Let V be an inner product space, S and $S_0 $be subsets of V, and W be a finite-dimensional subspace of V.

Prove:$S_0 \subset S$ implies $S^{\bot} \subset S_0^{\bot}$

Pf: Let $x \in S^{\bot}$, for all $y \in S$, the inner product $\langle x,y \rangle$ is given $\langle x,y \rangle=0$, which implies $y \in S_0$ Then $S_0 \subset S$, $x \in S_0^{\bot}$.Proved.

Prove: $W=(W^{\bot})^{\bot}$

Pf: Let $x \in W$ and $y \in W^{\bot}$. By definition, for all $y \in W^{\bot}$, the inner product is given by $\langle x,y \rangle=0$. Using $x \in W$, this gives $W \subset (W^{\bot})^{\bot}$ How to prove the other side?


2 Answers 2


Suppose that $w\in (W^\perp)^\perp$ then $\left<w,x \right>=0$ for all $x\in W^\perp$.

If $w\not\in W$, then $w=v+v'$ where $v\in W$ and $v'\in W^\perp\setminus \{0\}$. Then $$0=\left<w,x \right>=\left<v,x \right>+\left<v',x \right>=\left<v',x \right>$$ for all $x\in W^\perp.$ However, $v'\in W^\perp$ and $\left<v',v' \right>\neq 0$ so it is a contradiction.

  • $\begingroup$ can you explain the logic? I don't follow really well. Starting from $w=v+v'$ $\endgroup$
    – spruce
    Mar 15, 2020 at 4:08
  • 1
    $\begingroup$ @spruce So we note that $V=W+W^\perp$. Since $w\in V$ anyway, we should have such decomposition. And, especially, since $w\not\in W$, the decomposition should have a nonzero element in $W^\perp$. And the rest of them is just a calculations $\endgroup$
    – Lev Bahn
    Mar 15, 2020 at 4:11

The converse is trickier, and not necessarily true if $W$ is not finite-dimensional (actually, it is true if and only if $W$ is topologically closed in $V$). So, any proof will have to make use of the finite-dimensionality assumption.

Suppose $(e_1, \ldots, e_n)$ is an orthonormal basis for $W$. Suppose $x \in (W^\perp)^\perp$. Consider the projection $w$ of $x$ onto $W$, specifically: $$w = \langle x, e_1\rangle e_1 + \ldots + \langle x, e_n \rangle e_n \in W.$$ Then, $x - w$ is perpendicular to each $e_i$, since \begin{align*} \langle x - w, e_i\rangle &= \langle x, e_i \rangle - \sum_{j=1}^n\Big\langle \langle x, e_j \rangle e_j, e_i \Big\rangle \\ &= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \langle e_j, e_i \rangle \\ &= \langle x, e_i \rangle - \sum_{j=1}^n \langle x, e_j \rangle \delta_{ij} \\ &= \langle x, e_i \rangle - \langle x, e_i \rangle = 0. \end{align*} Since the linear map $\langle \cdot, x - w\rangle$ is constantly $0$ on the basis of $W$, it is constantly zero on all of $W$, hence $x - w \in W^\perp$.

But then, since $x \in (W^\perp)^\perp$, we have $\langle x - w, x \rangle = 0$, and since $\langle x - w, w \rangle = 0$, $$\|x - w\|^2 = \langle x, x - w \rangle - \langle w, x - w\rangle = 0,$$ i.e. $x = w \in W$, completing the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.