$M$ and $N$ are subspaces of a Hilbert space. If $M\subset N$, show that $N^{\perp}\subset M^{\perp}$. Show also that $(M^{\perp})^{\perp}=M$.

I know that the orthogonal complement of $X$ is the set $X^{\perp}=\{x\in H : x{\perp}X\}$ where $X$ is any subset of a Hilbert space $H$. I'm not sure how to proceed. Any hints or solutions are greatly appreciated.

  • $\begingroup$ You need closures somewhere to make any of these statements true. Given that, the inclusion follows directly from the definition. One direction of the equality is straightforward, the other benefits from the Hahn Banach theorem. $\endgroup$
    – copper.hat
    Jun 14, 2015 at 21:41
  • $\begingroup$ When you write $M\subset N$, do you allow $M=N$? $\endgroup$
    – egreg
    Jun 14, 2015 at 22:11
  • $\begingroup$ I think so, the book I'm using doesn't specify. $\endgroup$
    – 1233dfv
    Jun 14, 2015 at 22:15

1 Answer 1


If $x$ $\in$ $N^{\perp}$ then $x$ $\perp$ $N$ and since $M \subset N$ then $x \perp M$ therefore $x \in M^{\perp}$

Now let $x \in M$, then $x \perp M^{\perp}$ and therefore $x \in M^{\perp \perp}$. That means $M \subset M^{\perp \perp}$. Since $M^{\perp \perp}$ is a closed and linear subspace containing M that means $\overline{\lt M \gt} \subset M^{\perp \perp}$

Suppose now that there is a $x \in M^{\perp \perp} \setminus \overline{\lt M \gt}$. There exists a $y \in \overline{\lt M \gt}$ so that $x - y \perp \overline{\lt M \gt}$. Then $x - y \perp M$ and therefore $x - y \in M^{\perp}$. Also $x - y \in M^{\perp \perp}$ since it is a linear combination of elements of $M^{\perp \perp}$. This means that $x - y = 0$ and therefore $x = y$ and $x \in \overline{\lt M \gt}$ which is false.

  • 1
    $\begingroup$ Quite hard to read ! I strongly suggest you use mathjax to write math here. Click here for a basic tutorial $\endgroup$ Jun 15, 2015 at 0:51
  • $\begingroup$ Thanks Alonso. I corrected it $\endgroup$ Jun 15, 2015 at 1:53

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.