# 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. Aug 21, 2016 at 11:00
• Oh yea but how can I solve this question then ? Can you tell me how ? @heptagon Aug 21, 2016 at 11:02
• if $det(A)=0$, then rank is less than $n$ not full rank. Aug 21, 2016 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.... Aug 21, 2016 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