If $L: V\to V$ is a linear operator, and $V$ has bases of $n$ vectors, can we say rank$L = n$?

Question

If $L: V\to V$ is a linear mapping, and $\{v_1, ..., v_n\}$ and $\{L(v_1), ..., L(v_n)\}$ are bases for $V$, can we say that rank$L = n$ ?

I'm trying to prove that nullity$(L) = 0$, so my thought process is since $V$ has a basis of $n$ vectors, dim$V = n$ if I can assume that rank$L = n$ too because $\{L(v_1), ..., L(v_n)\}$ is a basis of $V$ with $n$ vectors, then by rank nullity theorem the proof becomes trivial.

Answer 1

We can see that $\dim \operatorname{range}(L)\ge n$ because the range of $L$ contains a linearly independent set of $n$ vectors. Then because $\dim V=n$, we get $\operatorname{range}(L)=n$. So, $\operatorname{rk}(L)=n.$ Thus, by rank-nullity, $n=\operatorname{rk}(L)+\dim\operatorname{null}(L)=n+\dim\operatorname{null}(L).$ Hence, $\operatorname{null}(L)=\{0\}.$

Answer 2

The rank of $L$ is the dimension of its image so it is $dim(L)-dim(\ker(L))$. We will show that $\ker(L) = 0$, let $v\in V$ s.t. L(v)=0, then

$$0=L(v)=L\left(\sum_{i=1}^n \lambda_iv_i\right) = \sum_{i=1}^n \lambda_iL(v_i)$$

Since the $L(v_i)$ form a basis we have $\lambda_i = 0$ for all $i$. Hence $v = 0$ and $\ker(L)=0$.

This implies $dim(L)= n$.