Let $A: n\times n$, I want to prove that if $N(A)=\{0\}$ if, and only if, $A$ has $n$ linearly independent columns.
Important: You cannot use $dim(N(A)) + dim(R(A)) = n$.
I thought of doing so, as $N(A)=\{0\}\Rightarrow Ax = 0 \Rightarrow x = 0$ so A is invertible, so there is a single inverse matrix $A^{-1}$ and the system $Ax=b \Rightarrow x=A^{-1} b$ soon, this system has a single solution so each column of A has a pivot so the $n$ columns are linearly independent and $rank(A)=n$.
I need to formalize better but I would like to know if the ideas are good.
Any tips on how to start the other implication?
Hint: If the columns of $A$ are, in order, $\mathbf c_1,\mathbf c_2,\dots, \mathbf c_n$ then:
$$A\begin{pmatrix}a_1\\a_2\\\vdots\\a_n\end{pmatrix} = a_1\mathbf c_1+a_2\mathbf c_2+\cdots + a_n\mathbf c_n.$$
Now use the definition for "linearly dependent" to show the equivalent: $$\operatorname{Ker}(A)\neq \{0\}\iff \operatorname{rank}(A)<n.$$
That is there is a nonzero vector $\mathbf x$ such that $A\mathbf x=0$ if and only if the columns are not linearly dependent.