# Let $A$ be an $n \times n$ matrix and suppose $\det (A) = 0$. What is the maximum value that $\mbox{rank} (A)$ can be?

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 ?

• 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

## 2 Answers

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.