# $(V_1 \cap V_2)^{0}=V_1^0+V_2^0$ for infinite dimensional spaces.

Let $V$ be a vector over a field and $V_1, V_2$ are linear subspaces of V.

$$(V_1 \cap V_2)^{0}=V_1^0+V_2^0$$

is what I'm supposed to show, but I do not believe this to be true. Let's assume $V_1\neq V_2$ and then we can find w.l.o.g a $v_1 \in V_1$ that's not in $V_2$.

Let then $\psi \in (V_1 \cap V_2)^{0} = \{ \psi \in V^{*}| \psi(v)=0 \quad \forall v\in V_1 \cap V_2 \}$. $\psi(v_1)$ is not necessarily 0, since $\psi$ only has to be zero for vectors in the intersection. But surely $\psi$ has to fullfill that condition to be an element of $V_1^0+V_2^0$ since that is the very definition of all the elements of $V_1^0$. Hence, the sets are not necessarily equal.

Where is my mistake? As this is a book exercise I'm quite sure it's true, so I'd appreciate somebody convincing me of the hypothesis

• $\psi\left(v_1\right)$ is not necessarily $0$, but it could be $0$. Also, there exists $\phi\in V_1^0+V_2^0$ such that $\phi\left(v_1\right)=0$, but not all of them satisfy this. Hence your argument does not hold. – Guy May 10 '17 at 16:34
• I phrased that wrongly, I edited it. But still I can't see how an element of $(V_1 \cap V_2)^{0}$ needs to be in $V_1^0+V_2^0$ – Jonathan May 10 '17 at 16:36
• – Arnaud D. May 10 '17 at 16:37
• Also : $V^0_1+V_2^0$ is larger than $V^0_1$, so even if $\psi(v_1)\neq 0$, it can still be in $V^0_1+V_2^0$. – Arnaud D. May 10 '17 at 16:40
• Okay, I see the mistake. I was able to prove the other inclusion now, but I'm still stuck on the first one! – Jonathan May 10 '17 at 16:49

Here's a proof of the forward inclusion that works for infinite dimensional $V$.

Let $B_{1,2}$ be a basis of $V_1 \cap V_2$. Since $V_1 \cap V_2$ is a subspace both of $V_1$ and of $V_2$, we can (assuming choice) extend this basis both to a basis $B_1$ of $V_1$ and a basis $B_2$ of $V_2$. Moreover the union $B_1 \cup B_2$ is linearly independent (easily verified). This means that $B_1 \cup B_2$ can be extended to a basis $B$ of $V$.

With this framework established, we let $\psi \in (V_1 \cap V_2)^0$ arbitrary.
We'll define $\psi_1$ and $\psi_2$ by their actions on the basis $B$ of $V$ so that

• $\psi_1 \in V_1^0$
• $\psi_2 \in V_2^0$
• $\psi = \psi_1 + \psi_2$

For any $b \in B$, define:

$$\psi_1(b) = \begin{cases} 0 & b \in B_1 \\ \psi(b) & \text{else} \end{cases} \\ \psi_2(b) = \begin{cases} \psi(b) & b \in B_1 \setminus B_{1,2} \\ 0 & \text{else} \end{cases}$$

From these definitions $\psi_1$ clearly annihilates the basis of $B_1$ of $V_1$, and since $(B_1 \setminus B_{1,2}) \cap B_2 = \emptyset$, $\psi_2$ annihilates the basis $B_2$ of $V_2$.

Furthermore we have the following breakdown of $\psi_{1} + \psi_2$:

$$\psi_1(b) + \psi_2(b) = \begin{cases} 0 + 0 & b \in B_{1,2} \\ 0 + \psi(b) & b \in B_1 \setminus B_{1,2} \\ \psi(b) + 0 & \text{else} \end{cases}$$

Thus the desired decomposition of $\psi$ into the sum of annihilators of $V_1$ and $V_2$ does exist, showing the inclusion $(V_1 \cap V_2)^0 \subset V_1^0 + V_2^0$.