# Confused about double dual spaces

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.

• 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 $V^{\star\star}$. – AdLibitum Jan 14 '17 at 21:02

## 2 Answers

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 $$v$$.

• I understood the words "weird" and took comfort in "that's fine". +1 – Antoni Parellada Jan 14 '17 at 21:36
• Best answer I’ve seen on this site in a long time. – PossumP Apr 4 '19 at 22:34
• @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) – C.F.G Nov 23 '20 at 18:46
• @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 – Ben Grossmann Nov 23 '20 at 18:52

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 $E''$)

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