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!

  • 2
    $\begingroup$ What is 8your* definition of "unipotent matrix" ? Because in my definition it follows from it at once the matrix is invertible.. $\endgroup$ – Timbuc Jan 17 '15 at 18:59
  • $\begingroup$ Well, the definition I'm given is: a matrix A is unipotent, if there is a natural number k so that $(A-E_n)^k=0 $ where $ E_n $ is the identity matrix. Given this, I don't exactly see why a unipotent matrix must always be invertible. $\endgroup$ – moran Jan 17 '15 at 19:05
  • $\begingroup$ @SMM Maybe you should write this as an answer. $\endgroup$ – user84413 Jan 17 '15 at 19:14
  • $\begingroup$ @user84413 OK. Thanks for suggestion. $\endgroup$ – SMM Jan 17 '15 at 19:17
  • $\begingroup$ A unipotent matrix is similar to an upper-triangular matrix with $1$'s on the diagonal, hence has determinant $1$, i.e., it is invertible. $\endgroup$ – Dietrich Burde Jan 17 '15 at 19:40

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


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 $k$.


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