Annihlators of Intersects and Sums

Question

Let $V$ be an $n$-dimensional vector space over a field $\mathbb{F}$, and let $V'$ be the vector space of linear maps from $V$ to $\mathbb{F}$. For each subspace $X$ of $V$, let \begin{align*} X^0:=\{f \in V' : f(x) = 0 \textrm{ for all } x\in X\}. \end{align*}

Let $U$ and $W$ be subspaces of $V$.

(a) dim $(U^0)$ = $n-$dim($U$).

(b) $U^0 \cap W^0$ = $(U+W)^0$.

(c) $(U\cap W)^0$ = $U^0 + W^0$.

Would appreciate some feedback on my proofs.

(a) Since $U$ is a subspace of $V$, there exists some basis $\mathcal{U} := \{u_1,u_2,\dots,u_m\}$ of $U$. Now, extend this basis to a basis $\mathcal{B}$ for by adding the vectors $v_1,v_2,\dots, v_\ell$ where $\ell + m = n$. Now, let $\varphi_i$ be the function in $V'$ such that $\varphi_i u_k = 1$ if $i = k$ and $0$ otherwise. Also, let $\varphi_i v_k = 0$ for all $k$. Similarly, let $\psi_i v_k = 1$ if $i=k$, 0 otherwise, and let $\psi u_k = 0$ for all $k$. Then the set, $\mathcal{C} = \{ \varphi_1,\dots,\varphi_n,\psi_1,\dots,\psi_\ell \}$ is a basis for $V'$. Now, if $\rho \in V'$ is a functional such that $\rho u = 0$ for some $u \in U$, $\rho$ must be a linear combination of $\{\psi_1,\dots, \psi_\ell\}$ else $\rho$ would not annihilate $u$. Thus there are $\ell$ many vectors in the basis of $U^0$. But $n = \ell + m$, so $\ell = n-m$ as desired.

(b) Let $\varphi \in U^0 \cap W^0$. Let $x \in U+W$. Then $x = u + w$ for some $u \in U, w \in W$. Now, $\varphi(x) = \varphi(u + w) = \varphi(u) = \varphi(w) = 0$, so $\varphi \in (U+W)^0$. Conversely, suppose $\psi \in (U+W)^0$. Let $x \in U+ W$, with $x = u +w$ for some $u \in U$ and $w \in W$. Now, $\psi (x) = 0$, but $\psi(x) = \psi(u+w) = \psi (u) + \psi(w)$, so $\psi(u)$ and $\psi(w)$ must both be 0, hence $\psi \in (U^0 \cap W^0)$.

(c) Let $\varphi \in (U \cap W)^0$. If $x \in u \cap w$ it can be written $\sum_{i = 0 }^m\alpha_i u_i$ for $\alpha_i \in \mathbb{F}$, and $u_i$ basis vectors for the subspace. Similarly, it can be written as $\sum^n_{i = 0}\beta_i w_i$, with $\beta_i \in \mathbb{F}$ and $w_i$ basis vectors of $W$. Now, $\varphi(x) = 0$ which means $\varphi(\sum^m_{i = 0}\alpha_i u_i) = 0$ and $\varphi(\sum^n_{i = 0}\beta_i w_i) = 0$. This implies $\varphi \in U^0 + W^0$. Now let $\rho \in U^0 + W^0$, then $\rho = \varphi + \psi$ for some $\varphi \in U^0$ and $\psi \in W^0$. Using the same $x$ as before we get, \begin{align*} (\varphi + \psi)x = \varphi x + \psi x = \varphi (\sum^n_{i = 0}\beta_i w_i) +\psi( \sum^m_{i = 0} \alpha_i u_i). \end{align*} These two terms are only 0 if $\varphi \in W^0$ and $\psi \in U^0$ which would imply $\rho \in (U\cap W)^0$.

In particular, I'm the least confident about the forward containment in part (c). I feel like I may have made some logical jumps without proper justification.

Answer 1

I think your (a) and (b) "$U^0 \cap W^0 \subseteq (U+W)^0$" part are fine.

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For (b) your proof of "$U^0 \cap W^0 \supseteq (U+W)^0$", $\psi (u) + \psi(w)=0$ does not imply both of them are zero, and hence your proof does not work.

A proof for $U^0 \cap W^0 \supseteq (U+W)^0$: Assume $\psi\in (U+W)^0$, i.e. $\psi(x)=0$ for all $x\in U+W$. Given any $u\in U$, since \begin{equation*} u=u+0\overbrace{\in}^{0\in W} U+W \end{equation*} and (by hypothesis) $\psi(x)=0$ for all $x\in U+W$, we have $\psi(u)=0$. Therefore by the definition of $U^0$, we infer $\psi\in U^0$. The same argument gives $\psi\in W^0$ $\implies$ $\psi\in U^0\cap W^0$.

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For (c) "$(U\cap W)^0 \subseteq U^0 + W^0$": (I have written in the comment, but for the completeness I repeat it)

Let $\{e_i\}$ be a basis of $U\cap W$. Extend it and let $\{e_i\}\cup \{u_i\}$ and $\{e_i\}\cup \{w_i\}$ be a basis of $U, W$ respectively. Assume $\{e_i\}\cup \{u_i\}\cup \{w_i\}\cup \{f_i\}$ is a basis of $V$. Consider $\varphi=\psi+\tau$, where $\psi(e_i)=\psi(u_i)=0$, $\psi(w_i)=\varphi(w_i)$, $\psi(f_i)=\varphi(f_i)$ and $\tau(e_i)=\tau(w_i)=\tau(f_i)=0$, $\tau(u_i)=\varphi(u_i)$. (The values of $\psi,\tau$ on basis vectors are defined, and so $\psi,\tau$ are well-defined.) It is immediate that $\psi\in U^0$ and $\tau\in V^0$.

For "$(U\cap W)^0 \supseteq U^0 + W^0$" your proof is overcomplicated.

A shorter proof: Select $\omega\in U^0 + W^0$. Assume $\omega=\psi_U+\psi_W$, where $\psi_U\in U^0,\ \psi_W\in W^0$. Then for all $x\in U\cap W$, $\psi_U(x)=0$ since $x\in U\cap W\subseteq U$ and $\psi_U\in U^0$. The same argument gives $\psi_W(x)=0$. Therefore \begin{equation*} \omega(x)=\psi_U(x)+\psi_W(x)=0\implies \omega\in (U\cap W)^0. \end{equation*}