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

Question

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?

Answer 1

Hint: Add all other rows to the first row. What row you'll obtain after that?

Answer 2

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 $\lambda = n$.