# Isomorphisms and Linear Maps

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?

• Perhaps have a look at this answer of mine: math.stackexchange.com/a/2411743/403337 – Chris Custer Feb 28 '19 at 3:10
• $\dim \left (\Bbb M_n (\Bbb R) \right ) = \dim\left (\Bbb {R}^{n^2} \right ) = n^2$ over $\Bbb R.$ Are the two spaces $\Bbb M_n (\Bbb R)$ and $\Bbb {R}^{n^2}$ equal? – Dbchatto67 Feb 28 '19 at 3:38

$$\Rightarrow$$ 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.