Show that all unipotent matrices are invertible. Also, specify a formula for the inverse of a unipotent matrix.
Now, I've tried to approach the problem using the determinant: a matrix is unipotent, if it results in a nilpotent matrix when the identity matrix is subtracted, and a nilpotent matrix must obviously have the determinant 0 (because it will result in 0 when you cube it enough times).
I'm also aware that the determinant of the identity matrix is 1, but as there are no general rules to how the determinant behaves when adding/subtracting matrices (at least none that I know of), this didn't lead to anything so far. In order to be invertible, a unipotent matrix must have a determinant different than zero… but how to prove that using what I'm given?
Any help would be appreciated!
Best Answer
Write: $$0= (A-E)^k= \sum_{i=0}^k{k\choose i}(-1)^iA^{k-i}= \sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i}+ (-1)^kE=$$ $$= A\sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i-1}+ (-1)^kE.$$ It follows that $A^{-1}= (-1)^{k+1} \sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i-1}$.