Invertibility conditions for $I_n + aJ_n$ where $I_n$ identity and $J_n$ is matrix of all ones

Question

Let $I_n$ denote the $n \times n$ identity matrix and $J_n$ the $n\times n$ matrix with all entries equal to 1. Determine for which real numbers $a$ the matrix $I_n + aJ_n$ is invertible.

My attempt at a solution.

One approach - you can calculate the first few determinants to see that it's $na+1$ so the answer is $a \ne -1/n$. How can I make this rigorous (through induction)?

Second approach - It's clear that $(1,1, \dots,1)^T$ is an eigenvector with eigenvalue $n$ (multiplicity 1) and 0 is an eigenvalue (multiplicity) $n-1$. The solution seems to imply the conclusion that invertible iff $a \ne -1/n$ follows immediately. How?

Answer 1

Think of the eigenvalues of $aJ_n$, which are simply the eigenvalues of $J_n$ (which you seem to know already) times $a$. Then think of the eigenvalues of $I_n + aJ_n$, which are quite easy to derive, since its eigenvectors are the same. Really, if $aJ_n\cdot\vec v=\lambda\vec v$, then $(I_n+aJ_n)\cdot\vec v=?$

When none of the eigenvalues is 0, the matrix is invertible.

Answer 2

A matrix is invertible if and only if zero is not it's eigenvalue, i.e $\det (I_n+aJ_n)\ne 0$. As $\det(I_n+aJ_n)=na+1$ the condition is $a\ne-\frac1n$