What is the difference between the rowspace and the columnspace in linear algebra?

The question is essentially all in the title, What is the difference between the rowspace and the columnspace?

Additionally, do they have the same solution space?

I had hoped that the other answer would answer all of your questions, but as I'm pretty sure my last answer to your last question was the source of this question (which is the proper way for things to go - much better than trying to fit everything into one question).

In general, row spaces and column spaces can be very different. For non-square matrices, they will be in different dimension ambient spaces (although the dimension of the subspace spanned by each will be the same - the dimension of this subspace is usually called the rank of the transformation). And so this is very, very different.

Now, when you refer to 'the values that you can set the syste equal to and have a valid solution for,' you imply more than just a matrix. In particular, you assume a whole equation, and this is important. I suspect that you refer to cases like $\left( \begin{array}{cc} a & b \\ c&d \\ e&f \end{array} \right) \left( \begin{array}{cc} x\\ y \\ \end{array} \right) = \left( \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right)$. In these cases, the 'sets of $\left( \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array} \right)$ ' that have solutions, known as the Image of the transformation (this is an important word - remember Image) is the column space. I used a non-square matrix to show it very easily: the dimension is even different.

But here is something confusing: the Image is not affected by elementary row operations. Whoa - that's cool. Perhaps nonintuitive. In other words, the set of vectors that can be 'reached' by this matrix do not change. And here's something else important: these row operations do not even change the solution for each particular element in the Image. That's why they're so great - but this is weird. Changing the rows doesn't affect the column space, and though the coefficients of the x and y might change, these direct solution for any $x_1, x_2, x_3$ vector do not change. This is, I think, the confusing part, because column operations do change these solutions, even though the Image itself isn't effected.

So how does one get a case where column operations don't change the solutions themselves? By left multiplication. $\left( \begin{array}{ccc} x & y & z \end{array} \right) \left( \begin{array}{cc} a & b \\ c&d \\ e&f \end{array} \right) = \left( \begin{array}{cc} x_1 & x_2\end{array} \right)$. But this is just something that we usually don't touch - it looks funny to me, even. Why? Usually we deal with the familiar right multiplication and transpose things if we need to.

Alright, so I gave a quick guide through some of the ideas here. Now I want to recommend something to you, so that it feels more real. Just like in the last question, I advised you to directly solve the system before and after a couple changes, I advise you to try new and different changes, and see if vectors can still be hit (are still in the image) and if they can, see if the solutions are the same. Then you'll have quick reference for your thoughts on different actions you might try.

I didn't quite answer your question proper, but I tried to hit at what I think is the heart of the issue. Is that right?

I think this : http://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra should answer your question and more.