(1) In each case, justify your answer. (a) Is U = {f e F([a,b])|f(a) = f(b)} a s
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Question
(1) In each case, justify your answer.(a) Is U = {f e F([a,b])|f(a) = f(b)} a subspace of F([a, b]), where F([a, b]) is the vector space of real-valued functions defined on the interval [a,b]?
(b) Is R2 a subspace of R3?
(c) Consider Pn the vector space of polynomials of at most degree n with the standard addition and scalar multiplication. Is P2 a subspace of P3?
(1) In each case, justify your answer.
(a) Is U = {f e F([a,b])|f(a) = f(b)} a subspace of F([a, b]), where F([a, b]) is the vector space of real-valued functions defined on the interval [a,b]?
(b) Is R2 a subspace of R3?
(c) Consider Pn the vector space of polynomials of at most degree n with the standard addition and scalar multiplication. Is P2 a subspace of P3?
Explanation / Answer
(a) YES: U is a set of functions f from the set F([a,b]) who happen to have the same value for a and b.
Example 1:
function f1(x) = 2x for a=< x =< b could be a function from set F([a,b]),
But f1(a)=2a, and f1(b) = 2b, meaning f1(a) not equal to f1(b) so that function f1 is not in set U.
Example 2:
Let f2 be the function f2(x) = (x-a)(x-b) + 3 = x2 - (a+b) x + ab + 3 for a=< x =< b
This function is such that f2(a)=3 and f2(b)=3 so f2(a) = f2(b)
f2 is part of F([a,b]) and is also in set U because it meets the condition f(a)=f(b).
(b) NO. Elements of R2 are of the form (x1, x2) with two coordinates, while elements of R3 have 3 coordinates (x1, x2, x3). Elements of R2 are not elements of R3, hence R2 is not a sub-set of R3. One could establish a mapping of elements from R2 into R3 by adding to them a third coordinate, zero as an example, but that is not the same as being part of R3.
(c) YES. P2 includes all functions of degree 0, 1 or 2, and P3 includes all functions of degree 0, 1, 2 and 3. So elements of P2 (functions of degree 0, 1, and 2) are elements of P3, and P2 is included in P3.
Good luck.
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