The Polar Express: Optimal Matrix Sign Methods and their Application to the Muon Algorithm

• p. 16 Implementation 1 code

  X = X / (X.norm(dim=(-2, -1), keepdim=True) * 1.01 +1e-7)

• p. 7 Algorithm 1 line 10

  Set X0 = M/(∥M∥F + 10−2).

• p. 8 Section 3.5

  to M/(∥M∥F + 10−2).

Only the authors can say what is right here.

Judge's note: Checked on the p.16 code and the p.7 algorithm. App. G calls the +10^-2 normalisation 'the third change … reflected in line 10 of Algorithm 1'. The text says Implementation 1 'gives a Python implementation of the online stage of Algorithm 1 … which we use in our numerical experiments'. The code multiplies the norm by 1.01 and adds 1e-7, separately from the 1.01 safety factor already built into the coefficients. Nothing in the paper explains the difference.

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• p. 31 Figure 11(b) caption

  adamw (lr = 0.001): 4.197

• p. 9 Figure 4 caption

  adamw(lr =0.0005): 4.197

• p. 30 H.2 Additional Metrics

  The upper rows of each subfigure are identical to Figure 1 and Figure 4

Judge's note: Page 30 (H.2) states that the upper rows of Fig. 11 are identical to Figs 1 and 4. The other three methods have matching learning rates and losses in both captions. In both the Fig. 4 and Fig. 11b top-left panels, the AdamW minimum (about 4.2) is the second point, to the left of the 10^-3 tick, i.e. about 5e-4. The point at 10^-3 is higher. Fig. 13b also lists adamw at lr = 0.0005. The '0.001' in the Fig. 11b caption does not match the plot.

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• p. 25 Appendix F, paragraph after (19)

  is the [q/0] Padé approximant to (1 − x)^{1/2}

• p. 23 Appendix D proof

  [q/0] Padé-approximant to (1−x)^{−1/2}

• p. 25 Appendix F, general-degree limiting polynomial

  the [q/0] Padé approximant to (1 − x)^{1/2}

• p. 25 Appendix F, paragraph after (19)

  is the [q/0] Padé approximant to (1 −x)1/2

Judge's note: The page 25 image shows exponent 1/2 with no minus sign; page 23 shows −1/2. For q=2, the [2/0] approximant of (1−y)^{−1/2} is 1 + y/2 + 3y²/8, and x·h(1−x²) = (15x − 10x³ + 3x⁵)/8. This matches the polynomial named in the same paragraph. With (1−y)^{1/2} the result is (3x + 6x³ − x⁵)/8 instead. So the exponent sign on page 25 is wrong.

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• p. 25 Appendix F, inequality chain

  ≤ 2 (1 − ℓ/u)^3 .

• p. 25 Appendix F, displayed bound preceding Algorithm 2

  ≤ 2 (1 − ℓ/u)³

• p. 25 Appendix F, displayed chain bounding max|p⋆−p|

  ≤2 (1 −ℓ/u)3 .

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Judge's note: Page 25 image. With s = x/u, 1−p(s) = (1−s)³(3s²+9s+8)/8, which is ≈ 2.5(1−s)³ near s = 1. At s = 0.9, p = 0.997684, so 1−p = 0.002316 > 0.001 = (1−0.9)³. The cited Kenney–Laub bound is stated in terms of 1−s², not 1−s. A valid replacement is 2.5(1−ℓ/u)³, which gives about 5ε_double rather than 2ε_double. The qualitative conclusion (agreement near machine precision) survives, but the printed constant does not follow.

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• p. 34 Figure 15 caption and left panel

  Adamw (lr = 0.01): 0.643

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Judge's note: In the left panel of Fig. 15 (page 34), the blue AdamW curve peaks at about 0.643 at the 10^-3 tick. It then falls to about 0.62 near 10^-2 and about 0.45 at 10^-1. The caption's learning rate of 0.01 does not match the plotted maximum; 0.001 appears to be meant. This verdict rests on reading the plot, so the authors should confirm.

