(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
(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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![(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 <j < 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 subspaces](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0282592e-adf8-40c1-9d31-7bca34586c75%2Fd16d0fad-2e72-408e-948a-eb1184ad3b4b%2Fp451xg.png&w=3840&q=75)
Transcribed Image 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 <j < 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 subspaces
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