# Prove $\det(A - nI_n) = 0$.

Problem: Prove that $\det(A - n I_n) = 0$ when $A$ is the $(n \times n)$-matrix with all components equal to $1$.

Attempt at solution: I tried to use Laplace expansion but that didn't work. I see the matrix will be of the form \begin{align*} \begin{pmatrix} 1-n & 1 & \cdots & 1 \\ 1 & 1-n & \cdots & 1 \\ \vdots \\ 1 & 1 & \cdots & 1-n \end{pmatrix} \end{align*} I want to somehow get two equal rows or columns here, or a row/column of zero using elementary operations. But I don't see what I should do?

• Are you sure it's equal to zero? In the 2x2 case, you get $n^2-2n$ Commented Jun 26, 2015 at 15:39
• @man_in_green_shirt No, you get $2^2 - 2*2$, since $n = 2$.
– MT_
Commented Jun 26, 2015 at 15:40
• Yeah, sorry, ignore me :) Commented Jun 26, 2015 at 15:41

The eigenvalue of the $n \times n$ matrix in which every entry is $1$ is $n$ (of multiplicity $1$) and $0$ (of multiplicity $n-1$). Proof here
So, for such a matrix $A$, we have $\det(A - \lambda I_n) = 0$, and $\lambda = n$.
• @anderstood It's a typo I think, 1 should say 0 since $\det(A - \lambda I_n) = x^n - nx^{n-1}$.