#math/linear-algebra #TODO For $U\subseteq V$, the `annihilator` of $U$, denoted by $U^{0}$, is defined by $ U^{0}=\{ \varphi \in V':\varphi(u)=0 \text{ for all }u\in U\}. $ The set of all linear functionals, $\varphi:V\to \mathbf{F}$, where $\varphi$ maps all of $U$ to $0$, hence $U^{0}\subseteq V'$. Since $U^{0}$ is dependent on $V$, $U^{0}_{V}$ would be better notation, but we opt to use the simpler notation. # $U^{0}$ is a Subspace Suppose $U\subseteq V$, then $U^{0}$ is a subspace of $V'$. **Proof:** 1. $0\in U^{0}$ where $0$ is the linear functional on $V$ 2. Suppose $\varphi_{1},\varphi_{2}\in U^{0}$. Then $\varphi_{1},\varphi_{2}\in V'$ and $\varphi_{1}(u)=\varphi_{2}(u)=0$ for all $u\in U$. Thus, if $u\in U$, then $(\varphi_{1}+\varphi_{2})(u)=\varphi_{1}(u)+\varphi_{2}(u)=0+0=0$. Thus, $(\varphi_{1}+\varphi_{2})\in U^{0}$. 3. Evidently, $U^{0}$ is also closed under scalar multiplication. Therefore, $U^{0}$ is a subspace of $V'$. # Dimension of $U^{0}$ ==TODO==: Think more about the why and reason the proof. For now accept and move on. Suppose $V$ is finite-dimensional and $U$ a subspace of $V$. Then, $ \text{dim }U^{0}=\text{dim }V-\text{dim }U. $ We're saying that the dimension of the set of all linear maps which maps each $u\in U$ to $0$ to be the dimension of $V$ minus the dimension of $U$. Suppose $\text{dim }V=5$ and $\text{dim }U=2$, then $\text{dim }U^{0}$ is $3.$ # Conditions of the Annihilator Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then $ \mathbf{(1)}\quad U^{0}=\{ 0 \}\iff U=V $ $ \mathbf{(2)}\quad U^{0}=V'\iff U=\{ 0 \} $