Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Prove that $$\dim V = \dim U \ +\dim U^{0}$$
I had a go with it but I am new to proofs so I would appreciate if someone could see if my argument is justified.
I tried the following approach.
First, suppose $\big\{u_{1},u_{2},..u_{m}\big\}$ is a basis of $U$. We can extend this as $\big\{u_{1},u_{2},..u_{m},u_{m+1},...u_{n}\big\}$ to form a basis of $V$.
then, $\dim \ V =n$ and $\dim \ U =m$
Note that $U^0$ is the annihilator of $U$ where $U^0 = \big\{\varphi \in V^*| \varphi(u)=0 \big\}$ for $u \in U$
Now, let $\big\{\varphi_{1},\varphi_{2},...\varphi_{n}\big\} $ be a dual basis of dual space $V^*$
Then, $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\} \in U^0$ since, $$\varphi_{m+1}(u)=\varphi_{m+2}(u)=...\varphi_{n}(u)=0$$ for any $u \in U$ where u is a linear combination of $\big\{u_{1},u_{2},..u_{m}\big\}$
(this is a direct consequence of the definition of dual basis).
Now, to show that $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$ spans $U^0$,
Let $\varphi_{i} \in U^0$, where $\varphi_{i} $ is any dual basis of $U^0$
then $\varphi_{i}$ is an element of $\big\{\varphi_{1},\varphi_{2},...\varphi_{n}\big\} $, since $ U^0 \in V^*$
Now suppose that $\varphi_{i}$ is not in $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$, then $ i<m+1$. So
$$\varphi_{i}(u)=a_{i}$$ where $a_{i}\neq 0$
since $\varphi_{i}(u)\neq 0$,
$$\varphi_{i}(u)\notin U^0$$
Now this is a contradiction so $i\geq m+1$ which means that $\varphi_{i}$ must be in $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$
This implies that $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$ spans $U^0$. But we know that $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$ is linearly independent, so $\big\{\varphi_{m+1},\varphi_{m+2},...\varphi_{n}\big\}$ is a basis of $U^0$
$\therefore \dim U^0 =n-m$
We already know that $\dim V =n$ and $\dim U =m$
$\therefore \dim V = \dim U \ +\dim U^{0}$
$\text{QED}$
To add, I appreciate any advice on how I can improve this proof.
Your proof is fine, and almost fully correct.
However, when showing that $(\varphi_{m+1}, \dots, \varphi_n) $ spans $U^\circ$, we have to consider all elements of $U^\circ$, as it might happen in general that basis vectors $b_1$ and $b_2$ are not in a given subspace but e.g. their sum $b_1+b_2$ is.
Now this is not the case: assume $\varphi=a_1\varphi_1+\dots+a_n\varphi_n\ \in U^\circ$. Then, by the dual basis, we have $\varphi(u_i) =a_i$, where the left hand side is $0$ if $i\le m$.
This shows $\varphi\in span(\varphi_{m+1}, \dots, \varphi_n) $.