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Define::= $I_n$ -- the $n \times n$ identity matrix.

Let $A$ be an $n \times n$ real matrix.

Define::= Nilpotent matrix -- an $n \times n$ real matrix $X$ such that $X^n = $ the zero matrix for some $n$ in the positive integers.

Define::= unipotent matrix $U$ -- $A - I$ which is nilpotent.

I'd like an example of a unipotent matrix which is not upper triangular

Frentos
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luysii
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  • Silly examples: lower triangular with 1's on the diagonal. And, then, in effect, everything between. Any matrix with all eigenvalues $1$ is unipotent, as follows from Jordan form. – paul garrett Nov 01 '17 at 22:15
  • Paul Garrett -- I'm using "From Groups to Geometry and Back" and unfortunately on p. 212 the author says that A is upper triangular, in the definition of unipotent p. 213 he uses A again saying that is a square matrix. I'm assuming that the meaning of A did not change, if it did change then of course you are correct. Thanks for trying to help. – luysii Nov 02 '17 at 02:14
  • Paul Garrett -- I'm using "From Groups to Geometry and Back" and unfortunately on p. 212 the author says that A is upper triangular, in the definition of unipotent p. 213 he uses A again saying that is a square matrix. I'm assuming that the meaning of A did not change, if it did change then of course you are correct. Thanks for trying to help. – luysii 20 secs ago edit – luysii Nov 02 '17 at 02:15
  • It is true that (square) upper-triangular matrices with 1's on the diagonal are unipotent. Also, (square) lower-triangular matrices with 1's on the diagonal. Also any (square) matrix $A$ whose characteristic poly is $(x-1)^n$, or, whose minimal poly is $(x-1)^m$ for some $m$. Either of the latter two has the property that any sufficiently large power of $(A-1)$ is $0$. – paul garrett Nov 02 '17 at 16:59

2 Answers2

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$$A=\left[ \begin{array}{ccc}1&1&0\\ 0&1&0\\ 0&1&1\end{array}\right]$$is unipotent because $(A-I)^2=0$ and $(A-I)$ is nilpotent where $I$ is the identity (or one can also argue that the characteristic polynomial of $A$ is $(\lambda-1)^3$) whereas A is neither upper nor lower triangular.

üzeyir saçıkay
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What you're asking is whether there can be a non-triangular nilpotent matrix. For example, consider a matrix $A = I_n + uv'$, where $u,v\in(\mathbb{K}\setminus\{0\})^n$ such that $\langle{u,v}\rangle = 0$. Since $u,v \neq 0$, $uv'$ and by extension $A$ are not necessarily triangle matrices.

Furthermore, $A$ is unipotent due to: $$(A - I_n)^n = (uv')^n = u(v'u)^{n-1}v' = 0$$

Lucas Aschenbach
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