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I am a first year student, studying linear algebra. In the lecture we briefly discussed double dual spaces and I am not sure if I understood it correctly: we take a function f that is an element of the dual space and we evaluate the function on a vector from V and its value is an element of the double dual space?

Thank you very much in advance.

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  • $\begingroup$ That's right. The bidual space $V^{\star\star}$ is the the space of linear functions that assign a scalar value to linear functionals on $V$. Fixing a vector $v$ the assignment $V^\star\ni\lambda\mapsto\lambda(v)$ defines an element in $V^{\star\star}$. $\endgroup$ – AdLibitum Jan 14 '17 at 21:02
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Nope, you're missing a bit here. Long story short: the elements of the double dual space are functions that take a function $f$ from the dual space, and evaluate the function on a vector from $V$. That is, an element of the double dual space is a function of the form $f \mapsto f(v)$. It's nice to think of this weird function as just being that vector $v$ from $V$.


Long story long: for simplicity, I'll talk about vector spaces over $\Bbb R$, but the same applies over arbitrary fields.

The first thing to understand is that the set $\mathcal L(U,V)$ of linear transformations between two vector spaces $U$ and $V$ forms a vector space. For example, if $U = \Bbb R^n$ and $V = \Bbb R^m$, then $\mathcal L(U,V)$ is canonically identified with the space $\Bbb R^{m \times n}$ of $m \times n$ matrices. In general, $\dim (\mathcal L(U,V)) = \dim(U) \cdot \dim(V)$. Dimension is important because any vector spaces of the same (finite) dimension are isomorphic.

Now, for any vector space $V$, $V^* = \mathcal L(V,\Bbb R)$ is the dual space of $V$. The elements of $V^*$ are called linear functionals; they are linear transformations that take vectors and produce numbers. Notably, $\dim(V^*) = \dim(V) \cdot \dim(\Bbb R) = \dim(V) \cdot 1 = \dim (V)$. So, any (finite dimensional) space is isomorphic to its dual space.

The double dual space is the dual of the dual. That is, $V^{**} = \mathcal L(V^*, \Bbb R) = \mathcal L(\mathcal L(V,\Bbb R),\Bbb R)$. The elements of this space are linear transformations that take linear functionals and produce numbers. If that seems weird and unintuitive, that's fine: it should.

Just like $V^*$, $V^{**}$ is isomorphic to $V$, since $\dim(V^{**}) = \dim(V^*)\cdot 1 = \dim(V)$. However, it turns out that $V^{**}$ is canonically isomorphic to $V$. That is (for our purposes), it is isomorphic in a "really nice way". In particular, there is a really nice invertible linear map that takes us from $V$ to $V^{**}$, and it's so slick that we can think of $V$ and $V^{**}$ as being "essentially the same space".

Let's describe that map $\alpha:V \to V^{**}$. For any vector $v \in V$, we want an element $\alpha(v) = \alpha_v \in V^{**}$, which is to say that $\alpha_v$ takes in functionals $f \in V^*$, and produces a number. So, we define $$ \alpha_v(f) = f(v) $$ In other words, the question of "is $V$ canonically isomorphic to $V^{**}$?" can be roughly translated as "is there a natural way to use $v$ to make an element $f \in V^*$ into a number?" Our answer is, "yes: plug $v$ into $f$". For any vector $v \in V$, $\alpha_v$ is the element of $V^{**}$ that tells you to plug in $v$.

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  • $\begingroup$ I understood the words "weird" and took comfort in "that's fine". +1 $\endgroup$ – Antoni Parellada Jan 14 '17 at 21:36
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    $\begingroup$ Best answer I’ve seen on this site in a long time. $\endgroup$ – PossumP Apr 4 '19 at 22:34
  • $\begingroup$ @BenGrossmann: Although it is $V^{**}$ canonically isomorphic to $V$ but I don't see it naturally, i.e. $V\subset V^{**}$ or $V^{**}\subset V$. (the embedding of a vector space in its double dual) $\endgroup$ – C.F.G Nov 23 '20 at 18:46
  • $\begingroup$ @C.F.G Pinning down the definition of "natural" in this context is a bit tricky. Of course, it is not technically correct to say that $V \subset V^{**}$. However, it suffices for an introductory context to note that the injective map $v \mapsto \alpha_v$ can be defined using only the fact that $V$ is a vector space, and no "choices" (e.g. a selection of basis) need to be made $\endgroup$ – Ben Grossmann Nov 23 '20 at 18:52
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Yes, $\psi_x : E' \to \Bbb{R}$ defined by $\psi_x(f) = f(x)$ is an element of $E''$.

That even give us a canonical injection from $E$ to $E''$ ( $x\mapsto \psi_x$)

But note that often not all elements of $E''$ can be written as such (there's not always a bijection between $E$ and $E''$)

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  • $\begingroup$ This doesn't address the fundamental confusion of what a double dual space is $\endgroup$ – Ben Grossmann Jan 14 '17 at 21:04

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