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√ d for some

First note that we just need to choose the degree of our Chebyshev polynomial q ≥ c ∗ log

√ d is an integer

constant c. We will always choose q to be odd. If q is odd then (q − 1)/2 = Θ log

and 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 .

(1)

We can prove this via induction. For the base case when i = 1 we have:

def

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

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

(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 .

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

K = A1 Π, A3 Π, A5 Π, ..., Aq Π

(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 .

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.

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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 = T1+γ)α

. Recall that we have chosen q to be odd, so by our above claim about

q (1+γ)

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 x

ˆ ∈ Rqk is given by x

ˆ = [b1 x, b3 x, ..., bq x] .

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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