What is the relationship between onto/ one to one and rank? Let A ∈ R^m×n.
What is the rank if A is onto? What about not onto?
What is the rank if A is one to one? what about not one to one? 
I understand what rank is but i dont understand how you could find the rank just from knowing if a matrix is onto/one to one. And i dont get how it changes if its not onto/one to one 
 A: Let $A\in R^{m×n}$
1) To be onto it is necessary that $n\geq m$ and that $rank(A)=m$
Infact to be onto (i.e. surjective) means that the columns of $A$ contain a basis of $\mathbb{R^m}$ thus you need exactly $m$ columns of A to be linearly independent thus must be rank(A)=m and of course $n \geq m$.
2) To be one to one it is necessary that $n \leq m$ and that $rank(A)=n$
Infact to be one to one (i.e. injective) means that the columns of $A$ are linearly independent (but not necessarly contain a basis of $\mathbb{R^m}$) thus you can't have more than $m$ columns that is $n \leq m$ and must be rank(A)=n. 
A: $A \in \mathbb R^{m \times n}$ is a function $\mathbb R^n \to \mathbb R^m$.
Rank is the dimension of the image as a vector space.
If it is injective (one to one), then a linear map can't map to a vector space of lesser degree. Likewise, if it is surjective, then its rank should be $m$, since otherwise, the image cannot be all of $\mathbb R^m$ as a linear subspace.
A: If the rank is m, it is onto (any less, and it fails to be). If the rank is n, it is one-one (any less, and it fails to be). Note that the rank is at most $\min(m,n)$.
