I know that Rank is addition of column/row space and null space. However I don't know how to find this as I don't have the values in the matrix. So i guess that I should prove it but I am pretty much a newbie at it.

So if my nxn is a 3x3 matrix is with some random values with a zero determinant, will the rank be like 3+2 = 5 ?

  • 2
    The rank is the dimension of the column space. So it can't be greater than three for a 3x3 matrix. – heptagon Aug 21 '16 at 11:00
  • Oh yea but how can I solve this question then ? Can you tell me how ? @heptagon – James Rodricks Aug 21 '16 at 11:02
  • if $det(A)=0$, then rank is less than $n$ not full rank. – user252783 Aug 21 '16 at 11:07
  • @user252783 And can you explain in simpler terms. I didn't understand som of what you said about linear maps and finites etc.... – James Rodricks Aug 21 '16 at 11:10

If the determinant is $0$, the matrix is not invertible. Since surjective linear maps between finite dimensional vector spaces are invertible, the matrix cannot have a $3$ dimensional column space. Hence the maximal rank the matrix can have is 2

Considering that the rank is the number of linearly indipendent vectors in the matrix, the matrix in the example has less than n linearly indipendent vector. So the maximum rank you can have in this case is n-1. You can also consider the rank as the highest order of a minor in your matrix with not null determinant. And again if the nxn determinant is null, the maximum determinant different from zero that you can find is in a minor of order n-1.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.