Let $V$ be a $k$-dimensional subspace of $\mathbb R^n$ equipped with Euclidean inner product. Let $W$ be another $k$-dimensional subspace of $\mathbb R^n$ such that $W \cap V^{\perp} = (0).$ Then does there exist any $x \in V$ such that $\left (x + V^{\perp} \right ) \cap W = \varnothing\ $?
This question appeared in my mind just for the sake of curiosity while I checked few examples. I don't find any example where there is an empty intersection. So I conjectured that the intersection should be non-empty but I can't able to prove it explicitly.
Any help would be appreciated.
By considering the decomposion of $\mathbb{R}^n$ as $V \oplus V^{\perp}$, we may write any vector $x\in \mathbb{R}^n$ as $x = v + v_{\perp}$. Now define a linear map $\phi : W \rightarrow V$ by $\phi(w) = v$, where $w = v+v_{\perp}$. As $W\cap V^{\perp} = 0$, and $dim(W) = dim(V)$, $\phi$ is an isomorphism.
Now let $x$ be any vector. Then $x = v_x + y_{\perp}$. Now, $v_x\in \left( x+V^{\perp}\right)$. As $\phi$ is an isomorphism, there exists a $w_x$ such that $\phi(w_x) = v_x$. Hence, $w_x = v_x +z_{\perp} = x-y_{\perp} + z_{\perp}$. As $V^{\perp}$ is a vector space,$-y_{\perp} + z_{\perp}\in V^{\perp}$. Hence $w_x\in W\cap \left(x+V^{\perp}\right)$.