Column rank of a rectangular matrix with number of columns is greater than the number of rows? If I have the SVD decomposition of a rectangular matrix $A$ of dimension ($m\times n$) with $n>m$, how can I know the column rank of that matrix? Actually my aim is to know if the columns of that matrix $A$ are linearly independent? Is the column rank related to the dimensions of any of the matrices of the SVD decomposition?
 A: $\operatorname{rank}(A)\le \operatorname{min}(m,n)$, so in your case $\operatorname{rank}(A)< m$, the number of column. The columns of this matrix are clearly not linearly independent.
The $\operatorname{row-rank}$ of a matrix is equal to its $\operatorname{column-rank}$ of the matrix (you can see, for example, https://en.wikipedia.org/wiki/Rank_(linear_algebra)), which should be the number of nonzero singular values.
A: The first answer basically answers your question already however I want to go a bit further and maybe add some things about the meaning of the rank.
The row-rank of a matrix is equal to the dimension of the subspace that is generated by the rows of a matrix. The same for the column-rank. As such one says a matrix has "full"-(row/column) rank if the rank is equal to the number of rows/columns.
Although I am not particularly familiar with the SVD decomposition, it seems to me as if since $U$ and $V*$ in $A= U \sum V$ are both unitary matrices or inverses of unitary matrices, they have determinant $\neq 0$ and as such are of full rank.
From this one can deduce, that one should be able to get the rank of A from the matrix $\sum$ (so $Rank(A)=Rank(\sum)$).
Further information can also be found on the linked wikipedia page at the chapter 4.4 on Range, null space and rank.
Hope this helps
