Question

all polynomial a+a1x+a2x^2+a3x^3 for which a+a1+a2+a3=0 is this a subspace of p3

Answer 1

p3?

Answer 2

p3 shows polynomial3

Answer 3

As in the previous problem you need to see if it satisfies all of the vector space axioms.

A subspace must have a zero vector and must be closed under addition and scalar multiplication.

Answer 4

In order to show that the set S of cubic polynomials, the sum of whose coefficients os 0, is a subspace, we must show that S is closed under addition. If we find even one counterexample, i.e., two polynomials in S whose sum is not in S, that is sufficient to prove S is not a subspace of \[p _{3}.\]

Answer 5

1) is the 0 polynomial in the space? well, 0+0x+0x^2+0x^3 is the 0 polynomial, 0+0+0+0 = 0, so yes, it is in the space :)

Answer 6

Just looking at the coefficients, you can see that its closed under addition.

a_1 + a_2 + ... = 0
b_1 + b_2 + ... = 0
a_1 + a_2 + .. + b_1 + b_2 + .. = 0

Answer 7

2) if :\[a+a_1x+a_2x^{2}+a_3x^{3}, b+b_1x+b_2x^{2}+b_3x^{3} \]
are 2 polynomials in the space, is their sum still in the space?
adding them gives:
\[(a+b)+(a_1+b_1)x+(a_2+b_2)x^{2}+(a_3+b_3)x^{3}\]
where
\[(a+b)+(a_1+b_1)+(a_2+b_2) + (a_3+b_3) = (a+a_1+a_2+a_3)+(b+b_1+b_2+b_3) = 0+0 = 0\]
so it is still in the space, and is closed under addition.

Answer 8

oops, got cut off >.<
last lines are
0 + 0 = 0

Answer 9

Also obviously closed under scalar multiplication.

Answer 10

3) if:
\[a+a_1x+a_2x^2+a_3x^{3} \]
is in the space, and i multiply it by a scalar, is it still in the space?
multiplying by c we get:
\[ca+ca_1x+ca_2x^2+ca_3x^{3}\]
where:
\[ca+ca_1+ca_2+ca_3 = c(a+a_1+a_2+a_3) = c(0) = 0\]
so its closed under scalar multiplication.