Question: If $T:\Bbb R^n \to\Bbb R^m$ (where $n\neq m$) is a linear transformation, can $T$ be both one to one and onto? 

My first instinct was it can, but after thinking about it, it seems the set will either be dependent or there would be a row without a pivot. Is there a way to prove this for all examples?

Answer: 
It's a ready consequence of [Kernel Extension Theorem (for vector spaces)](http://en.wikipedia.org/wiki/Dimension_theorem#Kernel_extension_theorem_for_vector_spaces) that such a transformation must either fail to be one-to-one or fail to be onto. (You may have heard "kernel" referred to as "null space" instead.) If $n<m$, then since the image (range) of the transformation has dimension at most $n$, it can't be the whole space $\Bbb R^m$, and so fails to be onto, even if it's one-to-one (that is, has a kernel of dimension $0$). If $n>m$, then since the image of the transformation has dimension at most $m$ (as a subspace of $\Bbb R^m$), it follows that the kernel of the transformation has positive dimension, and to the transformation fails to be one-to-one, even if it's onto (that is, has an image of dimension $m$).

If you'd rather think of things in terms of the matrix corresponding to the transformation, see the [Rank-Nullity Theorem](http://en.wikipedia.org/wiki/Rank-nullity_theorem).
Suppose an operator $T$ on a finite dimensional space is one to one and onto.

Then $v\_1,...,v\_k$ are linearly independent iff $T v\_1, ..., T v\_k$ are linearly independent.

(Since $T$ is onto, it follows from this that $u\_1,...,u\_k$ are linearly independent iff $T^{-1} u\_1, ..., T^{-1} u\_k$ are linearly independent.)

Hence a basis for the domain can be transformed into a basis for the range (of the same cardinality) and vice versa. It follows that the dimension of the range and domain are the same.