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• p. 7 Section 3.4

  Another issue, identified by Nakatsukasa & Higham (2013), is due to the non-monotonicity of pt.

• p. 26 Appendix G

  The second issue was identified in Nakatsukasa & Higham (2012)

• p. 7 Section 3.4

  Nakatsukasa & Higham (2013), is due to the non-monotonicity of pt.

• p. 7 Section 3.4

  identified by Nakatsukasa & Higham (2013), is due to the non-monotonicity

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Judge's note: Checked on the p.7 and p.26 images. The reference list has two different entries: 2012 is 'Backward stability of iterations for computing the polar decomposition' and 2013 is 'Stable and efficient spectral divide and conquer algorithms...'. App. G cites 2012 both for this issue and for the p_t(σ_i)/σ_i condition, so one of the two attributions for the same claim is inconsistent. The authors should confirm which entry is meant.

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• p. 31 Figure 11(b) bottom-right panel / caption

  Bottom right: training loss vs. time for best learning rate.

Judge's note: On page 31, the bottom-right panel of Fig. 11b is labelled 'Validation Loss' on the y-axis. The matching panels of Fig. 11a and Fig. 13 are labelled 'Loss', which fits the caption. Either the axis label or the plotted quantity is wrong.

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• p. 23 Appendix D proof

  ≤ |1 − ℓ²|^{(d+1)^T}

• p. 22 Theorem 3.3, Eq. (12)

  ≤ |1 − ℓ²|^{(q+1)^T}

• p. 23 Appendix D, displayed convergence proof

  |1 − x²|^{(d+1)^T}/(1 + f(x))

• p. 23 Appendix D, proof of Theorem 3.3, displayed chain

  ≤ |1 − ℓ2|(d+1)T

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Judge's note: Checked the page 22–23 images. The [q/0] Padé iteration has order q+1, so exponent (d+1)^T = (2q+2)^T is too strong. Counterexample to the third line (q=1, T=1, ℓ=1/2): f=(3x−x³)/2 and f(1/2)=11/16, so max|1−f| = 5/16. The RHS is maximised at x=1/2 and gives (3/4)^4/(27/16) = 3/16 < 5/16. Counterexample to the last line (ℓ=0.9, d=3, closed form (17)): 1−p⋆(0.9) ≈ 0.00208 > 0.19^4 ≈ 0.0013. With (q+1)^T the chain holds: (3/4)^2/(27/16) = 1/3 ≥ 5/16. The stated theorem (12) is therefore unaffected and the proof needs q in place of d. The proof also writes sign(M) where the theorem has polar(M).

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• p. 20 Proof of Lemma C.2

  We first rule out the trivial case that p⋆ ≠ 0

• p. 20 Proof of Lemma C.2, first paragraph

  We first rule out the trivial case that p⋆ ̸= 0, since p(x) = 2/(ℓ+u) x would then be a better approximation.

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Judge's note: The p.20 image shows ≠. My reworking: the zero polynomial has error max|1−0| = 1. The polynomial 2x/(ℓ+u) has error (u−ℓ)/(u+ℓ) < 1. So the case excluded is p⋆ = 0, which matches the next sentence: 'Hence, p⋆ cannot be the zero polynomial.' The step is wrong as printed, but the lemma still holds.

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• p. 20 Proof of Lemma C.2, Step 1

  be the number of stationary points and n_stat be the number of endpoints

• p. 20 Proof of Lemma C.2, Step 1

  Let next be the number of stationary points and nstat be the number of endpoints

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Judge's note: Checked on the p.20 image. n_ext is never used again. As defined, 'n_stat ≤ q' would bound the number of endpoints by q, which is not what the preceding paragraph proves. With n_end = endpoints (≤2) and n_stat = stationary points (≤q), the counting q+2 ≤ n_end + n_stat ≤ q+2 is valid, so the lemma stands.

