Let \(F\) be a field such that \(n\neq0\) in \(F\), and let \(M\in M_n(F)\) be the matrix whose entries are all \(1\). Find its Jordan canonical form (over a splitting field, if necessary).
Solution
Let \[ \mathbf 1=(1,\ldots,1)^t. \] Then \[ M\mathbf 1=n\mathbf 1, \] so \(n\) is an eigenvalue.
For every vector \(v=(v_1,\ldots,v_n)^t\) with \[ v_1+\cdots+v_n=0, \] each coordinate of \(Mv\) is this same sum, so \(Mv=0\). Hence the hyperplane \[ H=\left\{v\in F^n:\sum_i v_i=0\right\} \] is contained in the \(0\)-eigenspace and has dimension \(n-1\).
Because \(n\neq0\) in \(F\), the vector \(\mathbf1\) does not lie in \(H\). Therefore \[ F^n=H\oplus F\mathbf1 \] is a direct sum of eigenspaces. Thus \(M\) is diagonalizable, with eigenvalue \(0\) of multiplicity \(n-1\) and eigenvalue \(n\) of multiplicity \(1\).
Consequently \[ J(M)=\operatorname{diag}(\underbrace{0,\ldots,0}_{n-1},n). \] Equivalently, since \(M^2=nM\), its minimal polynomial is \(x(x-n)\).