Let $\mathbb{W}$ be a finite dimensional subspace of an inner product space $\mathbb{V}$. Prove or disprove the following: the double orthogonal complement of $\mathbb{W}$ is equal to itself, or, in other words, $(\mathbb{W}^\perp)^\perp = \mathbb{W}.$

We might begin a potential proof as follows. If $\vec{v} \in \mathbb{W}$, then $\langle \vec{v}, \vec{w}\rangle = \vec{0}$ for any vector $\vec{w} \in \mathbb{W}^\perp.$ Therefore, by definition, $\vec{v} \in (\mathbb{W}^\perp)^\perp$, and so $\mathbb{W} \subseteq (\mathbb{W}^\perp)^\perp.$

Now, it's easy to prove equality if $\mathbb{V}$ is finite dimensional. For instance, we can use the fact that $\dim \mathbb{W} + \dim \mathbb{W}^\perp = \dim \mathbb{V}$ to show that $\dim \mathbb{W} = \dim (\mathbb{W}^\perp)^\perp$ which implies the desired result. However, in this case, $\mathbb{V}$ is not necessarily finite dimensional and so this step is not valid.

In the same vein, if we were to relax the condition that $\mathbb{W}$ must be finite dimensional, then the statement is false. Let $\mathbb{V}$ be the inner product space of all polynomials with real coefficients under the inner product $$ \langle a_0 + a_1x + \dots + a_n x^n, b_0 + b_1x + \dots + b_k x^k\rangle = a_0b_0 + a_1b_1 + \dots + a_jb_j $$ where $j$ is the lesser of $n$ and $k$. Then, the subspace $$ \mathbb{W} = \{ f(x) \in \mathbb{V} \mid f(1) = 0\} $$ of $\mathbb{V}$ has orthogonal complement $\{ \vec{0}\}$ which shows that $(\mathbb{W}^\perp)^\perp = \mathbb{V} \neq \mathbb{W}$.

The question, then, is: does there exist a counter example where $\mathbb{V}$ is infinite dimensional, but the subspace $\mathbb{W}$ is not? My intuition tells me yes, but I'm stuck.


2 Answers 2


Hint: Generally in a Hilbert space $H$ we have that for a linear subspace $W \subset H$ that $$ (W^\perp)^\perp = \overline{W} $$ where the bar denotes the closure.

  • $\begingroup$ @HughDenoncourt Of course it is rather trivial for Hilbert spaces. I was trying to hint at using the completion of the inner product space, but maybe I was being too subtle. In any case, my apologies. $\endgroup$
    – Demophilus
    Jun 12, 2017 at 18:24
  • $\begingroup$ No problem; that strategy sounds like it works. $\endgroup$ Jun 12, 2017 at 19:48
  • $\begingroup$ @jacer21 Do you see how this might work? $\endgroup$
    – Demophilus
    Jun 12, 2017 at 20:39
  • $\begingroup$ @Demophilus Since $W$ itself is closed, it must be equal to its closure, correct? $\endgroup$
    – Jacob
    Jun 12, 2017 at 21:55
  • 1
    $\begingroup$ @jacer21 Correct, and in case you only have inner product space you can always complete it. Its completion will then be a complete inner product space or a Hilbert space. You have to be a little careful though, because the orthogonal complement with respect to the completion will be much larger than the orthogonal complement with respect to the original inner product space. But there's an easy relation between the two. If you're unfamiliar with completing a metric space, take a look at proofwiki.org/wiki/Completion_Theorem_(Metric_Space). $\endgroup$
    – Demophilus
    Jun 12, 2017 at 22:38

And another hint: finite-dimensional subspaces are closed.


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