First-order methods almost always avoid strict saddle points


We establish that first-order methods avoid strict saddle points for almost all initializations. Our results apply to a wide variety of first-order methods, including (manifold) gradient descent, block coordinate descent, mirror descent and variants thereof. The connecting thread is that such algorithms can be studied from a dynamical systems perspective in which appropriate instantiations of the Stable Manifold Theorem allow for a global stability analysis. Thus, neither access to second-order derivative information nor randomness beyond initialization is necessary to provably avoid strict saddle points.

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    This line of work assumes that f is a random function with a specific distribution.

  2. 2.

    For the purposes of this paper, strict saddle points include local maximizers.

  3. 3.

    The determinant is invariant under similarity transformations, so is independent of the choice of basis.


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This paper extends upon the special case of gradient descent dynamics developed in the conference proceedings of the authors [32, 42].

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Proof of Claim 4


We assume that \(\alpha |e_j^{T}Hz_j| < \delta \left\| z_j \right\| _2\) for all \(j \in \{1,\ldots ,n\}\), for some \(\delta \) to be chosen later. For the base case \(j=2\), it holds that \(\left\| y_t-z_2 \right\| _2 = \left\| z_1-z_2 \right\| _2 = \alpha |e_1^{T}Hz_1|< \delta \left\| z_1 \right\| _2 < 2\delta \left\| y_t \right\| _2\) and \(\left\| z_2 \right\| _2 < (1+2\delta )\left\| y_t \right\| _2\). Suppose for \(j \ge 2\) that \(\left\| y_t-z_{j} \right\| _2 < 2(j-1)\delta \left\| y_t \right\| _2\) and thus \(\left\| z_j \right\| _2 < [1+2(j-1)\delta ] \left\| y_t \right\| _2.\) Using induction and triangle inequality we get

$$\begin{aligned} \left\| y_t - z_{j+1} \right\| _2&\le \left\| y_t - z_{j} \right\| _2+ \left\| z_{j} -z_{j+1} \right\| _2\\&= 2(j-1)\delta \left\| y_t \right\| _2 + \alpha |e_j^{T}Hz_j| \\&< 2(j-1)\delta \left\| y_t \right\| _2 + \delta \left\| z_j \right\| _2 \\&< 2(j-1)\delta \left\| y_t \right\| _2+ \delta [1+2(j-1)\delta ] \left\| y_t \right\| _2\\&\le 2j\delta \left\| y_t \right\| _2, \end{aligned}$$

where we assume \(\delta < \frac{1}{2n}\) so that \(2(j-1)\delta <1\) for all \(j \in [n]\). Using the above calculation,

$$\begin{aligned} \alpha |e_i^{T}Hy_t|&< \alpha |e_i^{T}Hz_i|+\alpha |e_i^{T}H(y_t - z_i)| \\&< \delta \left\| z_i \right\| _2+ \alpha \left\| He_i \right\| _2\left\| y_t - z_i \right\| _2\\&< \delta \big (1 +2(i-1)\delta \big ) \left\| y_t \right\| _2 + \alpha \left\| He_i \right\| _2 \big (2(i-1)\delta \big ) \left\| y_t \right\| _2\\&\le \delta \big ( 1+2n \delta + 2n \alpha L\big )\left\| y_t \right\| . \end{aligned}$$

Thus \(\alpha \left\| Hy_t \right\| _2 < \sqrt{n}\delta \big ( 1+2n \delta + 2n \alpha L\big ) \left\| y_t \right\| _2\), and

$$\begin{aligned} \sigma _{\min ^+} (H) \left\| y_t \right\| _2 \le \left\| Hy_t \right\| _2 < \frac{\sqrt{n}}{\alpha }\delta \big ( 1+2n \delta + 2n \alpha L\big ) \left\| y_t \right\| _2, \end{aligned}$$

where \(\sigma _{\min ^+}\) is the smallest non-zero singular value of H. Thus by choosing \(\delta \) small enough such that

$$\begin{aligned} \sigma _{min^+} (H) \ge \frac{\sqrt{n}}{\alpha }\delta \big ( 1+2n\delta + 2n\alpha L\big ), \end{aligned}$$

we have obtained a contradiction. \(\square \)

Proof of Proposition 7


Let \(H= \nabla ^2 f(x^*)\), \(J =\mathrm {D}g(x^*) = \prod _{i=1}^b (I- \alpha P_{S_{b-i+1}} H)\), and \(y_0\) be an eigenvector of the Hessian at \(x^*\).

The proof technique is very similar to that of the proof of Proposition 5. We shall prove that \(\left\| J^t y_0 \right\| _2 \ge c(1+\eta )^t \). Hence by Gelfand’s theorem J must have at least one eigenvalue with magnitude greater than one.