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• p. 21 Proof of Theorem 3.1, arbitrary competing composition

  We will prove that a/b ≤ ℓT/uT by contradiction.

• p. 22 Proof of Theorem 3.1, range-ratio argument

  Hence, we must have a/b ≤ ℓT/uT.

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Judge's note: Checked pages 21–22. The proof places no sign restriction on f = p̃_{T−1}∘…∘p̃_1. Counterexample, d=1, T=2, [ℓ,u]=[1/4,1]: p1(x)=8x/5 (c=2/(ℓ+u)), image [2/5,8/5], so ℓ2/u2 = 1/4. f = −p1 is admissible, has image [−8/5,−2/5] and gives a/b = 4 > 1/4, so the claimed inequality is false. Its contradiction argument also breaks: with c = 2/(a+b) = −1, the true error is 3/5, but (b−a)/(a+b) = −3/5. The theorem still holds, because p̃_T can absorb a sign flip and ranges containing 0 give a/b ≤ 0. The step needs separate handling of negative ranges.

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• p. 22 Proof of Theorem 3.1, final line of chain

  = max_{x∈[ℓT,uT]} |1 − pT(g(x))|

• p. 22 Proof of Theorem 3.1, final line of the displayed inequality chain

  max over x ∈ [ℓT,uT] |1 − pT(g(x))|

• p. 22 Appendix C, proof of Theorem 3.1, last line of the displayed chain

  = max x∈[ℓT ,uT ] |1 −pT (g(x))| = 1 −pT (ℓT ) = 1 −ℓT +1

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Judge's note: Page 22 image: the previous line is min_p max_{x∈[ℓ,u]} |1−p(g(x))|, and the last line switches the domain to [ℓT,uT] while keeping g. Check with d=1, T=2, ℓ=1/4, u=1: g(x)=8x/5, ℓ2=2/5, u2=8/5, p2(x)=x (c = 2/(ℓ2+u2) = 1). At x=8/5 the printed expression is |1−64/25| = 39/25, but 1−ℓ3 = 3/5. So the equality fails as printed. With the corrected domain the theorem stands; this is an index slip.

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• p. 25 Algorithm 2, linear system

  (x(1) 2 )5

• p. 25 Algorithm 2, third row of the inverted matrix

  (x₂⁽¹⁾)⁵

• p. 24 Equation (19), corresponding interpolation entry

  (x₂⁽ᵏ⁾)⁵

• p. 24 Eq. (19)

  (x(k) 2 )5

Judge's note: The crop and page 25 image show (x_2^{(1)})^5 in the third row, while every other entry uses superscript (k). Eq. (19) on page 24 uses (x_2^{(k)})^5. As printed, for k ≥ 2 the system does not impose p_k(x_2^{(k)}) = 1 − E_k. This is an index slip with an evident fix. The ϵ_d subscript is a minor notational inconsistency. 1.1 vs 1.11 ×10^-16 is just rounding.

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• p. 29 Figure 9 caption

  Computations are performed in bfloat32.

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Judge's note: The page 29 image shows 'bfloat32' in the caption. Everywhere else the paper uses bfloat16: Section 4.2 says the GPT-2 runs use bfloat16, and Appendix J also mentions bfloat16. The intended precision could be float32 or bfloat16, and only the authors can say which.

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• p. 36 Appendix J, numerical issues paragraph

  Thus, QT ≈ V^⊤Σ^{−1}V

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Judge's note: The page 36 image shows V^⊤Σ^{−1}V. UΣV^⊤·VΣ^{−1}V^⊤ = UV^⊤, whereas V^⊤Σ^{−1}V does not satisfy this unless V is symmetric. Footnote 11 says Q_T → Y^{−1/2} = VΣ^{−1}V^⊤. This is a transposition slip; Algorithm 3 is unaffected.

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