Let $A$ be an $m \times n$ matrix of real numbers and $cl(A)$ denotes the column space of $A$, $R(A)$ denotes the row space of $A$ and $N(A)$ denotes the Null space of $A$.
I want to prove this statement. If $m=n$ and $det(A) \neq 0$ then $dim(N(A))=0$
My attempt.
An $m \times n$ matrix of real numbers is a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$
let $m=n=k \implies$ $T: \mathbb{R}^k \to \mathbb{R}^k$ Now since the matrix $A$ is invertible $\implies$ $T$ has an inverse $T^{-1}$ $\implies$ $T$ is injective thus the Null space $N(T)=\{0\} \implies $ $dim(N(T))=0$