1. The problem statement, all variables and given/known data Find all 10x10 Matrices such that the column space is equal to the null space. 2. Relevant equations Choose Function: n!/k!(n-k)! where n is the total number of elements and k is the number k-cominations of the set. rankA+dimNulA=n for a matrix in R^n 3. The attempt at a solution There is a short and simple proof that we can use to determine that the rank of such a matrix should be 5. Proof: Let A be a 10x10 matrix and suppose that ColA=NulA => rankA=dimNulA => rankA+dimNulA=10 => rankA+rankA=10 => 2rankA=10 => rankA=5 Ok so from this I can deduce that I will have 5 linearly independent vectors in R^10 that will make up my matrix. Now what I was thinking about this is that this must mean that to find ALL matricies I simply have to take permuations of my linearly independent vectors. For example a possibility for a matrix would be <e1,e2,e3,e4,e5,0,0,0,0,0> where e1,...,e5 are the basis vectors for R^10 and those 0's are the 0 vectors in r^10. Another possibility would be <0,e2,e4,e5,0,0,0,e1,e3,0> and so on. This seems like it'd be A LOT of possibilites so what I thought about doing was utilizing the choose function and from it's utilization I determined that because I have 10 columns total, with 5 vectors as a possibility to make up those columns, then I should have 252 possible combinations which are simply permutations of a 10x10 matrix with the vectors e1,...e5 as a basis. Is my line of thinking correct or am I way off? Thanks for the help!