I think the answer is No. Am I missing something here?:
Note: What I'm saying is that if A is an mxn matrix that is Linearly Independent, A MUST span R m. However, if A spans R m, A isn't necessarily linearly independent. One entails the other, but not vice versa.
My reasoning:
If A is an mxn matrix that spans R m, it seems to me that by spanning R m , A merely needs a pivot in every row (where A is a matrix of mxn size, that is m rows by n columns). This means you can have a case where n > m, in which a pivot exists in each row, but since n > m, there exists at least 1 free variable which entails that at least one non-trivial solution exists for Ax = 0. And thus if A is an mxn matrix which spans R m, A is not necessarily linearly independent. That is, A can (emphasis on the word can) be linearly dependent (and thus not independent), yet still span R m.
Note: By non-trivial I mean that for Ax = 0, there exists at least one element in x which is not 0.
Best Answer
You need to be more precise with your statements. When you say $A$ is independent, do you mean the rows are independent or the columns are independent?
Just some background which I'm putting just to clarify something: For a non square matrix $A\in \mathbb{R}^{m\times n}$ it is impossible for both the rows to all be independent and the columns to all be independent. If $m>n$, then it is impossible for all the rows to be independent, but the columns may be independent. Also, the rank of $A$ must be between 0 and $n$ inclusive.
To answer your question (as I understand it), if $A\in\mathbb{R}^{m\times n}$ , i.e. $A$ has $m$ rows and $n$ columns, and the columns of $A$ span $\mathbb{R}^m$ (it doesn't really make sense to say that the matrix spans $\mathbb{R}^m$), then it must be that $m\leq n$. But this doesn't mean that $Ax=0$ does not have a non trivial solution, because we have not showed that the rank of the row space (which is $m$) is less than $n$. If $n=m$, then the row space spans the whole of $\mathbb{R}^n$ and so there is no non trivial solution to $Ax=0$. But, if, as you said, $n>m$, then the null space is non trivial and there are $n-m$ linearly independent solutions to $Ax=0$