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Problem Let $V$ and $W$ be finite dimensional vector spaces over the field $F$. Prove that $V$ is isomorphic to $W$ iff $\operatorname{dim}V=\operatorname{dim}W$. \operatorname{dim} Attempt

$\Rightarrow$ Define a linear transformation $T$ from $V$ to $W$. Suppose $V$ is isomorphic to $W$ but $\operatorname{dim}V\neq \operatorname{dim}W$.Let $\operatorname{dim}V=m$ and $\operatorname{dim}W=n$ provided $m\neq n$. If $m<n$ then $T$ is not onto and if $m>n$ then $T$ is not one-one. Contradiction ,Thus $\operatorname{dim}V=\operatorname{dim}W$.

$\Leftarrow$ Suppose $\operatorname{dim}V=\operatorname{dim}W$. Let $(a_1,...,a_n)$ and $(b_1,...,b_n)$ be basis of $V$ and $W$ respectively. Define a linear transformation $T:V\rightarrow W$ such that $T(a_i)=b_i$ ,where $1\leq i\leq n$.

  1. $T$ is injective iff $T$ sends linearly independent set to linear independent. Let $a_1,...,a_n$ be vectors in $V$ and $a\in V$, then $$a=c_1a_1+...+c_n a_n$$

therefore $$c_1T(a_1)+...+c_n T(a_n)=0$$

$$T(c_1a_1+...+c_n a_n)=0=T(0)$$ Thus ,$c_1a_1+...+c_na_n=0$ and $c_1=...=c_n=0$. Thus the image set of $T$ linearly independent.

  1. $T$ is onto. Since the nullity of $T$ is $0$.

  2. $T$ is linear transformation: $$T(ca_i+a_j)=cb_i+b_j=cT(a_i)+T(a_j)$$.

Q.E.D. Is the proof correct?

  • @surb See my recent edit. –  Jun 24 '20 at 11:33
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    The idea is here, but not well written. You want to show that $\ker(T)=0$. Let $x\in V$, then $$x=x_1a_1+...+x_na_n.$$ Then $$0=T(x)=T(x_1a_1+...+x_na_n)=x_1T(a_1)+...+x_nT(a_n)=x_1b_1+...+x_nb_n.$$ Since $(b_1,...,b_n)$ is a basis of $W$, we get $x_i=0$ for all $i$, and thus $x=0$. Therefore $\ker(T)={0}$. – Surb Jun 24 '20 at 11:59
  • Tip: use \operatorname{dim}. It produces $\operatorname{dim}$. – K.defaoite Jun 24 '20 at 13:51
  • @defaoite People can understand I think. I feel a bit lazy to write more. –  Jun 24 '20 at 13:54

2 Answers2

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Proof seems OK to me. Only the lack of precise writing. Further

For one-one, we need to prove that $Ker T= \{0\}$ (as $T(a_i) = T(a_j)$ provided $i\neq j$) is same as $a_i-a_j \in Ker T$). Assume the contrary. Suppose $a\neq 0 \in Ker T$. Let $a = c_1 a_1 + \dots + c_n a_n$ not all $a_i$s zero.

$T(a) = T(c_1 a_1 + \dots + c_n a_n) = c_1 T(a_1) + \dots + c_n T(a_n) = c_1 b_1 + \dots + c_n b_n = 0$ which implies $c_i = 0$ for all $1 \leq i \leq n$. Since $b_1,...,b_n$ are basis for $W$.

For more details .You can see here

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See this is more precise.

$(\Rightarrow)$ Let $m=\operatorname{dim}(V)$ and $n=\operatorname{dim}(W)$. Let $T$ be a isomorphism from $V$ onto $W$. Since $T$ is one one, $\operatorname{Nullity}(T)=0$. Next, since $T$ is onto, so $\operatorname{Im}(T)=W$, which implies $\operatorname{rank}(T)=n$. Therefore from Rank Nullity theorem we get, $m=n$.

$(\Leftarrow)$ Let $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ be bases of $V$ and $W$ respectively. Let $T:V\rightarrow W$ be the linear transformation induced by the map $v_i\mapsto w_i$ for all $i=1,2,\dots,n$. Then $T$ is an isomorphism from $V$ to $W$.