There exists a nonzero element $x \in W$ such that $x \perp V$ if $\dim(W) > \dim(V)$

Question

Consider an $n$-dimensional Euclidean space $U$. Let $V, W \subseteq U$ be two subspaces with dimensions $\dim(V) = k$ and $\dim(W) = k+1$ for some $k < n$.

How can we prove that there exists a nonzero element $x \in W$ such that $x \perp V$, i.e., the inner product $x^T y = 0$ for any $y \in V$?

If $V$ is a subspace of $W$, the above is obvious. But what if it is not? Since $W$ has a higher dimension, we know there must exist a nonzero element $x \in W$ that does not belong to $V$. But I failed to go any further...

Answer 1

We have $\dim (V^{\perp} + W) = \dim V^{\perp}+\dim W - \dim (V^{\perp}\cap W) $.

Then $\dim V^{\perp}=n-k$, $\dim W=k+1$, and $\dim (V^{\perp}+W) \leq n$.

Thus, $\dim (V^{\perp} \cap W) \geq 1$.

Answer 2

Write

$$V+W = V_1 \oplus (V \cap W ) \oplus W_1$$

where $V = V_1 \oplus (V \cap W) $ and $V_1 \perp (V \cap W)$ and similar equalities for $W_1$. With finally also $V_1 \perp W_1$. For this, we can take for $V_1$ the orthogonal complement of $V \cap W $ in $V$. And for $W_1$ the orthogonal complement of $V_1 \oplus (V \cap W)$ in $V +W$.

Based on the given dimensions for $V,W$, we have $\dim W_1 \ge 1$ and any non zero element $x \in W_1$ satisfies the desired result.