Orthogonality Loss

[luckyxiaoqi]

I can't understand the sentence " Since the loss is a symmetric function of those eigenvalues, it can be rewritten as a fraction w.r.t. the coefficients of the characteristic polynomial of (I + A)T(I + A) by Viète’s theorem. " in your VTN paper,it's like a math's question.I thought a lot ,but can't find the answer.Could you tell me the reason.

[zsyzzsoft]

The eigenvalues are the roots of the characteristic polynomial. Then, for example, the sum of those eigenvalues is exactly the sum of roots of the characteristic polynomial, and thus -a_{n-1}/a_n by Viète’s formulas (suppose the characteristic polynomial can be written as P(x) = an x^n + a{n-1]^ x^{n-1} + ... + a_0). The sum of the inverse eigenvalues can be similarly calculated if you rewrite it as a single fraction term. Detailed formula can be found in the code.

[Algolzw]

@zsyzzsoft I'm confused about how to get the formula in Orthogonality Loss: ortho_loss = s1 + (1 + eps) * (1 + eps) * s2 / s3 - 3 * 2 * (1 + eps) , and what's the mean of the function elem_sym_polys_of_eigen_values? Singular values？

[zsyzzsoft]

s1, s2, s3 correpond to a+b+c, ab+bc+ca, abc respectively, where a, b, c are the eigenvalues of (I+A)^T(I+A). Then orthonagonality loss = -6 + (a + b + c) + (1/a + 1/b + 1/c) = s1 + s2 / s3 - 6.