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 $\endgroup$ Commented Nov 30, 2018 at 0:02
  • $\begingroup$ Use $\operatorname{text}$ for $\operatorname{text}$, whenever "text" is, say, "span". $\endgroup$
    – Shaun
    Commented Nov 30, 2018 at 0:17
  • 1
    $\begingroup$ @Shaun thank you. $\endgroup$
    – Dima
    Commented Nov 30, 2018 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 $T(u)$..." $\endgroup$ Commented Dec 2, 2018 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. $\endgroup$ Commented Dec 3, 2018 at 4:04

1 Answer 1

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
    Commented Dec 6, 2018 at 22:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .