# Show false: If $V$ is a vector space of finite dimension over $F$ and $V=A \oplus B$, then $V^* = A^* \oplus B^*$.

Let $V$ be a finite dimensional vector space over a field $F$, $A$ and $B$ subspaces of $V$ such that $V=A \oplus B$. I want to show that it does not follow that $V^* = A^* \oplus B^*$.

Proof attempt: Assume that $A\neq V$. Let $f\in A^*$, then $f:A\to F$. If $g\in V^*$, then $g:V\to F$. Since $f$ and $g$ have different domains, we can't conclude that $f\in V^*$. So $A^*$ is not a subset of $V^*$ and the result follows.

• But if $f \in A^*$ and $g \in B^*$ then $(f,g) \in V^*$. – mechanodroid Sep 2 '18 at 19:55
• What does $=$ stand for? Equality as sets? Just a note, $A$ is not a subset of $V$ either. – Ennar Sep 2 '18 at 22:01

It depends on what you mean by that equality, and how do you define the direct sum (internal or external constructions may change the result, if you want an equality of sets). What is certainly true is that $V^* \simeq A^* \oplus B^*$.

Recall that a linear map $h : A \oplus B \to W$ is uniquely determined by maps $f: A \to W$ and $g: B \to W$. That is, given two such maps, we can construct $h(a,b) := f(a)+g(b)$, and this is the only such linear function that verifies $h(a,0) = f(a)$ and $h(0,b) = g(b)$.

Now, every element of $V^* = (A \oplus B)^*$ is a linear functional $\phi : A \oplus B \to \mathbb{F}$ and so it is in bijective correspondence with maps $\phi(-,0): A \to \mathbb{F}$ and $\phi(0,-) : B \to \mathbb{F}$. That is, the function

$$\Gamma: V^* \rightarrow A^* \oplus B^* \\ \ \phi \mapsto (\phi(-,0),\phi(0,-))$$

is bijective. Moreover, this application is linear and so it is an isomorphism.

• Do you have any insight into why the equality does not hold? Also, I should have been more specific: Take this to be equality of sets (or subspaces) and the external direct sum. – JJH Sep 3 '18 at 7:38
• I don't know if this counts as insight but, the elements of $V^*$ are linear functions from $V$ to the field and the elements of $A^* \oplus B^*$ are tuples of elements, with each coordinate having a linear functional with the corresponding domain. Clearly these two don't have the same 'type' of elements, even though they have the same structure as vector spaces. It is 'just' a technical problem, like the one that makes $A \times (B \times C) \neq A \times B \times C$ because the latter has $3$-uples as elements and the former has tuples, where the second coordinate is a tuple itself. – Guido A. Sep 3 '18 at 19:04
• The analogy with the Cartesian product of sets is much appreciated. Marking this as the answer. – JJH Sep 3 '18 at 21:10

It depends on how you interpret $\oplus$.

The dual $V^*$ is indeed isomorphic to the (external) direct sum $A^*\oplus B^*$. Indeed, if $(f,g)\in A^*\oplus B^*$, we can define $$\overline{(f,g)}\colon V\to F \qquad \overline{(f,g)}(a,b)=f(a)+g(b)$$ and $\overline{(f,g)}$ is linear. Suppose $\overline{(f,g)}$ is the zero map. Then, for all $a\in A$, $$\overline{(f,g)}(a,0)=f(a)=0$$ and so $f=0$; similarly, $g=0$. Therefore the map $(f,g)\mapsto\overline{(f,g)}$ is an injective linear map $A^*\oplus B^*\to V^*$. Since domain and codomain have the same dimension, the map is also surjective.