Linear transformation onto and one to one? (1)If a linear transformation $T:\mathbb{R}^n\rightarrow\mathbb{R^m}$ maps $\mathbb{R}^n$ onto
$\mathbb{R^m}$ what is the relation between m and n?
(2)If T is one to one what is the relationship between m and n.
For the first question I think $m>n$ because if you try to do for example $R^3 \rightarrow R^4$
you have a matrix $A=4x3$, 4 rows and three columns and this cannot span $R^4$. As you cannot have 4 pivot.
For the second I think $n<m$ because if $R^4 \rightarrow R^3$ because if you have a A=3x4 matrix then one will always be linear independnet. 
But I am not sure if my justfication is correct.
 A: If $T: \mathbb{R}^n \to \mathbb{R}^m$ is onto then $m\leq n$.
If $T: \mathbb{R}^n \to \mathbb{R}^m$ is one-to-one then $m\geq n$.
Note in particular that it is possible to have $m$ and $n$ equal.
To prove this, you can make use of the fact (you might know this, might now) that the dimension of the image of a linear map is always less than or equal to the dimension of the domain. (To prove this fact, try to start with a basis and see what happens.)
A: You can represent $T$ as an $m \times n$ matrix. The columns of $T$ span the range. 
You can recast your questions as follows:
Under what conditions can $n$ vectors span $\mathbb R^m$? Under what conditions can $n$ vectors be linearly independent in $\mathbb R^m$?
A: (1) There's no relation, other than just n and m should be finite.
And for Question 1, m doesnt have to be > n since the A you choose can be 3x4 too. So youll have v(1x3) * A(3x4) = w(1x4) v and w belong to R^n and R^m respectively.
(2) If one-one, the rank of the transformation must be = n and yes therefore, n < m.
Any transform will have an image in R^m. ie the transform T has a 'range' in R^m, which is a subspace of R^m. A subspace will always have dimension less than the parent space ie rank of T (dimension of the range) must be < m.
Secondly, for it to be one-one, the 'nullity' of R^n must be zero. And by rank nullity theorem, n = rank of T + nullity of T, in this case, n = rank of T which is < m. So, n < m. Dont confuse with linear independance. n and m are determined by linear independance, because they're the dimensions for the spaces, so that's taken care of.
