S7

Question 1

If S₁ and S₂ are subspaces of a vector space V , then S₁ ∩ S₂ is also a subspace.

(S₁ ∩S₂ is the intersection of S₁ and S₂, the set of vectors that are in S₁ and S₂.)

Answer 1

True

Each of the 3 condition holds up under taking the intersection: The zero vectoris in S₁ and in S₂,soitisinS₁ ∩ S₂; if u and v are in S₁ ∩ S₂,thenthey are both in S₁ and both in S₂, so their sum and any scalar multiple are in both, hence also in S₁ ∩ S₂.

Question 2

If S₁ and S₂ are subspaces of a vector space V, then S₁ ∪ S₂ is also a subspace. (S₁ ∪ S₂ is the union of S₁ and S₂, the set of vectors that are in S₁ or S₂.)

Answer 2

False

If u is in S₁ but not in S₂, and v is in S₂ but not in S₁, then u, v ∈ S₁∪S₂, but u+v need not be.

Question 3

Let V be the vector space of functions f : ℝ → ℝ. Then the set of functions such that *f *(3) = 0 is a subspace.

Answer 3

True

The zero function (f (x) = 0 for all x ∈ ℝ) satisfies the condition. If f and g are in V, then so is f+g because (f+g)(3) = f(3) + g(3) = 0+0 = 0. If *f (3) = 0 and c ∈ R then (cf *)(3) = c · *f *(3) = c · 0 = 0.

Question 4

Let V be the vector space of functions f : ℝ → ℝ. Then the set of functions such that f (t) ≥0 for all t ∈ ℝ is a subspace.

Answer 4

False

If f (t) is in this set and has some f (t₀) > 0, then (−1) · *f *(t) is not in this set, because (−1) · f (t₀) = −f(t₀) < 0.

Question 5

Let V be the vector space of functions f : ℝ → ℝ. Then the set of functions such that *f *(3) · *f *(6) = 0 is a subspace.

Answer 5

False

The zero vector is there and it contains scalar multiples, but it does not contain all sums.

For instance, let f₁(3) = 1 and f₁(t) = 0 for all t ≠ 3, and let f₂(6)=1 and f₂(t)=0 for all t ≠ 6.

Then f₁(3)·f₁(6)=1·0=0 and f₂(3)·f₂(6)=0·1=0, so both are in this set. But f₁ +f₂ has (f₁ +f₂)(3)·(f₁ +f₂)(6) = (f₁(3)+f₁(6))·(f₂(3)+f₂(6)) = (1+0)·(0+1) = 1, so is not in this set.