Sum of two subspaces is a subspace

Question

I am wondering if someone can check my proof that the sum of two subspaces is a subspace:

1) First show that $0 \in W_1 + W_2$:

Since $W_1, W_2$ are subspaces, we know that $0 \in W_1, W_2$.

So if $w_1, w_2 = 0, w_1 + w_2 = 0 + 0 = 0 \in W_1 + W_2$.

2) Show that $cu+v \in W_1 + W_2$.

Let $W = W_1 + W_2$

Let $cu \in W$. We can find some $u_1 \in W_1, u_2 \in W_2, c \in \mathbb{F}$ such that $cu = u_1 + u_2$.

Similarly, since $v \in W$, we can find $v_1, v_2 \in W_1, W_2$ such that $v = v_1 + v_2$.

So $cu+v = c(u_1 + u_2) + (v_1 + v_2)$

$= cu_1 + cu_2 + v_1 + v_2$

$= cu_1 + v_1 + cu_2 + v_2$

And so $cu_1 + v_1 \in W_1$ and $cu_2 + v_2 \in W_2$, and therefore $cu+v \in W_1 + W_2$.

Is my proof correct? Please let me know!

Answer 1

Your proof is not rigorously correct.

The first part 1) is perfect.

For the part 2), there are some points that can be improved.

• Your started with "let $cu\in W$". You need to start with "let $u\in W$." Why can you conclude that $cu\in W$, for any $c\in\mathbb{F}$?
• Your conclusion $cu+v=cu_1+v_1+cu_2+v_2$ is correct. But why $cu_1+v_1\in W_1$ and $cu_2+v_2\in W_2$?

You should focus in these details in order to obtain a logically correct proof.

• Minor detail: the notation $v_1,v_2\in W_1,W_2$, despite well understable, is not standard. It is better to write $w_1\in W_1$ and $w_2\in W_2$.

Answer 2

Yes it is correct, indeed you have verified that all the properties hold even if usually they are stated as

$1) \ \vec{0} \in W= W_1 + W_2\\ 2) \ \vec{v}+\vec{w} \in W= W_1 + W_2\\ 3) \ \vec{cv}\to c \cdot \vec{v}\in W= W_1 + W_2 \ ,c \in \mathbb{R}$

Thus you could improve and make more clear in my opinion as follow

2) $\ \vec{v}+\vec{w} \in W= W_1 + W_2\\$

• $\vec v=v_1+v_2$ for some $v_1\in W_1$ and $v_2 \in W_2$
• $\vec w=w_1+w_2$ for some $w_1\in W_1$ and $w_2 \in W_2$

then

• $v+w=(v_1+w_1)+(v_2+w_2)$

3) $\ \vec{cv}\to c \cdot \vec{v}\in W= W_1 + W_2 \ ,c \in \mathbb{R}$

• $c\cdot \vec v =c(v_1+v_2)=cv_1+cv_2$