Question

Prove that the set W = x ∈ Rⁿ: ax = 0 is a subspace of Rⁿ, where a is a fixed m × n matrix.

Answer 1

The set W is proven to be a subspace of R^n by demonstrating it satisfies non-emptiness, closure under addition, and closure under scalar multiplication. The zero vector confirms non-emptiness, and algebraic properties of matrix multiplication confirm closure under addition and scalar multiplication.

Step-by-step explanation:

To prove that the set W = {x ∈ R^n: ax = 0} is a subspace of R^n, where a is a fixed m × n matrix, we must verify three properties: non-emptiness, closure under addition, and closure under scalar multiplication.

First, we show non-emptiness by noting that the zero vector — denoted by 0 — satisfies a0 = 0, which belongs to W. Hence, W is non-empty.

Next, to show closure under addition, let x and y be elements of W. This implies ax = 0 and ay = 0. For any such vectors x and y, it follows that a(—x + y) = ax + ay = 0 + 0 = 0, which means x + y is also in W.

Lastly, for closure under scalar multiplication, let c be any scalar in R and x be an element of W, then a(cx) = c(ax) = c⋅ 0 = 0, which indicates that cx is also in W.

Therefore, the set W satisfies all the required properties and is a subspace of R^n.