Inverse of a unipotent matrix

Question

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!

Answer 1

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 $A^{-1}= (-1)^{k+1} \sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i-1}$.

Answer 2

Another explanation for the invertibility of uni-potent matrices you might want to consider, is that $(A-E_n)^k = 0$ means that the generalized eigenvectors of order $k$ to the eigenvalue $\lambda = 1$ span the whole vector space. Hence $\lambda = 1$ is the only eigenvalue of $A$ and hence, $A$ is invertible, since the nullspace is empty.

Furthermore $(A-E_n)^k = 0$, by the same logic implies there exists a change of basis, where your matrix $A$ transforms into a Jordan matrix, where every Jordan-Block has $1$'s on the diagonal and has a size less or equal to $k$.