We fix some arbitrary iteration t and let \(y_t = J^t y_0\). We will first show that there exists an \(\epsilon >0\),

$$\begin{aligned} y_{t+1}^{T} Hy_{t+1} \le (1+\epsilon ) y_{t}^{T}Hy_{t}, \end{aligned}$$

for all \(t\in {\mathbb {N}}\). Let \(z_1 = y_t\) and \(z_{i+1} = (I- \alpha P_{S_i} H)z_i = z_i - \alpha \sum _{j \in S_i}(e_j^{T}Hz_i)e_j \), so that \(y_{t+1} = Jy_t = z_{b+1}\). We get that

$$\begin{aligned}&z_{i+1}^{T}Hz_{i+1} = \left( z_i^{T} - 2 \alpha \sum _{j \in S_i}(e_j^{T}Hz_i)e_j^{T}\right) H \left( z_i - \alpha \sum _{j \in S_i}(e_j^{T}Hz_i)e_j\right) \\&= z_i^{T}Hz_i -2\alpha \sum _{j \in S_i}(e_j^{T} H z_i)^2 + \alpha ^2 \left( \sum _{j \in S_i}(e_j^{T}Hz_i)e_j\right) ^{T} H \left( \sum _{j \in S_i}(e_j^{T}Hz_i)e_j \right) \\&< z_i^{T}Hz_i - 2\alpha \sum _{j \in S_i}(e_j ^{T} H z_i)^2 + \alpha ^2 L_{b} \left\| \sum _{j \in S_i}(e_j^{T}Hz_i)e_j \right\| _2^2 \quad { \text { (using } \left\| H_{S_i} \right\| _2 \le L_b)}\\&= z_i^{T}Hz_i - \alpha (2 -\alpha L_b )\left\| \sum _{j \in S_i}(e_j^{T}Hz_i)e_j \right\| _2^2 \\&\le z_i^{T}Hz_i - \alpha \left\| \sum _{j \in S_i}(e_j^{T}Hz_i)e_j \right\| _2^2 \quad { \text {(using } \alpha L_b <1)}. \end{aligned}$$

Thus \(z_{i}^T H z_i \) is a decreasing (non-increasing) sequence.

We shall prove that there exists an \(i \in [b]\) so that \(z_{i+1}^{T}Hz_{i+1} \le (1+\delta ) z_{i}^{T}Hz_{i}\) for some global constant \(\delta \) to be chosen later.


Let \(y_t\) be in the range of H. There exists an \(i \in [b]\) so that \(\alpha \sum _{j \in S_i} \left| e_j^{T}Hz_i\right| \ge \delta \left\| z_i \right\| _2\) for some \(\delta >0\).

To finish the proof of the lemma, suppose that Claim B applies. Then by Cauchy–Schwarz, there exists an index i such that

$$\begin{aligned} z_{i+1}^{T}Hz_{i+1}< & {} z_i^{T}Hz_i - \alpha \sum _{j \in S_i}(z_i^{T}He_j)^2 \\< & {} z_i^{T}Hz_i - \frac{\alpha }{n} \left( \sum _{j \in S_i} \left| z_i^{T}He_j\right| \right) ^2 < z_i ^{T}Hz_i - \frac{\delta ^2}{n\alpha } \left\| z_i \right\| ^2_2. \end{aligned}$$

However, \(w^{T}Hw \ge \lambda _{\min }(H)\left\| w \right\| ^2_2 \ge - L \left\| w \right\| ^2_2\), hence we get that

$$\begin{aligned} z_{i+1}^{T}Hz_{i+1} < \left( 1+ \frac{\delta ^2}{\alpha L n}\right) z_i ^{T}Hz_i . \end{aligned}$$

By choosing \(\epsilon =\frac{\delta ^2}{\alpha L n}\) we showed that \(y_{t+1}^{T}Hy_{t+1} \le (1+\epsilon )y_t^{T}Hy_t\) as long as \(y_t\) is in the range of H.

Assume that \(y_t = y_{{\mathcal {N}}} + y_{{\mathcal {R}}}\). It is easy to see \(y_t^{T}Hy_t = y_{{\mathcal {R}}}^{T}Hy_{{\mathcal {R}}}\) and also \(y_{t+1} = Jy_t = y_{{\mathcal {N}}} + Jy_{{\mathcal {R}}}\), hence \(y_{t+1}^{T}Hy_{t+1} =(Jy_{{\mathcal {R}}})^{T} H (Jy_{{\mathcal {R}}})\). Therefore from Inequality (14) proved above, if the starting vector is \(y_{{\mathcal {R}}}\), which Claim B applies too, then \((Jy_{{\mathcal {R}}})^{T}HJy_{{\mathcal {R}}} \le (1+\epsilon )y_{{\mathcal {R}}}^{T}H y_{{\mathcal {R}}} = (1+\epsilon )y_t^{T}Hy_t\).

