Just want to throw in that every Carleman matrix has a natural LU-decomposition into the Carleman matrix of a translation - which is lower triangular - and the Carleman matrix of the development without its constant coefficient - which is upper triangular.
In formulas \( f(x)=f_0 + f_1 x + f_2 x^2 + \dots \),
\( \tau(x)=f_0+x \)
\( g(x)=f_1 x + f_2 x^2 + \dots \)
\( f = \tau \circ g \)
\( C[f] = \underbrace{C[\tau]}_{L} \underbrace{C[g]}_{U} \).
Though however this does not help for finding a LU-decomposition of \( C-I \) (with first row removed). The most I can get:
\( C[f] - I = LU - I = (L-U^{-1})U \)
@Gottfried:
Be aware that not (C-I) (or its transpose) is inverted but (C-I) with the first row (or column) removed!
In formulas \( f(x)=f_0 + f_1 x + f_2 x^2 + \dots \),
\( \tau(x)=f_0+x \)
\( g(x)=f_1 x + f_2 x^2 + \dots \)
\( f = \tau \circ g \)
\( C[f] = \underbrace{C[\tau]}_{L} \underbrace{C[g]}_{U} \).
Though however this does not help for finding a LU-decomposition of \( C-I \) (with first row removed). The most I can get:
\( C[f] - I = LU - I = (L-U^{-1})U \)
@Gottfried:
Be aware that not (C-I) (or its transpose) is inverted but (C-I) with the first row (or column) removed!