Prove that if $T:U \to V$ is an isomorphism and $\dim_F (V)=n\in \mathbb{N}$, show that $\dim_F(U)=n\in \mathbb{N}$ 
Prove that if $T:U \to V$  is an isomorphism and $\dim_F (V)=n\in \mathbb{N}$, show that $\dim_F(U)=n\in \mathbb{N}$

my attempt: [any of these is true? please tell me if there exists any mistake, My prof is so careful - and sorry I don't speak English well. Thanks]


*

*First proof:


Let $\{T(u_1),...,T(u_n)\}$ is a basis for $V$.
it is obvious that $span\{u_1,...,u_n\} \subseteq U$.
if $u\in U$, then there exists $T(u) \in rang(T)=V$ [since $T$ is onto, then $rang(T)=V$]
Hence, $T(u)$ can be written as a linear combination of $T(u_1),...,T(u_n)$.
$T(u)=\alpha_1 T(u_1)+...+\alpha_n T(u_n)$, where $\alpha_1, ..., \alpha_1 \in F$
by linearity of $T$
$T(u)=T(\alpha_1 u_1)+...+T(\alpha_n u_n)=T(\alpha_1 u_1+...+\alpha_n u_n)$
since $T$ is one-to-one, then 
$u=\alpha_1 u_1+...+\alpha_n u_n$
Thus, $u \in \operatorname{span} \{u_1,...,u_n\}$.
Therefore, $U \subseteq \operatorname{span}\{u_1,...,u_n\}$
As a result, $U=\operatorname{span}\{u_1,...,u_n\}$.
If $\beta_1 u_1+...+\beta_n u_n=0_U$, where $\beta_1, ...,\beta_n \in F$
Then $T(\beta_1 u_1+...+\beta_n u_n)=\beta_1 T(u_1)+...+\beta_n T(u_n)= 0_V$
since $T(u_1),...,T(u_n)$ are linearly independent, so
$\beta_1=...=\beta_n=0$.
Therefore, $u_1,...,u_n$ are linearly independent.
As a result, $\{u_1,...,u_n\}$ is a basis for $U$, and $\dim_F U =n$.


*

*Second proof:


Since $T$ is isomorphism, then $T$ invertible, $\operatorname{Ker(T)}=\{0_U\}$ and $\operatorname {Rang(T)}=V$.
Thus $dim_F \operatorname{Ker(T)}=0$ and $dim_F \operatorname{Rang(T)}=dim_F V=n$
Therefore, $dim_F U=dim_F \operatorname{Ker(T)} + dim_F \operatorname{Rang(T)}=0+n=n$.
 A: Your proof is basically right. You could just make it simpler and clearer.
Let $\{T(u_1),T(u_2),\dots,T(u_n)\}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. $\{u_1,u_2,\dots,u_n\}$ is linearly independent.
Indeed, if $\alpha_1u_1+\alpha_2u_2+\dots+\alpha_nu_n=0$, then also
$$
0=T(\alpha_1u_1+\alpha_2u_2+\dots+\alpha_nu_n)=
\alpha_1T(u_1)+\alpha_2T(u_2)+\dots+\alpha_nT(u_n)
$$
forcing $\alpha_1=\alpha_2=\dots=\alpha_n=0$.
Second fact. $\{u_1,u_2,\dots,u_n\}$ spans $V$.
Let $v\in V$; then $T(v)=\alpha_1T(u_1)+\alpha_2T(u_2)+\dots+\alpha_nT(u_n)$ for some scalars $\alpha_1,\alpha_2,\dots,\alpha_n$. This can be rewritten as
$$
T(v)=T(\alpha_1u_1+\alpha_2u_2+\dots+\alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=\alpha_1u_1+\alpha_2u_2+\dots+\alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $\dim\operatorname{range}(T)=\dim V=n$; since $T$ is injective, $\dim\ker(T)=0$. The rank-nullity theorem says
$$
\dim U=\dim\ker(T)+\dim\operatorname{range}(T)=0+n=n
$$
