Linear independence of columns and rows of a matrix. I read somewhere in my Algebra 1 notes something about how if a matrix has $n$ linearly independent rows (or it might have been columns) then it's rank is $n$. If this is the correct statement can someone please explain this, I can't seem to understand why this is true. 
   However, if this statement has been quoted incorrectly, please can you correct it. Apologies if there is an error in my quotation, I can't find my notes. 
I refer to rank as dim(Im(A)) i.e. the dimension of the A's image space.
 A: The image of the matrix will be the span of the images of the basis vectors, i.e. the columns of the matrix. If there are $n$ linearly independent columns then the rank will evidently be $n$. 
To see why the first line holds take a vector x=$\sum_{i=1}^n x_i \mathbf{e_i}$. First see that the image of a basis vector is the column of the matrix M since for the $k^{th}$ basis vector $(M\mathbf{e_k})_i =M_{ij}e_j=M_{ij}\delta_{jk}=M_{ik}$ which is the $i^{th}$ element of the $k^{th}$ column. Now we can apply $M$ to each component of x in the basis so $M\mathbf{x}$ would be in the span of the columns of M.
You can proceed similarly for the rows.
A: The rank is the dimension of the column space (a.k.a. range or image) of a matrix. The column space is the span of the columns of the matrix. So one way to think of the rank is the size of a basis of the column space.
Some consequences:


*

*If the matrix has $n$ columns, and $k$ of the columns are linearly independent, then the rank is at least $k$, since the span of these $k$ columns (which has dimension $k$) is a subspace of the column space.

*If all $n$ columns of the matrix are linearly independent, then the columns themselves are a basis of the column space, so the rank is $n$.

