2
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ rank nullity thm $\endgroup$ – mathworker21 Nov 30 '18 at 0:02
  • $\begingroup$ Use $\operatorname{text}$ for $\operatorname{text}$, whenever "text" is, say, "span". $\endgroup$ – Shaun Nov 30 '18 at 0:17
  • 1
    $\begingroup$ @Shaun thank you. $\endgroup$ – Dima Nov 30 '18 at 0:27
  • 1
    $\begingroup$ 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 $T(u)$..." $\endgroup$ – Andrés E. Caicedo Dec 2 '18 at 19:32
  • 1
    $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Dec 3 '18 at 4:04
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Thank you so much. $\endgroup$ – Dima Dec 6 '18 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.