# 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, $$dim_F U=dim_F \operatorname{Ker(T)} + dim_F \operatorname{Rang(T)}=0+n=n$$.

• rank nullity thm – mathworker21 Nov 30 '18 at 0:02
• Use $\operatorname{text}$ for $\operatorname{text}$, whenever "text" is, say, "span". – Shaun Nov 30 '18 at 0:17
• @Shaun thank you. – Dima Nov 30 '18 at 0:27
• In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..." – Andrés E. Caicedo Dec 2 '18 at 19:32
• Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof. – Andrés E. Caicedo Dec 3 '18 at 4:04

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 $$\dim U=\dim\ker(T)+\dim\operatorname{range}(T)=0+n=n$$

• Thank you so much. – Dima Dec 6 '18 at 22:15