If $U$ is spanned by $e_1,\ldots,e_k$ and $V$ is a subspace with $U^\perp\cap V=\emptyset$, then $v=e_k+w$ for some $v\in V$ and $w\in U^\perp$

Question

Let $H$ be a $\mathbb R$-Hilbert space, $k\in\mathbb N$, $e_1,\ldots,e_k$ be mutually orthonormal, $U$ be the subspace of $H$ spanned by $e_1,\ldots,e_k$ and $V$ be an arbitrary subspace of $H$ with $\dim V=k$.

Let $\pi$ denote the orthogonal projection of $H$ onto $U$. Assume $U^\perp\cap V=\{ 0 \}$ so that $\left.\pi\right|_V$ is bijective.

Why can we find $v\in V$ with $$v=e_k+w\tag1$$ for some $w\in U^\perp$?

This should be trivial, since we know $H=W\oplus W^\perp$ for all closed subspaces $W$ of $H$. But how do we need to apply this result here?

Answer 1

Let $v\in V$. Since $H=\mathbb Re_k\oplus(\mathbb Re_k)^\perp$, $$v=\alpha e_k+x$$ for some $\alpha\in\mathbb R$ and $x\in(\mathbb Re_k)^\perp\subseteq U^\perp$. Since $U^\perp\cap V=\{0\}$, $$\alpha\ne 0$$ and hence $$\frac v\alpha\in V$$ and $$\frac x\alpha\in U^\perp$$ with $$\frac v\alpha=e_k+\frac x\alpha.$$