Proving Subspaces

Question

by checking closure under the zero vector, closure under addition and closure under scalar multiplication; investigate if     W = { (w₁, w₂, w₃) ∈ R³ : 6w1 - 5w2 + 4w3 = 0 } is a subspace of R³ ?

Answer 1

Okay well first we'll check if (0,0,0) ∈ W.

6*0 - 5*0 + 4*0 = 0; so yes, (0,0,0) ∈ W.

 

Now we'll check closure under addition.

Let u, v ∈ W where u = (u₁, u₂, u₃) and v = (v₁, v₂, v₃).

Since u, v ∈ W, we know that 6u₁ - 5u₂ + 4u₃ = 0 and 6v₁ - 5v₂ + 4v₃ = 0.

Consider u + v = (u₁ + v₁, u₂ + v₂, u₃ + v₃).

Notice that 6(u₁ + v₁) - 5(u₂ + v₂) + 4(u₃ + v₃) = 6u₁ + 6v₁ - 5u₂ - 5v₂ + 4u₃ + 4v₃

                                                                       = (6u₁ - 5u₂ + 4u₃) + (6v₁ - 5v₂ + 4v₃)

                                                                       = 0 + 0 = 0.

Thus, u + v ∈ W.

 

Finally, we'll check closure under scalar multiplication.

Let w ∈ W and let c ∈ R where w = (w₁, w₂, w₃).

Since w ∈ W, we know that 6w₁ - 5w₂ + 4w₃ = 0.

Consider cw = (cw₁, cw₂, cw₃).

Notice that 6(cw₁) - 5(cw₂) + 4(cw₃) = c(6w₁) - c(5w₂) + c(4w₃) = c(6w₁ - 5w₂ + 4w₃) = c(0) = 0.

Hence, cw ∈ W.

 

Therefore, W is a subspace of R³.