Proof about double orthogonal complements

What I had to prove was:

Let $W$ be a subspace of an inner product space $V$. Show that $W\subset W^{\perp\perp}$ and that $W=W^{\perp\perp}$ if dim$V$ is finite.

There are two areas where I'm having trouble labelled with **.

My attempt is:

If $w\in W$, then $\forall v\in W^{\perp}$, $\langle w,v\rangle=0$. **Then somehow $w\in W^{\perp\perp}$ which would show that $W\subset W^{\perp\perp}$ but I'm not sure why $w\in W^{\perp\perp}$.

Let $V$ have finite dimension $n$, and let dim$W=m$. Then dim$W^{\perp}=$dim$V-$dim$W=n-m$, and dim$W^{\perp\perp}=$dim$V-$dim$W^{\perp}=n-(n-m)=m=$dim$W$. **Then I'm not sure how this shows or helps show that $W=W^{\perp\perp}$.

Could someone help me with these two areas of the proof that I'm struggling with?

• If you can show that $W\subset W^{\perp\perp}$ and $\dim W = \dim W^{\perp\perp}$, does that tell you anything about the relationship between them? – Mark Oct 31 '16 at 8:34
• I'm guessing that $W=W^{\perp\perp}$ but how do you know that these two conditions is enough to conclude this? – user342661 Oct 31 '16 at 9:48

It might help to rename, say $F=W^\perp$
Then for all $w\in W$ and $f\in F$ you have: $\langle w,f\rangle =0$ which shows that $W$ and $F$ are orthogonal to each other. In particular, $w\in F^\perp = (W^\perp)^\perp$. In other words $W\subset (W^\perp)^\perp$.
If $V$ has finite dimension, then there are different ways to go: Either construct orthonormal bases for $W$ and $W^\perp$ and show that $V=W\oplus W^\perp$ (from which the conclusion follows) or use that the orthogonal projection $P:V\rightarrow W$ has $W^\perp$ as kernel.