The complement of identity matrix has full rank

Question

I am try to understand why $rank(J_n - I_n) = n$ over any field that does not divide $n-1$ or any infinite field (in particular over $\mathbb{R}$).

$J_n$ is the $n\times n$-Matrix with all ones $J_n[i][j]=1$ for all $i,j\in[n]$.

$I_n$ is the $n\times n$-identity matrix.

Thanks for helping.

Answer 1

The determinant of $J_n - I_n$ is nonzero in the base field if and only if $J_n - I_n$ is invertible: so let's compute $\operatorname{det}(J_n - I_n)$. Recall that for a diagonalizable matrix its determinant is just the product of its eigenvalues (with multiplicity): we will show that $J_n - I_n$ is diagonalizable and compute these eigenvalues.

First I claim that there are $n - 1$ linearly independent eigenvectors of $J_n - I_n$ all with eigenvalue $-1$ given by (check this): $$ \begin{bmatrix} -1\\1\\0\\0\\\vdots\\0 \end{bmatrix}, \begin{bmatrix} -1\\0\\1\\0\\\vdots\\0 \end{bmatrix}, \cdots, \begin{bmatrix} -1\\0\\0\\0\\\vdots\\1 \end{bmatrix}. $$ Next I claim that the vector $$ \begin{bmatrix} 1\\1\\1\\\vdots\\1 \end{bmatrix} $$ of all ones is an eigenvector with eigenvalue $n - 1$ (check this too).

This pins down all of the eigenvalues with multiplicity of $J_n - I_n$: we have $\det(J_n - I_n) = -1 \cdot -1 \cdot \cdots \cdot -1 \cdot (n - 1) = (-1)^{n-1}( n - 1)$. In particular $J_n - I_n$ is invertible if and only if $n - 1$ is nonzero in the base field. This is always the case as long as the characteristic of the base field does not divide $n - 1$, as you say.

Answer 2

The matrix $J_n-I_n$ has the determinant $(-1)^{n-1} \cdot (n-1)$. Thus if the field has characteristic $0$ or positive characteristic $p$ not dividing $n-1$, then the determinant is nonzero and so the matrix is invertible.