(b) Let f V -» V be any linear map of vector spaces over a field K. Recall that, for any polynomial p(X) = 0 X e K[X] and any v E V, p(X) vp(v) = ef*(v). i-0 The kernel of p(X) is defined to be...

Extracted text: (b) Let f V -» V be any linear map of vector spaces over a field K. Recall that, for any polynomial p(X) = 0 X e K[X] and any v E V, p(X) vp(v) = ef*(v). i-0 The kernel of p(X) is defined to be Ker(p(X)) {v e V : p(X) v 0}. (b.1) Show that Ker(p(X)) is a linear subspace of V. When p(X) = X - A where E K, explain that Ker(P(X)) is the eigenspace of f with respect to A. (b.2) Let p(X) and q(X) be polynomials in K[X] so that gcd(p(X), q(X)) = 1. Show that Ker(p(X)q(X)) — Ker(p(X)) + Кer(4(х)) is a direct sum of subspaces. (Hint: Use the fact that if gcd(p(X), q(X)) = 1 then there exist a(X), Ь(X) € К|X] so that a(X)p(X) + b(X)q(X) %3D 1.) (b.3) Let c(X) e K[X] be any nonzero polynomial so that c(f) = 0 (for example c(X) is the characteristic polynomial of f). Suppose с (X) — р1 (X)р:(X)... Pm(X) where each p,(X) e K[X] and gcd(pg(X), p3(X))= 1 for all pairs 1 i < m.="" then="" (b.4)="" let="" ,="" i="">< t,="" be="" different="" eigenvalues="" of="" f.="" let="" bi="{uj" :="" 1="">< j="">< m}="" be="" a="" basis="" for="" the="" eigenspace="" of="" a="" for="" 1="">< i="">< t.="" use="" (b.3)="" to="" show="" that="" b1u="" bbuub,="" is="" independent="" v="" ker(p1="" (x))="" ker(p2(x))="" .-ker(pm(x))="" is="" a="" direct="" sum="" of="">