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A is not square but rank(A) = rank(A') ? 
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#1
Dec612, 06:33 AM

P: 1,212

Hi
Can anyone help with understand a basic idea, I have a matrix A in MATLAB which is 100x3000. I have checked and there exist many columns of A that are all zeros. But apparently rank(A) = rank(A') = 100 Wikipedia states that the rank of an m x n matrix cannot be higher than m nor n... my interpretation of this sentence is that rank(A) ≤ min(m, n), is that correct? If so, surely that means my matrix is fullrank despite having zero columns? My understanding is that a fullrank matrix has fully orthogonal rows/columns, and since my matrix A clearly has multiple columns filled with only zeros, it cannot possibly be fullrank. So why is rank(A) = 100? Can someone tell me my interpretation of wikipedia is wrong and that this matrix is indeed really rankdeficient, it would make my day. Thanks edit bonus question... the nullspace of A is 3000x2900. So surely that would confirm the idea that there are only 100 linearly independent rows/columns in A? 


#2
Dec612, 07:25 AM

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P: 4,772

rank has nothing to do with orthogonality of rows/column. It is the dimension of the image (when the matrix is considered as a map R^{3000} to R^{100}).
As far as I can tell, your interpretation of wikipedia is correct. All you need to do in order to show that the matrix has rank 100 is to find, among the 3000 columns, 100 of them that are linearly independant. It does not matter that many of them are 0, as long as together they span R^{100}. 


#3
Dec612, 08:04 AM

P: 1,212

Wow, that's a really good answer. Thanks!



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