Prove that if $U$ and $W$ are subsaces of $V$ with $V = U+W$ then there exists a subspace $W_1$ of $W$so that $V=U \oplus W_1$ Prove that if $U$ and $W$ are subsaces of $V$ with $V = U+W$ then there exists a subspace $W_1$ of $W$so that $V=U \oplus W_1$
My attempt:
Let $W_1 = \big\{w \in W | w \notin W \cap U \big\}$. Then $W_1 \cap U = \{0\}$, but
i am not sure if $W_1$ is a subspace.  How do I proceed?
To address the comment: The spaces are vector spaces.
$U+W = \big\{u+w|v\in U, w \in W\big\}$. The symbol $\oplus$ refers to a direct sum. The sum $U+W_1$ is a direct sum if $U \cap W_1 = \{0\}$.
 A: Find the basis of $\frac VU$ like this form $\mathcal B=\{v_i+U \mid i\in I\}$, and there exists an isomorphism $L:\frac VU\to \frac W{W \cap U}$ by Second Isomorphism Thoerem.
Then you can find the basis of $\frac W{W \cap U}$ $\mathfrak B=L\mathcal B=\{w_i+{W \cap U}\mid i\in I\}$.
Then let $W_1=\mathrm{span}\mathfrak B$. It is trivial that $W_1$ is subspace of $W$, and $U\cap W_1=\{0\}$.
So $V=U\oplus W_1$.
A: Take a basis $(u_1, \dots , u_k)$ of $U$ and extend this to a basis $(u_1, \dots , u_k, v_{k+1}, \dots, v_n)$ of $V$. Since $V = U + W $,
$$(u_1, \dots , u_k, v_{k+1}, \dots, v_n) = (u_1, \dots , u_k, u_{k+1} + w_{k+1}, \dots, u_n + w_n)$$
for some $u_{k+i} \in U$ and $w_{k+i} \in W$. Since $u_{k+i} + w_{k+i} \in \text{span} \{u_1, \dots, u_k, w_{k+i}\}$, we see that
$$V = \text{span} \{u_1, \dots , u_k, u_{k+1} + w_{k+1}, \dots, u_n + w_n\} = \text{span} \{u_1, \dots , u_k, w_{k+1}, \dots,  w_n\}$$
so that $(u_1, \dots , u_k, w_{k+1}, \dots,  w_n)$ is a basis of $V$. Thus we can choose $W_1 = \text{span} \{w_{k+1}, \dots, w_n\}$
