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#1 h Mar 26, 2025, 9:49 AM

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The matrix $A = (a_{ij}) \in Mat_p(\mathbb{C})$ is defined by the conditions
$a_{12} = a_{23} = \dots = a_{(p-1)p} = 1$ and $a_{ij} = 0$ for a set of indices $(i,j)$ .
Prove that there do not exist nonzero matrices $B, C \in Mat_p(\mathbb{C})$ satisfying the equation
$(I_p + A)^n = B^n + C^n.$ $\forall$ $n$ is a postive integer.

This post has been edited 2 times. Last edited by hef4875, Mar 26, 2025, 9:51 AM

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#4 h Mar 26, 2025, 1:02 PM

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It's this problem.

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#5 h Mar 27, 2025, 4:18 PM • 1 Y

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Thanks @ Filipjack.
$\textbf{Proposition.}$ If $I+A=B+C$ and $(I+A)^2=B^2+C^2$ , then $B=I+A,C=0$ or $B=0,C=I+A$ .
$\textbf{Proof.}$ $I+A=B+C$ , $(I+A)^2=B^2+C^2=B^2+C^2+BC+CB$ ;
then $BC+CB=0$ . Since $C=I+A-B$ , we deduce that
(*) $2B-2B^2+AB+BA=0$ .
(*) is a Riccati equation; we associate the $2p\times 2p$ pseudo-Hamiltonian
$M=\begin{pmatrix}-A&2I_p\\0_p&2I_p+A\end{pmatrix}\in M_{2p}(\mathbb{C})$ . There is a one to one correspondence between
the set of solutions of (*) and the set of $p$ -dimensional $M$ -invariant subspaces $E$ s.t.
(1) $E\oplus span(e_{p+1},\cdots,e_{2p})=span(e_1,\cdots,e_{2p})$ .
The Jordan form of $M$ -in the new basis $(f_1,\cdots,f_{2p})$ - is $diag(J_p[2],J_p[0])$ , where
$J_p[\lambda]=\lambda I_p+J_p$ and $J_p$ is the nilpotent Jordan block of dimension $p$ .
$E$ is in one of those $p+1$ forms $span(f_1,\cdots f_q,f_{p+1},\cdots,f_{2p-q}),q=0,\cdots,p$ .
Only the following $2$ spaces verify condition (1):
$\ker(M^p)=span(e_1,\cdots,e_p)$ and $\ker((M-2I_p)^p)$ .
$\bullet$ It's complicated to write a proof of the previous line and I don't have any nice proof.
Thus (*) has $2$ solutions. Luckily, we have two obvious solutions: $B=0$ and
$B=I+A$ . $\square$

This post has been edited 3 times. Last edited by loup blanc, Mar 27, 2025, 8:32 PM

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#10 h Apr 3, 2025, 3:43 PM

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Here is a complete resolution of the above equation (*).
We put $A=J$ , the nilpotent Jordan block of dimension $n$ .
Let $U=I+J-B,V=B$ ; then $U+V=I+J,UV=-VU$ .
$(U+V)^2=(U-V)^2=U^2+V^2=I+2J+J^2$ . Thus $U+V,U-V$ commute with $2J+J^2$ , then with $J$ .
Therefore $U+V,U-V$ are polynomials in $J$ and $U,V$ too. In particular $UV=VU=0$ .
Finally $B$ is a polynomial in $J$ , $B=\sum_{i=0}^n b_iJ^i$ .
Case 1. $b_0\not= 1$ . Then $U$ is invertible and $V=B=0$ .
Case 2. $b_0= 1$ . Then $V=B$ is invertible and $U=0$ , that is, $B=I+J$ . $\square$

This post has been edited 2 times. Last edited by loup blanc, Apr 3, 2025, 3:47 PM

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