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We apologize if the following proof is overly formal – we wanted to err on the side of caution.
The claims about Chebyshev polynomials are well known, but we reprove from basic definitions for
completeness.
We mistakenly overloaded the definition of q in our paper, using it both in our pseudocode for
Algorithms 1 and 2 as an iteration count and then later as the degree of polynomial p1 (·). In this
note, take q to be the degree of p1 (·).
In the proof of our accuracy bounds, we just need to choose q ≥ c ∗

log
√d


for some constant c. We


√d
will always choose it to be odd. We run the Block Lanczos algorithm for (q − 1)/2 = Θ log

iterations. Note that since q is odd, this iteration count is an integer.
We have the following theorem:

Theorem 1. For any odd integer q, let K be the Krylov subspace computed by Algorithm 2 with
input A, random initialization matrix Π ∈ Rn×k , and iteration count (q − 1)/2. Let p1 (·) be the
degree q polynomial given in Lemma 5. Then:
range(p1 (A)Π) ⊆ range(K).


Proof. We have K = AΠ, (AAT )AΠ, (AAT )2 AΠ, ..., (AAT )(q−1)/2 AΠ . Writing A using
its SVD we claim that for any i ≥ 1,
def

(AAT )i A = UΣ2i+1 VT = A2i+1 .
We can prove this via induction. For the base case when i = 1 we have:
def

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And for any i > 1, assuming (1) holds for i − 1 by induction, we have:

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With equation (1) in hand we can rewrite K as:


K = A1 Π, A3 Π, A5 Π, ..., Aq Π

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(1)

(AAT )A = UΣVT VΣUT UΣVT = UΣIΣIΣVT = UΣ3 VT = A2·1+1 .

(AAT )i A = (AAT )(AAT )i−1 A = UΣVT VΣUT UΣ2(i−1)+1 VT
def

= UΣIΣIΣ2(i−1)+1 VT = UΣ2·i+1 VT = A2·i+1 .

(2)

We now claim that any Chebyshev polynomial of odd degree can be written as the sum of odd degree
monomials. Correspondingly, any even degree Chebyshev polynomial can be written as the sum of
even degree monomials. Formally, if i is an odd integer, Ti (x) can be written as:
a1 x + a3 x3 + a5 x5 + ... + ai xi
for some set of coefficients a1 , a3 , a5 , ..., ai . If i is even, Ti (x) can be written as:
a0 + a2 x2 + a4 x4 + ... + ai xi
for some set of coefficients a0 , a2 , a4 , ..., ai .
We prove this by the recursive definition of Chebyshev Polynomials (See our proof of Lemma 5 in
supplemental). In the base cases i = 0 and i = 1 we have T0 (x) = 1 and T1 (x) = x so the above
claim is clearly true. Now for even i ≥ 2, assume by way of induction that the claim holds for all
0 ≤ j < i. Since i − 1 is odd and i − 2 is even:
Ti (x) = 2xTi−1 (x) − Ti−2 (x)


= 2x a1 x + a3 x3 + ... + ai−1 xi−1 − a0 + a2 x2 + ... + ai−2 xi−2
= −a0 + (2a1 − a2 )x2 + ... + (2ai−3 − ai−2 )xi−2 + (2ai−1 )xi .
1

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This gives us the claim, since we have written Ti (x) as the sum of even degree monomials of x.
Similarly, for odd i ≥ 3, since i − 1 is even and i − 2 is odd,
Ti (x) = 2xTi−1 (x) − Ti−2 (x)


= 2x a0 + a2 x2 + ... + ai−1 xi−1 − a1 x + a3 x3 + ... + ai−2 xi−2
= (2a0 − a1 )x + (2a2 − a3 )x3 + ... + (2ai−3 − ai−2 )xi−2 + (2ai−1 )xi .
So we have written Ti (x) as the sum of odd degree monomials. Overall, we have the claim by
simultaneous induction on the odd and even integers.
Now, consider p1 (x), the polynomial given in Lemma 5. p1 (x) is defined as:
p1 (x) = (1 + γ)α

Tq (x/α)
= c · Tq (x/α)
Tq (1 + γ)

where we just set c = T(1+γ)α
. Note that Tq (1 + γ) is a constant – γ is a parameter that does
q (1+γ)
not depend on x. Recall that we have chosen q to be odd, so by our above claim about Chebyshev
polynomials, we can write p1 (x) as:

p1 (x) = c a1 (x/α) + a3 (x/α)3 + a5 (x/α)5 + ... + aq (x/α)q
= (ca1 /α)x + (ca3 /α3 )x3 + (ca5 /α5 )x5 + ... + (caq /αq )xq .
That is, we can write p1 (x) as the sum of odd degree monomials. For ease of notation define
bi = cai /αi and write: p1 (x) = b1 x + b3 x3 + ... + bq aq . So:
p1 (A)Π = Up1 (Σ)VT Π

= b1 UΣVT + b3 UΣ3 VT + ... + bq UΣq VT Π

= b1 AΠ + b3 A3 Π + ... + bq Aq Π .
For y ∈ Rn , y ∈ range(p1 (A)Π) iff we can write y = p1 (A)Πx for some x ∈ Rk . That is if
y ∈ range(p1 (A)Π) we can write:

y = b1 AΠ + b3 A3 Π + ... + bq Aq Π x


= AΠ, A3 Π, ..., Aq Π x
ˆ

T
where the vector x
ˆ ∈ Rqk is given by x
ˆ = b1 xT , b3 xT , ..., bq xT .
Using (2), we have y = Kˆ
x so y ∈ range(K). Since this holds for all y ∈ range(p1 (A)Π) we
have: range(p1 (A)Π) ⊆ range(K).

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