TITLE: Is a linear transformation onto or one-to-one?
QUESTION [30 upvotes]: The definition of one-to-one was pretty straight forward. If the vectors are lin.indep the transformation would be one-to-one. But could it also be onto? 
The definition of onto was a little more abstract. Would a zero-row in reduced echelon form have any effect on this? I just assumed that because it has a couple of free variables it would be onto, but that zero-row set me off a bit. Say if the matrix is 4 by 5. With two free variables. And the zero-row in echelon form. 
(Sorry for not posting the given matrix, but that is to specific. And I don't want to get a ban from uni for asking online. The task is determine the onto/one-to-one of to matrices)
I'll check back after class and update the question if more information is desirable.
Update
The definitions in the book is this; 
Onto: 
$T: \mathbb R^n \to \mathbb R^m $ is said to be onto $\mathbb R^m  $ if each b in 
$\mathbb R^m  $ is the image of at least one x in $\mathbb R^n  $
One-to-one:
$T: \mathbb R^n \to \mathbb R^m $ is said to be one-to-one $\mathbb R^m  $ if each b in 
$\mathbb R^m  $ is the image of at most one x in $\mathbb R^n  $
And then, there is another theorem that states that a linear transformation is one-to-one iff the equation T(x) = 0 has only the trivial solution. 
That doesn't say anything about onto. 
Then there is this bit that confused be about onto: 
Let $T: \mathbb R^n \to \mathbb R^m $ be a linear transformation and let A be the standard matrix for T. Then: 
T maps $T: \mathbb R^n$ onto $\mathbb R^m $ iff the columns of A span $\mathbb R^m $.

REPLY [18 votes]: let $T(x)=Ax$ be a linear transformation.
$T(x)$ is one-to-one if the columns of $A$ are linearly independent.
$T(x)$ is onto if every row of $A$ has a pivot.