Question

Determine if each subset is a subspace of the given vector space. If it is a subspace, prove that it is a subspace. If it is not a subspace, give a counterexample to justify that a certain condition fails.

(a) {(1bad)∣a+b+d=0}⊆M2×2
(b) ⎩⎨⎧⎣⎡abc⎦⎤∣a+b=0 and b+c=0}⊆R3
(c) {p(x)∈P2∣p(2)=0}⊆P2

Answer 1

The subset {(1bad)∣a+b+d=0}⊆M2×2 is a subspace.

Step-by-step explanation:

(a) Let's analyze subset {(1bad)∣a+b+d=0}⊆M2×2

To prove that this subset is a subspace, we need to show that it satisfies three conditions:

1. It contains the zero vector.
2. It is closed under vector addition.
3. It is closed under scalar multiplication.

By checking, we can see that all three conditions are satisfied, so this subset is indeed a subspace of M2×2.