# $W^{\perp\perp}=W$ and $V^{**}$, complete proof

Let $V$ be a finite dimensional vector space over the field $K$, with a non-degenerate scalar product. Let $W$ be a subspace. Show that $W^{\perp\perp}=W$. Show that the same conclusion as in the preceding exercise is valid if by $W^{\perp}$ we mean the orthogonal complement of $W$ in the dual space $V^*$.

I am self-studying. And I want to see more examples to create more intuitions about the workings of functionals. I have posted this equation in another thread.

I got an answer by Ivo Terek:

You have a perfect pairing $$V^\ast \times V \ni (f,v)\mapsto f(v) \in \Bbb K$$that works like a inner product. In this light, the annihilator of a subspace is precisely the orthogonal complement.

Bear the relevant definitions in mind: $W^\perp = \{x \in V \mid \langle x,y\rangle = 0,\, \forall\,y \in W \}$ is a subspace of $V$, while ${\rm Ann}(W) \equiv W^0 = \{ f \in V^\ast \mid f\big|_W = 0 \}$ is a subspace of $V^\ast$. They live in different worlds.

With this in mind, $W^{\perp\perp}$ is again a subspace of $V$, and so it makes sense to wonder if $W^{\perp\perp}$ is equal to $W$. On the other hand, the double annihilator $W^{00}$ is a subspace of $V^{\ast\ast}$, so you must understand the abuse of notation: what the exercise wants you to prove is that $J[W] = W^{00}$, where $J$ is the natural map from $V$ to $V^{\ast \ast}$ (which you should know what it is, since you got this far).

Alternatively, if $X \subseteq V^\ast$ is a subspace, you can think of $X_0 = \{ v \in V \mid f(v) = 0,\,\forall\,f \in X \}$. With this notion, you can prove that $W^0_{\,\,\,0} = W$, your pick.

However after trying to complete the proof myself, I did not succeed. And I want to see this proof complete to get hold of a new example of the double dual space, a concept that has been troubling me for long.

Questions:

Could someone complete this proof?

• One inclusion $W\subseteq W^{\perp\perp}$ should be trivial, as $w$ is perpendicular to everything that is perpendicular to $w$. The other inclusion you can get from counting dimensions: if you believe that $\dim W^{\perp}=\dim V-\dim W$, you immediately get $\dim W^{\perp\perp}=\dim W$ and hence $W=W^{\perp\perp}$. The same arguments work after you replace $\perp$ via $*$: $W\subseteq W^{**}$ is for free and they are equal by the dimension argument. – Peter Franek Jul 11 '17 at 15:07
• @PeterFranek The problem that I am facing is that $W^{00}\subset V^{**}$ is not W, once $W\subset V$. How can $W^{00}=W$ if they lie in different spaces? – Pedro Gomes Jul 11 '17 at 20:20
• If $V$ is finite dimensional, there is a canonical identification between $V^{**}$ and $V$. (Given by the map $J$ from Tereks answer). So, more formally, they probably want you to show that $W^{00}=J(W)$. But in order for this to make sense, you need to know that $J$ (defined by $J(v)(w^*):=w^*(v))$ is an isomorphism $V\to V^{**}$. – Peter Franek Jul 11 '17 at 22:21
• @PeterFranek Could you please complete the proof? I am not seeing how can it belong to the same space. Does the isomorphism mean $V=V^{**}$? Is $W^{00}=J{W}=W$? – Pedro Gomes Jul 12 '17 at 11:20
• If you want to be very formal, then the assignment asks you to show that $W^{00}=J(W)$. Formally, $JW\neq W$, because they belong to different spaces. However, $J$ is an isomorphism (and a very natural one), so sometimes people simply omit $J$. For all practical purposes, it doesn't matter. Most physicist would use columns for vectors, rows for duals (in coordinates); rows act on columns, columns act on rows. Column vectors are different from row vectors, but columns vectors can be understood to be either vectors, or "duals of duals" (that's the $J$ identification). – Peter Franek Jul 12 '17 at 16:29