Question

Let A be a fixed mxn matrix. Prove that the set W={x ∈ Rⁿ :A x = 0 } is a subspace of Rⁿ. Direct verification of definition style proof. Verify it is a subspace using the subspace test.

Answer 1

W = {x is an element of R^n : Ax = 0} is proven to be a subspace by showing it contains the null vector, is closed under addition, and is closed under scalar multiplication, fulfilling the subspace test.

Step-by-step explanation:

To prove that the set W = {x ∈ ℝn : A x = 0} is a subspace of ℝn, we must show that it satisfies three conditions: it contains the null vector, it is closed under vector addition, and it is closed under scalar multiplication.

Containing the null vector: The null vector in ℝn is the vector where all components are zero. Since A⋅0 = 0 (here '0' denotes the null vector), the null vector is in W.

Closed under vector addition: Take any two vectors u and v in W, so A⋅u = 0 and A⋅v = 0. If we add these two vectors, we get A⋅(u+v) = A⋅u + A⋅v = 0 + 0 = 0, which means u+v is also in W.

Closed under scalar multiplication: For any vector u in W and any scalar c, A⋅(cu) = c⋅(A⋅u) = c⋅0 = 0, showing that cu is also in W.

By verifying these properties, we have shown that W is indeed a subspace of ℝn using the subspace test.