Let V and W be n-dimensional vector spaces, and let $T:V \rightarrow W$ be a linear mapping. Suppose that $\beta$ is a basis for V. Prove that T is an isomorphism if and only if $T(\beta)$ is a basis for W.
My first question is since basis is a linearly independent set that spans V, why can it be written as $T(\beta)$, where it is in the input position?
Here is some work I have done:
$\Rightarrow$: We want to show that $T(\beta)$ is a linearly independent set and $T(\beta)$ spans $W$.
Let $x \in V$, then $x= \sum_{i=1}^n a_iv_i$, and let $w \in W$.
Since T is injective,N(T)={0},$\sum_{i=1}^n a_iv_i$ $\in N(T)$, so $\sum_{i=1}^n a_iv_i$=0.
Since {$v_i$} is a basis for V, we have got unique scalars $a_i=0$.
$T(\sum_{i=1}^n a_iv_i)$=0, by linearity, $\sum_{i=1}^n a_i T(v_i)$=0.
Since T is surjective, we have $\sum_{i=1}^n a_i T(v_i)$=w.
Since w is written as a linear combination of $T(v_i)$, span $(T(\beta))$=W. Hence $T(\beta)$ is a basis for W.
$\Leftarrow$: For this direction, is it trivial?
Best Answer
Write $\beta=\{b_1,...,b_n\}$. Suppose $T(x)=0$. Since $T(\beta)$ is a basis for $W$, there exists $c_1,...,c_n$ such that $c_1 T(b_1)+...+c_n T(b_n)=T(x)=0$. Clearly, $c_1=...=c_n=0$. Hence, $x=c_1b_1+...+c_nb_n=0$. This shows $T$ is injective.
By a similar method, let $y\in W$. Then there exists $d_1,...,d_n$ such that $d_1T(b_1)+...+d_nT(b_n)=y.$ Hence $x=d_1 b_1+...+d_nb_n$ solves $T(x)=y$. Therefore, $T$ is surjective.