Given finite-dimensional vector spaces $V$ and $W$ such that $\dim(V) = \dim(W)$, show that if {$v_1,…,v_n$} is the basis of $V$ such that {$T (v_1),…,T(v_n)$} is a basis of $W$, so $T$ is an isomorphism.
My attempt:
If $T$ is isomorphism then there are vector spaces $V$ and $W$ over a field $\mathbb K$, a linear transformation $T : V \to W$ bijection. Then we must show that $T : V \to W$ is a bijection.
$$\ker(T) =\lbrace v \in V \mid T(v) = 0\rbrace,$$
$$\operatorname{Im}(T) =\lbrace w \in W \mid \exists v \in V: \: T(v) = w \rbrace$$
Surjective
As per hypothesis {$T(v_1),…,T(v_n)$} is a base of $W$ so {$T(v_1),…,T(v_n)$} generates all $W$ and $\operatorname{Im} (T)$ is therefore surjective.
Injective
$$\ker(T) = \{0\}$$
($\Longrightarrow$)Suppose $v \in \ker (T)$, then $T(v) = 0$. However we know that $T(0)=0$ since $T$ is linear. Since $T$ is injective and $T(v) = T(0)$, we must conclude that $v = 0.$
($\Longleftarrow$)Suppose $T(v)=T(v’) ~\forall~ v,v’ \in V$, then $0 = T(v) − T(v’) = T(v − v’) \implies v − v’ \in \ker(T)(=\{0\})$.
How is it?
Thanks.