Converse of property of invertible matrix We know that if $A$ is invertible, then $N(A)=\{0\}$.
But I would like to ask that does the converse hold true? And how to explain it.
 A: If $A$ is $n \times n$ matrix over $\mathbb{F}$ and $N(A)$ = 0, the Rank-Nullity Theorem gives that
$$\text{rank }A = n - \dim N(A) = n,$$ so the image of the linear transformation $A$ defines is equal to all of $\mathbb{F}^n$, thus the transformation (and hence the matrix $A$ itself) is invertible.
A: Yes. We have the following statement: Suppose that $A$ is an $n\times n$ square matrix. $A$ is invertible if and only if $N(A)=\{0\}$. 
We already know that only if part is true. 
To prove the if part, suppose $N(A)=\{0\}$, then $\dim N(A)=0$. By dimension theorem, we have 
$$rank A+\dim N(A)=n$$
which  implies that $rank A=n$. Since $A$ is $n\times n$ matrix with rank $n$, $A$ must be invertible.
A: Yeah, the converse holds. That is if $N(A)=\{0\}$, then $A$ is indeed invertible. 
Suppose $A$ is a $n\times n $ matrix. Then $N(A)=\{0\}$ implies that the $n$ columns of $A$ are mutually linearly independent, because otherwise there would exist constants $c_1,c_2,\cdots,c_n$, not all $0$ such that $c_1a_1+\cdots+c_na_n=0$, where $a_1,\cdots,a_n$ are the columns of $A$. However then $c=(c_1,\cdots,c_n)^T\in N(A)$ because $Ac=0$ and $c\neq 0$. Contradiction. Hence the columns of $A$ are mutually independent. So they form a basis for $\mathbb{R}^n$. Hence there exists vectors $x_1,x_2\cdots,x_n\in\mathbb{R}^n$, such that
$Ax_k=e_k$ for $1\leq k\leq n$. ($e_k$ being the $n\times 1$ vector with all $0$'s except a $1$ as the $k$th component).
Hence the matrix $[x_1,x_2,\cdots,x_n]$ is the inverse of $A$, because 
$$A[x_1,x_2,\cdots,x_n]=[e_1,\cdots,e_n]=I_n$$
where $I_n$ is the $n\times n$ identity matrix. 
