This is a simple question in linear algebra but I just had a hard time thinking about the logic.
Let V and W be two vector spaces over a field F. For simplicity, we assume they have the same finite dimensions and any linear transformation $T:V \rightarrow W$, null space is {$0$}. Let {$v_1$, $v_2$, …, $v_n$} be a basis for V. Then by theorem, any linear transformation $T:V \rightarrow W$, {$T(v_1)$, $T(v_2)$, …, $T(v_n)$} should be a basis for W. However, theorem of existence of linear transformation states that there exists precisely one linear transformation
$T:V \rightarrow W$ such that
$$T(v_i) = w_i$$ where $w_i, i = 1,..,n$ is any vectors in W. If {$w_1$, $w_2$, …, $w_n$} are linear dependent, then one of two theorms is wrong. Did I omit something?
edit: the first theorem can be derived from $N(T)$ = {$0$}, then T is injective, then T is an isomorphism given dim(V) = dim(V) < $\infty$.
Best Answer
The image vectors $\{T(v_{1}),\ldots, T(v_{n})\}$ can form a basis only if the kernel of $T$ is zero. This is not true anymore if $\{w_{1},\ldots,w_{n}\}$ are linearly dependent, because by definition you can find scalars $\lambda_{i}$'s not all zero such that $$ \lambda_{1}w_{1}+\cdots+\lambda_{n}w_{n}=0 $$ Then $T(\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n})=0$, so we have a non-zero element in the kernel (if $\lambda_{1}v_{1}+\cdots +\lambda_{n}v_{n}=0$ then all $\lambda_{i}$'s would be zero by linear independence of the $v_{i}$'s).
So in the first theorem you need to assume that $T$ is an isomorphism (or equivalently that it has trivial kernel if both spaces have the same dimension).