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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.

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

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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, $x\in W$).

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