How to prove that $W=U_1 \oplus(U_2 \cap W)$ if $U_1 \subseteq W$, given that $V=U_1 \oplus U_2$ and $U_1,U_2,W$ are subspaces of $V$?

Question

Prove that $W=U_1 \oplus(U_2 \cap W)$ if $U_1 \subseteq W$, given that $V=U_1 \oplus U_2$ and $U_1,U_2,W$ are vector subspaces of $V$.

Here's what I have:

Let $U_1=Sp(S), U_2=Sp(T)$.

Because $U_1 \subseteq W$ then there exists a set of vectors {$x$} such that $U_1 \cup \{x\}=W.$

Then: $$ Sp(S) \cup \{x\} =Sp(S) \oplus(Sp(T) \cap(Sp(S) \cup \{x\}))\\ =Sp(S) \oplus (\{0\} \cup \{x\})\quad (*)\\ =Sp(S)+\{x\} $$

(*) because $V=U_1 \oplus U_2$ then $Sp(S) \cap Sp(T)=\{0\}$

At this point I'm stuck because {$x$} is a subset of vectors not a subspace.

Answer 1

It is better to use the definition of the direct sum.

The assumptions may be rewritten as:

a) any element of $V$ is the sum of an element of $U_1$ and an element of $U_2$.

b) The intersection of $U_1$ and $U_2$ is the null vector.

Now you want to prove to things:

1) any element of $W$ is the sum of an element of $U_1$ and an element of $U_2\cap W$.

2) The intersection of $U_1$ and $U_2\cap W$ is the null vector.

It seems pretty clear how to proceed now. You will need the assumption $U_1\subset W$ to get 1) from a).