Consider the $\mathbb{R}$-linear space $\mathbb{R}^n$ and $U, W \le \mathbb{R}^n$ with $\dim(U)=\dim(W)=n-1$.
Is it true that there exists $V\le \mathbb{R}^n$ such that $U\oplus V = W \oplus V = \mathbb{R}^n$?
Here is my attempt:
I claim that it is true.
Let $B_U=\{u_1,...,u_{n-1}\}$ and $B_W = \{w_1,...,w_{n-1}\}$ be basis of U and W respectively and take $u_n \in \mathbb{R}^n$ such that $B_U \cup \{u_n\}$ is a basis of $\mathbb{R}^n$.
Then for every choices of $a_1,...,a_n \in \mathbb{R}$, with $a_n \ne 0$, $B_U \cup \{\sum_{i=1}^{n}a_iu_i\}$ is a basis of $\mathbb{R}^n$ and $U \oplus \operatorname{span}(\sum_{i=1}^{n}a_iu_i)=\mathbb{R}^n$.
If $u_n \notin W$, then $W \oplus \operatorname{span}(u_n)=\mathbb{R}^n$; otherwise consider $u_1+u_n$, if this does not belong to W, then $W \oplus \operatorname{span}(u_1+u_n)=\mathbb{R}^n$.
If $u_1+u_n \in W$, then $u_1 \in W$, now repeat with $u_2$ instead of $u_1$ and so on.
In the end or we find $u_m+u_n, m\lt n$ such that $W \oplus \operatorname{span}(u_m+u_n)=\mathbb{R}^n$, or $u_1,...,u_{n-1}\in W$ that is $U=W$ and therefore $W \oplus \operatorname{span}(u_n)=\mathbb{R}^n$.
Is this correct? Thanks in advance