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WTS: $V\simeq$ W $\iff$ $dimV=dimW$

My proof: Assume $T: V \rightarrow W$ is an isomorphism, therefore the kernel for the linear map is the zero vector therefore $dim(ker(T))=0$ and since it is bijective, it is surjective therefore $imT=W$ hence $dimT=dimW$, by the rank-nullity theorem, dimV=dimW.

Assume $dimV=dimW$ since they are both finite dimensional, V=W hence defining a linear map $T:V \rightarrow V$ by $T(v) =v$ is an isomorphism.

Is my proof correct? May someone tell me how to improve it, please?

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  • $\begingroup$ Perhaps have a look at this answer of mine: math.stackexchange.com/a/2411743/403337 $\endgroup$ – Chris Custer Feb 28 '19 at 3:10
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    $\begingroup$ $\dim \left (\Bbb M_n (\Bbb R) \right ) = \dim\left (\Bbb {R}^{n^2} \right ) = n^2$ over $\Bbb R.$ Are the two spaces $\Bbb M_n (\Bbb R)$ and $\Bbb {R}^{n^2}$ equal? $\endgroup$ – Dbchatto67 Feb 28 '19 at 3:38
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$\Rightarrow$ Assume $T: V \rightarrow W$ is an isomorphism, let $v = \{v_1, v_2, ...,v_n\}$ the basis of $V$. Then $T(v)= \{T(v_1), T(v_2), ...,T(v_n)\}$ is a basis of $W$, thus dim $V$$=$ dim $W$.

$\Leftarrow$ Assume dim $V$$=$ dim $W$. Let $v = \{v_1, v_2, ...,v_n\}$ the basis of $V$, $w = \{w_1, w_2, ...,w_n\}$ the basis of $W$. For any $x = x_1v_1+x_2v_2+...+x_nv_n \in V$, define $T(x) = x_1w_1+x_2w_2+...+x_nw_n$. It is easy to check that $T$ is an isomorphism.

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    $\begingroup$ The restriction to finite dimensions is not needed $\endgroup$ – Hagen von Eitzen Feb 28 '19 at 4:51

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