About 3.F Exercise 21 on p.114 in "Linear Algebra Done Right" by Sheldon Axler.

Question

I am reading "Linear Algebra Done Right" by Sheldon Axler.

I cannot solve 3.F Exercise 21 on p.114.

Let $V$ be a vector space over $\mathbb{F}$.
Let $V'$ be the set of all linear maps from $V$ to $\mathbb{F}$.
Let $U^0 := \{\phi \in V' | \phi(u) = 0$ for all $u \in U\}$.

Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$ with $W^0 \subset U^0$.
Prove that $U \subset W$.

My attempt is here:

$\dim W^0 + \dim W = \dim V$
$\dim U^0 + \dim U = \dim V$

Since $W^0 \subset U^0$, $\dim W^0 \leq \dim U^0$.
So, $\dim U \leq \dim W$.

I cannot prove that $U \subset W$.

Please tell me a proof.

Answer 1

Let $\psi_1,\dotsc,\psi_m$ be a basis of $W^0$. Extend it to $\psi_1,\dotsc,\psi_m, \psi_{m+1},\dotsc,\psi_{m+n}$ a basis of $U^0$. Extend it to $\psi_1,\dotsc,\psi_m, \psi_{m+1},\dotsc,\psi_{m+n},\psi_{m+n+1},\dotsc,\psi_{m+n+k}$ to a basis of $V'$. Let $v_1,\dotsc,v_m, v_{m+1},\dotsc,v_{m+n},v_{m+n+1},\dotsc,v_{m+n+k}$ be the "primal" basis of $V$.

Let $v\in V$ be written as \begin{multline} v = a_1v_1+\dotsb+a_mv_m+a_{m+1}v_{m+1}+\dotsb+a_{m+n}v_{m+n}+\\ a_{m+n+1}v_{m+n+1}+\dotsb+a_{m+n+k}v_{m+n+k}. \end{multline} If $v\in W\subset V$ \begin{equation} \psi_1(v)=0=a_1 \end{equation} Similarly, using $\psi_2,\dotsc,\psi_m$, we conclude that $a_2=\dotsb=a_m=0$. Therefore \begin{equation} W\ni v = a_{m+1}v_{m+1}+\dotsb+a_{m+n}v_{m+n}+a_{m+n+1}v_{m+n+1}+\dotsb+a_{m+n+k}v_{m+n+k}. \end{equation} If $v\in U\subset V$, we have additionally that $a_{m+1}=\dotsb=a_{m+n}=0$. Therefore \begin{equation} U\ni v = a_{m+n+1}v_{m+n+1}+\dotsb+a_{m+n+k}v_{m+n+k}. \end{equation} Therefore, if $v\in U$, then $v\in W$. It follows that $U\subset W$.

Answer 2

Let $w_1, \cdots, w_m$ be a basis of $W$.
Let $w_1, \cdots, w_m, v_1, \cdots, v_n$ be a basis of $V$.
Let $u$ be an arbitrary element of $U$.

Then, we can write $u$ as $u = \alpha_1 w_1 + \cdots + \alpha_m w_m + \beta_1 v_1 + \cdots + \beta_n v_n$ for some $\alpha_1, \cdots, \alpha_m, \beta_1, \cdots, \beta_n \in \mathbb{F}$.

Let $\phi$ be an arbitrary element of $W^0(\subset U^0)$.

Then, $\phi(\beta_1 v_1 + \cdots + \beta_n v_n) = \phi(u) - \phi(\alpha_1 w_1 + \cdots, \alpha_m w_m) = 0$.

So, $\beta_1 v_1 + \cdots + \beta_n v_n = 0$ because if $\beta_1 v_1 + \cdots + \beta_n v_n \ne 0$, then we can find a $\psi \in W^0$ such that $\psi(\beta_1 v_1 + \cdots + \beta_n v_n) = 1$.

$\therefore$ $u = \alpha_1 w_1 + \cdots + \alpha_m w_m \in W$.