To sum up, we showed that \(y_t^{T}Hy_t \le (1+\epsilon )^t y_0^{T} H y_0\) and since \(y_0\) is an eigenvector of H (of norm one) with corresponding negative eigenvalue \(\lambda \), it follows that \(y_t^{T}Hy_t \le \lambda (1+\epsilon )^t.\) Finally using \(y_t^{T}Hy_t \ge \lambda _{\min }(H)\left\| y_t \right\| ^2_2\), we get \(\left\| y_t \right\| _2 \ge (1+\epsilon )^{t/2} \frac{\lambda }{\lambda _{\min }(H)}\). Observe that \(\frac{\lambda }{\lambda _{\min }(H)}>0\) is a positive constant, \((1+\epsilon )^{t/2} \ge (1+\epsilon /4)^t\) (since \(\epsilon \le 1/2\)) and the proof follows (the parameters as claimed in the beginning will be \(c = \frac{\lambda }{\lambda _{\min }(H)}\) and \(\eta = \epsilon /4\)). \(\square \)

Proof of Claim B

We assume that \(\alpha \sum _{j \in S_i}\left| e_j^{T}Hz_i \right| < \delta \left\| z_i \right\| _2\) for all \(i \in [b]\). For base case \(i=2\), it holds that \(\left\| y_t-z_2 \right\| _2 = \left\| z_1-z_2 \right\| _2 = \alpha |\sum _{j \in S_1}e_j^{T}Hz_1|< \alpha \sum _{j \in S_1} |e_j^{T}Hz_1|< \delta \left\| z_1 \right\| _2 < 2\delta \left\| y_t \right\| _2\) and \(\left\| z_2 \right\| _2 < (1+2\delta )\left\| y_t \right\| _2\). Suppose for \(i \ge 2\) that \(\left\| y_t-z_{i} \right\| _2 < 2(i-1)\delta \left\| y_t \right\| _2\) and thus \(\left\| z_i \right\| _2 < [1+2(i-1)\delta ] \left\| y_t \right\| _2.\) Using induction and triangle inequality we obtain

$$\begin{aligned} \left\| y_t - z_{i+1} \right\| _2&\le \left\| y_t - z_{i} \right\| _2+ \left\| z_{i} -z_{i+1} \right\| _2\\&= 2(i-1)\delta \left\| y_t \right\| _2 + \alpha \left| \sum _{j \in S_i}e_j^{T}Hz_i\right| \\&\le 2(i-1)\delta \left\| y_t \right\| _2 + \alpha \sum _{j \in S_i}\left| e_j^{T}Hz_i\right| \\&< 2(i-1)\delta \left\| y_t \right\| _2 + \delta \left\| z_i \right\| _2 \\&< 2(i-1)\delta \left\| y_t \right\| _2+ \delta [1+2(i-1)\delta ] \left\| y_t \right\| _2 \\&\le 2i\delta \left\| y_t \right\| _2, \end{aligned}$$

where we assume \(\delta < \frac{1}{2b}\) so that \(2(i-1)\delta <1\) for all \(i \in [b]\). Using the above,

$$\begin{aligned} \alpha \sum _{j \in S_i} \left| e_j^{T}Hy_t\right|&< \alpha \sum _{j \in S_i}\left| e_j^{T}Hz_i \right| +\alpha \sum _{j \in S_i}\left| e_j^{T}H(y_t - z_i)\right| \\&< \delta \left\| z_i \right\| _2+ \alpha \left( \sum _{j \in S_i}\left\| He_j \right\| _2 \right) \left\| y_t - z_i \right\| _2 \\&< \delta [1+2(i-1)\delta ]\left\| y_t \right\| _2 + \alpha [ 2(i-1)\delta ] \left\| y_t \right\| _2 \left( \sum _{j \in S_i}\left\| He_j \right\| _2 \right) . \end{aligned}$$

Since \(\left\| He_i \right\| _2 < \sigma _{\max }(H) \le L\), we get that \(\alpha \sum _{j \in S_i} \left\| He_j \right\| _2 < |S_i| \le n\) and we conclude

$$\begin{aligned} \alpha \sum _{j \in S_i}\left| e_j^{T}Hy_t\right| < 2n^2 \delta \left\| y_t \right\| _2. \end{aligned}$$

Finally, using Inequality (15) it follows that \(\alpha \left\| Hy_t \right\| _2 < 2n^2 \delta \sqrt{n} \left\| y_t \right\| _2\). Let \(w \in \text {Im}(H)\) be a vector that is orthogonal to \(\text {null}(H)\) (since H is symmetric). Then it holds that \(\left\| Hw \right\| _2 \ge \sigma _{\min ^+}(H) \left\| w \right\| _2\) where \(\sigma _{\min ^+}(H)\) denotes the smallest positive singular value of H (greater than zero). Assume that \(y_t \in \text {Im}(H)\) and we get \(\left\| Hy_t \right\| _2 < \frac{2n^2 \delta \sqrt{n}}{\alpha } \left\| y_t \right\| _2\). However, \(\left\| Hy_t \right\| _2 \ge \sigma _{\min ^+}(H) \left\| y_t \right\| _2\) thus by choosing \(\frac{2n^2 \sqrt{n}\delta }{\alpha } < \sigma _{\min ^+}(H)\) we reach a contradiction. \(\square \)

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Lee, J.D., Panageas, I., Piliouras, G. et al. First-order methods almost always avoid strict saddle points. Math. Program. 176, 311–337 (2019).

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  • Gradient descent
  • Smooth optimization
  • Saddle points
  • Local minimum
  • Dynamical systems

Mathematics Subject Classification

  • 90C26