Why the rank of a matrix is not full-rank, although all the row vectors are independent?

Question

A 48×36 matrix has 36 independent columns, but dose not have full-rank. I do not know what is going on.

the matrix: https://drive.google.com/file/d/1pVPidFDjbkIEnuhMiF4Ii7DyjpRlWhO0/view?usp=sharing

I checked the independency and the rank of the matrix with numpy.

Answer 1

To reiterate comments above, you are grossly misunderstanding the definition of linear independence.

A collection of vectors $v_1,v_2,\dots,v_n$ are considered to be linearly dependent if and only if there exist some collection of scalars $c_1,c_2,\dots,c_n$ with at least one of the $c_i$ nonzero such that $c_1v_1+c_2v_2+\dots+c_nv_n=0$. Otherwise, if the only collection of scalars that makes this zero is where $c_1=c_2=\dots=c_n=0$ then the vectors are said to be linearly independent.

Note, $c_1=c_2=\dots=c_n=0$ is always a solution, that is not in question here. The question is whether or not it is the only solution or if other less trivial solutions exist for your vectors.

Note that it is possible that you have a collection of vectors who when taken as subcollections at a time will have the subcollections be independent despite the fact that if you were to take them all at once that they would be dependent. For an example, note that $(1,0,0),(0,1,0),(0,0,1),(1,1,1)$ form a dependent set of vectors however any pair or even any triple of these will be independent. These are dependent, again, because we can find a linear combination of these that add up to the zero vector, for instance $(1,0,0)+(0,1,0)+(0,0,1)-(1,1,1)=(0,0,0)$.

To correctly check the independence of an entire collection of vectors, the entire collection must be simultaneously used... not just two at a time. Common techniques to test the independence would be to find the rank of the associated matrix formed using the vectors as the rows (or using them as the columns as you prefer). Finding the rank is commonly done by Gaussian Elimination.

Now... your matrix, which you say you found is of rank $33$ must then not have all $36$ of the columns as independent. It being rank $33$ means that you could find some independent subset of your columns of size $33$ but you can not find a subset of the columns of size greater than $33$ which is independent. With $34$ or more of the columns taken at once they necessarily become dependent.