Let $V$ and $W$ be n-dimensional vector spaces, and let $T: V \rightarrow W$ be a linear mapping. Suppose $\beta$ is a basis for $V$. Prove $T$ is an isomorphism if and only if $T(\beta)$ is a basis for $W$.
Let $\beta$ and $\gamma$ be basis for $V$ and $W$ respectively. $\beta = \{v_1,v_2,…v_n\}$, and let $\gamma=\{w_1,w_2,…w_n\}$. Since $T$ is an isomorphism, $T(\beta)$={$T(v_1), T(v_2), T(v_3),…, T(v_n)$} Since T is surjective, $T(\beta)$=W. Hence $T(\beta)$ is a basis for $W$.
Can someone check my forward proof?
Best Answer
$T(\beta)$ is not equal to $W$, the span is it. If $w\in W$, since $T$ is surjective, there exists $v\in V$ such that $w = T(v)$, and since $\beta$ is a basis for $V$, $v$ can be written as a linear combination of $v_1,\dots,v_n$. Use this to show that $w$ can be written as a linear combination of $T(v_1),\dots,T(v_n)$.