I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
I solved Exercise 3.F.21.
Is my solution ok?
If my solution is ok, I am afraid that my solution contains unnecessary arguments.
Does my solution contain unnecessary arguments?
Exercise 3.F.21
Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$ with $W^0\subset U^0.$ Prove that $U\subset W$.
3.94 Definition dual space, $V'$
The dual space of $V$, denoted $V'$, is the vector space of all linear functionals on $V$. In other words, $V'=\mathcal{L}(V,\mathbb{F})$.
3.102 Definition annihilator, $U^0$
For $U\subset V$, the annihilator of $U$, denoted $U^0$, is defined by $$U^0=\{\varphi\in V':\varphi(u)=0\text{ for all }u\in U\}.$$
3.106 Dimension of the annihilator
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then $$\dim U+\dim U^0=\dim V.$$
My solution:
Let $W_1:=\{v\in V : \varphi(v)=0\text{ for any }\varphi\in W^0\}$.
Let $U_1:=\{v\in V : \varphi(v)=0\text{ for any }\varphi\in U^0\}$.
Then, it is easy to check that $W_1$ and $U_1$ are subspaces of $V$.
Obviously, $W\subset W_1$ and $U\subset U_1$.We prove that $W^0=(W_1)^0$:
Proof:
Let $\varphi\in W^0$.
Then, $\varphi(v)=0$ for any $v\in W_1$ by the definition of $W_1$.
So, $\varphi\in (W_1)^0.$
Obviously, $(W_1)^0\subset W^0$ since $W\subset W_1.$By 3.106, $\dim W+\dim W^0=\dim V.$
By 3.106, $\dim W_1+\dim (W_1)^0=\dim V.$
So, $\dim W=\dim V-\dim W^0=\dim V-\dim (W_1)^0=\dim W_1.$
So, $W=W_1.$Similarly, $U=U_1.$
Obviously, $U_1\subset W_1$ since $W^0\subset U^0$.
So, $U=U_1\subset W_1=W$.
Best Answer
It looks ok without unnecessary arguments.
Let me describe another approach. Assume there is some $u_1 \in U $ outside $W$. There is a linear function $\psi$ such that $\psi (u_1)=1$ and $\psi(w)=0$ for all $w \in W$. So there is $\psi \in W^0$ outside $U^0$.