Is $L_B$ onto if $B$ is an $n$ by $p$ matrix with $\operatorname{rank}(B)=n$?

Question

Is $L_B$ onto if $B$ is an $n$ by $p$ matrix with $\operatorname{rank}(B)=n$?

definition from the book: Definition. If $A ∈ M_{m×n}(F)$, we define the rank of $A$, denoted $\operatorname{rank}(A)$, to be the rank of the linear transformation $L_A: F_n → F_m$.

By definition we have that $L_B: F_p\to F_n$ such that for $\forall x \in X$ $L_B$ maps it to $Bx \in F_n$. So we have $\operatorname{rank}(B) = R(L_B) = \dim(F_p)$. By dimension theory: $\operatorname{nullity}(L_B) + \operatorname{rank}(L_B) = \dim(F_p)$ so $\operatorname{nullity}(L_B) = 0$ so it is one-to-one.

Answer 1

If $B$ is a $n\times p$ matrix, then $\textsf{L}_B$ going from $\textsf{F}^p$ to $\textsf{F}^n$.

Now $\operatorname{rank} (B)=n$ means that $\dim \textsf{R}(\textsf{L}_B)=n$, and since $\textsf{F}^n$ is the only $n$-dimensional vector subspace of $\textsf{F}^n$, we have then $$\textsf{R}(\textsf{L}_B)=\textsf{F}^n$$ So, $\textsf{L}_B$ is onto.