Question

Linear algebra Section 4.2 Subspaces

1. Let W1 and W2 be subspaces of a vector space V . Define W to be the set of vectors in V oftheformw1+w2 wherew1 ?W1 andw2 ?W2.

(a) Show that W is a subspace of V .
(b) Show that W1 and W2 are subspaces of W

Answer 1

Proof

Explanation:

(a)

- We first have to show that
`0 \in W`. Since
`W_1, W_2` are subspaces, it holds that
`0 \in W_1, W_2`, hence
`0 \in W`.

- Second, we have to show that
`W` is closed under sum and scalar multiplcation. First, let us take
`a, b \in W`.

Since a, b belong to
`W=W_1+W_2`, we can find
`a_1,b_1 \in W_1` and
`a_2, b_2 \in W_2` such that

`a=a_1+a_2\\\\b=b_1+b_2`

Therefore,

`a+b=a_1+a_2+b_1+b_2=(a_1+b_1)+(a_2+b_2)`

since
`W_1, W_2` are vector subspaces of
`V`, it holds that

`a_1+b_1 \in W_1 \quad\text{and}\quad a_2 + b_2 \in W_2`, which shows us that W is closed under sum.

On the other hand, let us take
`a\in M` and
`\lambda \in \mathbb{R}`. We already know that

`a=a_1+a_2`

where
`a_1 \in W_1` and
`a_2 \in W_2`. Moreover,

`\lambda a_1 \in W_1 \quad \text{and} \quad \lambda a_2 \in W_2`,

hence

`\lambda a =\lambda(a_1 + a_2) =\lambda a_1 + \lambda a_2`

belongs to
`W`, which shows that
`W` is closed under scalar multiplication.

(b) It is clear that
`W_1, W_2 \subseteq W`. Moreover, since
`W_1, W_2` are subespaces of
`V`, they are closed under sum and scalar multplication, this propertie is remains true for
`W_1,W_2` as subsets of
`W`, which tells us that
`W_1,W_2` are subspaces of
`W`.