Question

Consider a nonempty subset W of${\mathbf{ℝ}}^{\mathbf{n}}$that is closed under addition and under scalar multiplication. Is W necessarily a subspace of${\mathbf{ℝ}}^{\mathbf{n}}$? Explain.

Short answer

W is a subspace of ${\mathrm{ℝ}}^n$

Step-by-step solution

Consider the set.

A subset W of the vector space ${\mathrm{ℝ}}^n$is called a (linear) subspace of ${\mathrm{ℝ}}^n$if it has the following three properties:

a. W contains the zero vector in ${\mathrm{ℝ}}^n$.

b. W is closed under addition: If ${\overrightarrow{w}}_1$and ${\overrightarrow{w}}_2$are both in W, then so is ${\overrightarrow{w}}_1+{\overrightarrow{w}}_2$.

c. W is closed under scalar multiplication: If $\overrightarrow{w}$is in W and k is an arbitrary scalar, then localid="1660801636209" $k\overrightarrow{w}$is in W.

The nonempty subset W should satisfy the above conditions.

If W is a subset in which the localid="1660801628244" $0\in W$, as a result W must be a subspace of localid="1660801632742" ${\mathrm{ℝ}}^n$.

Final answer.

W is a subspace of ${\mathrm{ℝ}}^n$.