Matrix with all off-diagonal elements equal 1

Question

Consider the matrix, of size $n \times n$, of the following form: \begin{bmatrix} a_1 & 1 & \cdots & \cdots1\\ 1 & a_2 & 1 &\cdots 1\\ \vdots & 1 & a_3 &\cdots 1\\ & \vdots & 1 & \ddots a_n & \\ \end{bmatrix} I must prove this assertion The matrix is nilpotent if and only if the $a_i -1$ are the zeros of the polynomial $P_n = X^n + (X^n)' + (X^n)'' + \cdots +n!$.

(For example, $P_4 = X^4 + 4X^3 + 12X^2 + 24X + 24$).

I have no idea how to approach this question. Could you help me? It seems that this property is not known see link

If we vote to close it, that's your right.

Answer 1

Let $d_i=a_i-1$ and $D=\operatorname{diag}(d_1,\ldots,d_n)$. The matrix is then $D+ee^T$ where $e$ is the vector of ones. Hence its characteristic polynomial is $$ \begin{aligned} p(x) &=\det(xI-D-ee^T)\\ &=\det(xI-D)-e^T\operatorname{adj}(xI-D)e\\ &=\prod_{i=1}^n(x-d_i)-\sum_{i=1}^n\prod_{j\ne i}(x-d_j).\\ \end{aligned} $$ Write $$ \prod_{i=1}^n(x-d_i)=\sum_{k=0}^n(-1)^kc_{n-k}x^{n-k} $$ where $c_n=1$ and $c_{n-k}=\sum_{i_1<i_2<\cdots<i_k}d_{i_1}d_{i_2}\cdots d_{i_k}$. Then $$ \begin{aligned} \sum_{i=1}^n\prod_{j\ne i}(x-d_j) &=\sum_{i=1}^n\sum_{k=0}^{n-1} (-1)^k\sum_{\substack{i_1<i_2<\cdots<i_k}\\ i_1,i_2,\ldots,i_k\ne i}d_{i_1}d_{i_2}\cdots d_{i_k}x^{n-1-k}\\ &=\sum_{k=0}^{n-1}(-1)^k \sum_{i=1}^n\sum_{\substack{i_1<i_2<\cdots<i_k}\\ i_1,i_2,\ldots,i_k\ne i}d_{i_1}d_{i_2}\cdots d_{i_k}x^{n-1-k}\\ &=\sum_{k=0}^{n-1}(-1)^k(n-k)c_{n-k}x^{n-1-k}.\\ \end{aligned} $$ Hence $$ \begin{aligned} p(x) &=\sum_{k=0}^n(-1)^kc_{n-k}x^{n-k} -\sum_{k=0}^{n-1}(-1)^k(n-k)c_{n-k}x^{n-1-k}\\ &=\sum_{k=0}^n(-1)^{n-k}c_kx^k -\sum_{k=0}^{n-1}(-1)^{n-1-k}(k+1)c_{k+1}x^k\\ &=x^n+\sum_{k=0}^{n-1}(-1)^{n-k}\big[c_k+(k+1)c_{k+1}\big]x^k.\\ \end{aligned} $$ That $D+ee^T$ is nilpotent is equivalent to $p(x)=x^n$, which in turn is equivalent to $c_k=-(k+1)c_{k+1}$ for $k=n-1,\,n-2,\ldots,0$. Since $c_n=1$, the latter recursive relation is equivalent to $c_{n-1}=-n,\,c_{n-2}=n(n-1),\,c_{n-3}=-n(n-1)(n-2),\ldots$ etc.. Hence it is equivalent to the condition that $c_{n-k}x^{n-k}=(-1)^k\frac{d}{dx^k}(x^n)$, or that $$ \prod_i(x-d_i) =\sum_{k=0}^n(-1)^kc_{n-k}x^{n-k} =\sum_{k=0}^n\frac{d}{dx^k}(x^n). $$

Answer 2

First follow either @user1551's answer or work slightly harder with elementary row operations, we get that $$\det(xI-A)=\prod_{i=1}^n (x-d_i) - \sum_{i=1}^n\prod_{j\not=i} (x-d_j)$$

Let $p(x):=\prod_{i=1}^n (x-d_i)$, then $$\det(xI-A)=p(x)-p'(x)$$ The matrix $A$ is nilpotent iff its characteristic polynomial is $\det(xI-A)=x^n$, therefore we have to show $p(x)-p'(x)=x^n$ iff $p(x)$ is the given polynomial $q(x):=x^n+(x^n)' + (x^n)'' + \cdots$

The $\Leftarrow$ direction is trivial by verifying $q(x)$ satisfies the differential equation. For $\Rightarrow$ direction, note that the first order linear ODE has general solution $p(x)=q(x)+Ce^x$ since $q(x)$ is a known special solution. Because $p(x)$ is a polynomial, $C$ has to be $0$, and $p(x)=q(x)$. (Or the differential operator $D$ is nilpotent on the space of polynomials of degree $\le n$, hence $(I-D)$ is invertible and $p(x)=(I-D)^{-1}x^n$ is unique.)