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?

Are you sure it's equal to zero? In the 2x2 case, you get $n^2-2n$ — man_in_green_shirt, Jun 26 '15 at 15:39

score 7 · Accepted Answer · answered Jun 26 '15 at 15:39

7

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

answered Jun 26 '15 at 15:39

Evgeny

5,895

score 3 · Answer 2 · edited Apr 13 '17 at 12:20

3

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 $\lambda = n$.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 26 '15 at 15:44

MT_

19,971

Haven't really studied eigenvalues yet properly, but good to know anyways. – Kamil Jun 26 '15 at 15:59
How can a matrix full of ones not have 0 as an eigenvalue? Its rank is 1. – anderstood Dec 29 '15 at 19:52
1

@anderstood It's a typo I think, 1 should say 0 since $\det(A - \lambda I_n) = x^n - nx^{n-1}$. – MT_ Dec 29 '15 at 22:24

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

2 Answers2