# Prove that there exists $y \in V$ such that $y \in W^\perp$, but $\langle x, y \rangle \not= 0$.

Let $$V$$ be an inner product space, and let $$W$$ be a finite-dimensional subspace of $$V$$. If $$x \not\in W$$, prove that there exists $$y \in V$$ such that $$y \in W^\perp$$, but $$\langle x, y \rangle\not= 0$$. Hint: Use Theorem 6.6.

Theorem 6.6: Let $$W$$ be a finite-dimensional subspace of an inner product space $$V$$, and let $$u \in V$$. Then there exists unique vectors $$u \in W$$ and $$z \in W^\perp$$ such that $$y = u +z$$. Furthermore, if $$\{ v_1, v_2, ... , v_k\}$$ is an orthonormal basis for $$W$$, then $$u = \sum_{i=1}^k \langle y, v_i \rangle v_i.$$

By Theorem, I first thought that $$x = u + v$$ for $$u \in W$$ and $$v \in W^\perp$$. But this is actually wrong because we do not know whether $$x \in V$$. I do not know how to solve this question, and I appreciate if you give some help.

• Isn't this just a straightforward consequence of $(W^\perp)^\perp=W$ (which is also a corollary of your theorem)? Dec 28, 2019 at 8:41

You can write $$x$$ as $$u+z$$, with $$u\in W$$ and $$z\in W^\perp$$. Now, take $$y=z$$. Then $$y\in W^\perp$$ and$$\langle x,y\rangle=\langle x,z\rangle=\langle u,z\rangle+\langle z,z\rangle=\lVert z\rVert^2\neq0,$$since $$z\neq 0$$ (otherwise, $$x\in W$$).