Jordan Canonical form of a matrix over rationals whose all entries are 1.

Question

How to compute the Jordan canonical form for the $n \times n$ matrix over $\mathbb{Q}$ whose entries equals to $1$.

Answer 1

I will call your matrix $A$.

Observe that the dimension of the null space of $A$ is $n-1$(Why?). So you know $n-1$ linearly independent eigenvectors (whose associated eigenvalue is zero). Further, the vector which has all co-ordinates equal to $1$ is clearly an eigenvector for $A$ (associated eigenvalue being $n$).

Can you fill in the gaps and guess the Jordan canonical form?

Answer 2

The characteristic equation is $x^{n-1}(x-n)$. There are $n-1$ Jordan blocks for eigenvalue 0 and only one for $n$.

hence Jordan canonical form is: $$ [n,0,0,...0;0 0 0 ,...,0;...;0 0 0 ,...0] $$