Question

Suppose m is a positive integer. Is the set consisting of 0 and all polynomials with coefficients in F and with degree equal to m a subspace ofP(F)?

Answer 1

Answer:

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements $a^m +1 , -a^m +1 \in S$ so both elements are in S, and if we apply the condition for the addition closed we got:

$(a^m +1) +(-a^m +1) = a^m -a^m +1+1 = 2$

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce

Step-by-step explanation:

For this case the answer would be no.

It can't be a subspace because we not satisfy the condition of closed under addition.

We need to remember that a subset U of V is a subspace of V if and only if U satisfies the following 3 conditions

1) Additive identity $0 \in U$

2) Closed under addition $u,v \in U \Rightarrow u+x \in U$

3) Cloases under scalar multiplication $a \in F, u \in U \Rightarrow au \in U$

Proof

For this case we don't have any problems for the conditions 1) and 3), but we have a problem with condition 2) since is not satisfied.

We just need to find a counterexample to show that the statement is False. If we find two elements in the subset provided S, and we show that the sum is not in S, then we have the counter example.

Let's say that we have two elements $a^m +1 , -a^m +1 \in S$ so both elements are in S, and if we apply the condition for the addition closed we got:

$(a^m +1) +(-a^m +1) = a^m -a^m +1+1 = 2$

And by definition of S, 2 is not in S so then since we can't satisfy the closed addition property then S can't be a subsapce