Rank of $2I_n-J_n$ where $I$ is the identity matrix and J is a matrix of ones

Question

I am to prove that the following matrix $2I_n - J_n$ has rank $n$. Where $I_n$ is the $n\times n$ identity matrix and $J_n$ is an $n\times n$ matrix of ones.

I cannot provide a formal definition for it where I was hoping for some help with.

Answer 1

The eigenvalues of $j_n$ are $n$ (multiplicity $1$) and $0$ (multiplicity $n-1$). (Why?)

So, the eigenvalues of $2I_n - j_n$ are $2-n$ (multiplicity $1$) and $2$ (multiplicity $n-1$). (Why?)

What does this tell you about the rank of $2I_n - j_n$? (It seems you need the assumption $n \ne 2$ for the original claim to be true.)

Answer 2

Note that $J^2=nJ$, hence

$$(2I-J)((n-2)I-J)=2(n-2)I-2J-(n-2)J+J^2=2(n-2)I-nJ+nJ=2(n-2)I$$

Hence $A=2I-J$ is regular iff $n\ne2$, and $A^{-1}=\frac1{2(n-2)}((n-2)I-J)$. And a regular matrix has full rank (that is, $n$ here). When $n=2$, $A$ has trivially rank $1$.