Let $f$ be a map $\mathbb R^n\to\mathbb R$, $$ f(x) = |x_1| + \cdots + |x_r| - |x_{r+1}| - \cdots - |x_{r+s}|, \quad\forall x = (x_1, \ldots, x_n) \in\mathbb{R}^n $$ with $r\geq s\geq0$ and $r+s\leq n$. Let $W_1,W_2$ be two linear subspaces of dimension $n-r$ satisfying $f(x)=0,\, \forall x\in W_1\cup W_2$. Show that $\dim(W_1\cap W_2)\geq n-(r+s)$.
For this problem, there is a hint that try to show that there exists a linear subspace $W$ such that $f(W)=0$ with $\dim W=n-(r+s)$. This is an easy part. I also notice $f$ is not a linear map since taking absolute values is not linear. How can we prove the desired inequality?
Let $\Bbb R^{i:j}$ be the linear subspace of $\Bbb R^n$ that is generated by $e_i, e_{i+1}, \cdots, e_{j}$, where $e_k$ is the $k$-th standard unit vector in $\Bbb R^n$. For example, $\Bbb R^{1:r+s}=\langle e_1, \cdots, e_{r+s}\rangle=\{(x_1,\cdots, x_{r+s}, 0,\cdots, 0)\in\Bbb R^n: x_i\in\Bbb R\}$. If $i=j$, then $\Bbb R^{i:j}$ is the zero subspace.
Lemma. Let $U$ be a linear subspace of $\Bbb R^{1:r+s}$ such that $f(U)=0$. Then $\dim(U)\le s$.
Proof. Towards a contradiction, suppose $\dim(U)> s$. Then there exists $v\in R^{r+1:r+s}$ and two different vectors $t_1, t_2$ in $\Bbb R^{1:r}$ such that $t_1+v, t_2+v\in U$. Since $U$ is a linear subspace, any linear combination of $t_1+v$ and $t_2+v$ is also in $U$. Hence we have the first equality below, where $\lambda$ is an arbitrary real number. $$\begin{aligned} 0&=f(\lambda (v+t_1) + (1-\lambda)(v+t_2))\\ &=f(v+(\lambda t_1+(1-\lambda)t_2))\\ &=f(v) + f(\lambda t_1+(1-\lambda)t_2)\\ \end{aligned}$$
This implies $f(\lambda t_1+(1-\lambda)t_2)$ is independent of $\lambda$, which is impossible since $t_1$ and $t_2$ are different vectors in $\Bbb R^{1:r}$.
Note $f(\Bbb R^{r+s+1:n})=0$ and $\dim \Bbb R^{r+s+1:n}= n-(r+s)$. "This is an easy part".
It looks the hint advises the following claim.
Claim. $\Bbb R^{r+s+1:n}\subseteq W_1$.
Proof. Let $V_1$ be the projection of $W_1$ to $\Bbb R^{1:r+s}$, i.e., $$V_1 =\{u\in\Bbb R^{1:r+s}: \exists v\in\Bbb R^{r+s+1:n} s.t. u+v\in W_1\}.$$ Similarly, let $V_2$ be the projection of $W_1$ to $\Bbb R^{r+s+1:n}$, i.e., $$V_2 =\{v\in\Bbb R^{r+s+1:n}: \exists u\in\Bbb R^{1:r+s} s.t. u+v\in W_1\}.$$ We have $W_1\subseteq V_1\oplus V_2$.
Since $f(\Bbb R^{r+s+1:n})=0$ and $f(W_1)=0$, we have $f(V_1)=0$. The lemma tells us that $\dim(V_1)\le s$. $$n-r=\dim(W_1)\le\dim(V_1)+\dim(V_2)\le s + \dim \Bbb R^{r+s+1:n} = n-r.$$ The end-to-end equality means all inequalities in-between must be equalities. So $ \dim(W_1)=\dim(V_1)+\dim(V_2)$ and $\dim(V_2)=\dim \Bbb R^{r+s+1:n}$. The former equality implies $W_1=V_1\oplus V_2$ and hence $V_2\subseteq W_1$ while the latter equality implies $V_2=\Bbb R^{r+s+1:n}$. Our proof of the claim is complete.
Similarly, we have $\Bbb R^{r+s+1:n}\subseteq W_2$. So $\Bbb R^{r+s+1:n}\subseteq W_1\cap W_2$.