - \widetilde {\beta} _ {t} ^ {2}) \geq 1 + \frac {\eta \lambda^ {\star}}{e ^ {i}} - \frac {\eta \lambda^ {\star}}{\log^ {2} n} \geq 1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e ^ {i}}.
$$
We used (C.5), (C.6), and the fact that  $\widetilde{\alpha}_t, \widetilde{\beta}_t \leq \sqrt{\lambda^\star}$  for all  $t$ . This implies
$$
\left(1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e ^ {i}}\right) ^ {N _ {i} - N _ {i - 1} - 1} x _ {N _ {i - 1} + 1} \leq x _ {N _ {i}}.
$$
From the lower and upper bounds provided by (C.6), we have
$$
\sqrt {\lambda^ {\star}} \sqrt {1 - \frac {1}{e ^ {i - 1}}} \left(1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e ^ {i}}\right) ^ {N _ {i} - N _ {i - 1} - 1} \leq \sqrt {\lambda^ {\star}} \sqrt {1 - \frac {1}{e ^ {i}}},
$$
$$
\left(1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e ^ {i}}\right) ^ {2 (N _ {i} - N _ {i - 1} - 1)} \leq \frac {e ^ {i} - 1}{e ^ {i}} \frac {e ^ {i - 1}}{e ^ {i - 1} - 1} = \frac {e ^ {i - 1} - \frac {1}{e}}{e ^ {i - 1} - 1} \leq 1 + \frac {1}{e ^ {i - 1}}.
$$
Taking  $\log$  on both sides and using the inequality  $\frac{1}{2} x < \log (1 + x) < x$  that holds for  $0 < x < 1$ , we get
$$
N _ {i} - N _ {i - 1} \leq 1 + \frac {1}{2} \frac {\log \left(1 + \frac {1}{e ^ {i - 1}}\right)}{\log \left(1 + 0 . 9 9 \frac {\eta \lambda^ {\star}}{e ^ {i}}\right)} \leq 1 + \frac {e}{0 . 9 9 \eta \lambda^ {\star}}.
$$
For  $t \leq N_1$ , we have
$$
\frac {\widetilde {\alpha} _ {t + 1}}{\widetilde {\alpha} _ {t}} \geq 1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e},
$$
$$
\sqrt {\lambda^ {\star}} \sqrt {1 - \frac {1}{e}} \geq \alpha_ {N _ {1}} \geq \left(1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e}\right) ^ {N _ {1}} \widetilde {\alpha} _ {T _ {2} ^ {\prime}} = \left(1 + 0. 9 9 \frac {\eta \lambda^ {\star}}{e}\right) ^ {N _ {1}} \sqrt {\frac {\lambda^ {\star}}{2 1 \log n}}.
$$
Taking log on both sides we get
$$
N _ {1} \leq 3 \frac {\log \log n}{\eta \lambda^ {\star}}.
$$
$$
\begin{array}{l} N _ {\log \log n + 1} + 1 \leq \left(1 + \frac {e}{0 . 9 9 \eta \lambda^ {\star}}\right) \log \log n + N _ {1} + 1 \\ \leq \left(1 + \frac {e}{0 . 9 9 \eta \lambda^ {\star}}\right) \log \log n + 3 \frac {\log \log n}{\eta \lambda^ {\star}} + 1 \\ \leq \frac {6 \log \log n}{\log (1 + \eta \lambda^ {\star})}, \\ \end{array}
$$
but at  $t = N_{\log \log n + 1} + 1$ , it holds that
$$
\widetilde {\alpha} _ {t} ^ {2} > \lambda^ {\star} \left(1 - \frac {1}{e \log n}\right),
$$
$$
\widetilde {\beta} _ {t} ^ {2} > \lambda^ {\star} \left(1 - \frac {1}{\log n}\right)
$$
as desired. Note that  $\widetilde{\beta}_t$  is also an increasing sequence as  $\widetilde{\alpha}_t$ .
It is implied from Lemma C.2 that  $T_{2} \leq \frac{1}{\log(1 + \eta\lambda^{\star})}\log \frac{\sqrt{\lambda^{\star}n}}{\beta_{0}} + \frac{6\log\log n}{\log(1 + \eta\lambda^{\star})} = (1 + o(1))\frac{1}{\eta\lambda^{\star}}\log \frac{\sqrt{\lambda^{\star}n}}{\beta_{0}}.$
The following corollary shows that  $\widetilde{\pmb{x}}^{(t)}$  is sufficiently close to  $\pmb{x}^{\star}$  at  $t = T_{2}$ .
Corollary C.3. At  $t = T_2$ , we have
$$
\min  \left\{\left\| \widetilde {\boldsymbol {x}} ^ {(t)} - \boldsymbol {x} ^ {\star} \right\| _ {2}, \left\| \widetilde {\boldsymbol {x}} ^ {(t)} + \boldsymbol {x} ^ {\star} \right\| _ {2} \right\} \lesssim \frac {1}{\sqrt {\log n}} \| \boldsymbol {x} ^ {\star} \| _ {2}, \tag {C.7}
$$
$$
\min  \left\{\left\| \widetilde {\boldsymbol {x}} ^ {(t)} - \boldsymbol {x} ^ {\star} \right\| _ {\infty}, \left\| \widetilde {\boldsymbol {x}} ^ {(t)} + \boldsymbol {x} ^ {\star} \right\| _ {\infty} \right\} \lesssim \frac {1}{\sqrt {\log n}} \| \boldsymbol {x} ^ {\star} \| _ {\infty}. \tag {C.8}
$$
Proof. When  $B^{(t)} > 0$ , from the decomposition
$$
\boldsymbol {x} ^ {(t)} - \boldsymbol {x} ^ {\star} = A ^ {(t)} \boldsymbol {x} ^ {(0)} + (B ^ {(t)} - \widetilde {\beta} _ {t}) \boldsymbol {u} ^ {\star} + (\widetilde {\beta} _ {t} - \sqrt {\lambda^ {\star}}) \boldsymbol {u} ^ {\star},
$$
$$
\left\| \boldsymbol {x} ^ {(t)} - \boldsymbol {x} ^ {\star} \right\| _ {2} \leq 2 \left\| \boldsymbol {x} ^ {(0)} \right\| _ {2} + \frac {\sqrt {\lambda^ {\star}}}{\sqrt {\log n}} \leq \frac {2 \sqrt {\lambda^ {\star}}}{\sqrt {\log n}} = \frac {2}{\sqrt {\log n}} \| \boldsymbol {x} ^ {\star} \| _ {2}.
$$
For the cases  $B^{(t)} < 0$  and  $\ell_{\infty}$ -norm, we may use similar technique.
D Phase I
D.1 Proof of Lemma 5.2