Refining uniform approximation algorithm for low-rank Chebyshev embeddings

Proof

Let us assume the opposite, that sign(D^ijDiD^kjDk)=−1. Since D^ij=(−1)i−j+1D^ji, we have

Let sign(DkD^ji)=δ, where δ = ± 1. Then

Since V is a Chebyshev matrix, the vector vⁱ can be uniquely represented as

where z₁ belongs to the linear hull of v¹, …, v^r+1, which do not include vⁱ, v^j, and v^k. We also represent

where z₂ belongs to the linear hull of v¹, …, v^r+1, v^r+2, which do not include vⁱ, v^j, and v^k.

From (A.1) and the properties of the determinant we obtain that

Similarly, from (A.2) and the properties of the determinant, we can deduce that

Hence,

Taking the signs from both sides we get that

whence

Note that the coefficients in (A.1) can be expressed by Cramer's formulas, and due to V being the Chebyshev matrix, we have α, γ ≠ 0. Then we can express the vector v^k from (A.1) as

Then

From (A.1) and (A.2) we have

hence

Note that βz₁ + z₂ is a linear combination of vectors v¹, …, v^r+1, v^r+2, which do not include vⁱ, v^j, and v^k. Since from the arguments above βγ ≠ 0, on the right-hand side of (A.4), there is a non-trivial linear combination of r linearly independent vectors, whence βγv^k + βz₁ + z₂ ≠ 0. Therefore, 1 – αβ ≠ 0. Then

and we can derive that

so we have from (A.3) and (A.5) that

Taking the signs from both sides we get that

therefore,

which is the contradiction.□

Proof

Let j = k. Then the condition of the lemma cannot be fulfilled, since

Let i = j ≠ k. Then the condition of the lemma can be written as

which by the definitions of D^ii and ŵⁱ writes as

Since the signs in the sequence

alternate,

Let us multiply the last two equations and get that

Since D^ik=(−1)i−k+1D^ki, we have that

It remains to note that w^kk=wr+2,w^ik=wi, and D^kk=Dk, whence

Thus, the first part of the lemma is proved. The second part can be obtained from the first one by rearranging the factors.

Let us prove the third part. Let i, j, and k be pairwise distinct. Then the condition of the lemma can be written as

Let us prove that in this case

On the contrary, let

Multiplying (A.6) and

we get that

Similarly, multiplying (A.7) and

we get that

It remains to note that (A.8) and (A.9) contradict Lemma A.1, whence

If we multiply (A.6) and (A.10), we get that

whence

Due to D^ki=(−1)k−i+1D^ik and D^kj=(−1)j−k+1D^jk, we get that

It remains to note that (A.10) and (A.11), up to notation, correspond to the statement of the lemma.□

Proof

Let us prove the lemma by induction. Let r = 2. If s₁₁ s₁₂ = –1, then the first row has the alternating signs. Otherwise, s₁₁s₁₂ = 1 and the condition of the lemma is fulfilled for i = 1, j = 1, k = 2. Then

and the second row has alternating signs.

Let us assume the statement is true for r – 1 and prove it for the matrix of size r × r. If the condition of the lemma is fulfilled for a matrix, it is also fulfilled for its principal submatrix. By the induction hypothesis, it contains a row t with the alternating signs. If s_t,r–1s_t,r = –1, then the signs alternate in the row t in the whole matrix. Let s_t,r–1s_t,r = 1. Then, due to the alternation of signs in the row t, we have

Then for i = t we have

and for i ≠ t, we have

so

which corresponds to the alternance of signs in the last row.□