A square matrix $n \times n$ is an invertible matrix iff the rank of the matrix is $n$.

Why does the following sentence is true?
A square matrix $n \times n$ is an invertible matrix iff the rank of the matrix is $n$.

Solutions Collecting From Web of "A square matrix $n \times n$ is an invertible matrix iff the rank of the matrix is $n$."

Rank equal to $n$ means that the range is all of $V^n.$ Since the dimension of the domain is $n$ (by assumption), the dimension of the kernel is $0.$ So, the matrix is $1-1$ and onto, so invertible.

It is a theorem that for $f:V\to W$, a linear transformation, $$\tag 1\dim\ker f+\dim{\rm im}\;f=\dim V$$ This in turns gives the result that when $\dim V=\dim W$; a transformations is onto iff it is one one iff it is a bijection. Thus, if the matrix has full rank it is an surjective transformation, whence it is one one. Conversely, a one one matrix has trivial kernel, so full rank by $(1)$, that is, its image has full dimension.