Determine which of the following are subspaces ?

Question

1. U∪V={x∈Rn | x∈U or x∈V}, where U and V are fixed (but arbitrary) subspaces of Rn
2.  w = {A^2 * x | x ∈ R2}, where A is a fixed, but arbitrary 2×2 matrix.
3. 3.3 X = {x ∈ Rn | Ax = Bx}, where A and B are fixed, but arbitrary m×n matrices.

Answer 1

Solution.

1. The subset U∪V={x∈Rⁿ | x∈U or x∈V} is not always a subspace of Rⁿ.

Counterexample: Let U be the x-axis and V be the y-axis in R². If is clear that U and V are subspace of R², but U∪V is not a subspace of R^2 because u=(1,0)∈U and v=(0,1)∈V but u+v=(1,1)∉U∪V.

2.  W = {A² * x | x ∈ R²}, where A is a fixed, but arbitrary 2×2 matrix, is a subspace of R².

Proof: We need to check the 3 conditions for a subspace.

(i) Clearly, 0=A²0∈W.

(ii) Suppose u∈W and v∈W. Then u=A²x and v=A²y for some x, y∈R². Then u+v=Ax+Ay=A(x+y)∈W.

(iii) Suppose u∈W and c∈R. Then u=A²x for some x∈R². Thus cu=cA²x=A²(cx)∈W.

Since all the 3 conditions hold, W is a subspace of R².

3. X = {x ∈ Rn | Ax = Bx}, where A and B are fixed, but arbitrary m×n matrices, is a subspace of Rn.

Proof. The matrix equation Ax = Bx is equivalent to (A-B)x=0. So X becomes the solution set of the homogeneous linear system (A-B)x=0 and so is a subspace of Rⁿ.

Remark: For part 3, we used the fact that the solution set of a homogeneous linear system in n variables is a subspace of Rⁿ. We can also prove part 3 by directly checking the 3 conditions (i), (ii), and (iiI) of a subspace, which is similar to part 2.