Basis for that $T: \mathbb{R}^3 \to \mathbb{R}^3$ is in rational canonical form

Question

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation such that

$T(x,y,z) = (x+y+z, x+y+z, x+y+z)$

Find a basis for $T$ such that your matrix$(A_T)$ is in rational canonical form.

I know since $\textit{Dummit and foot}$-$theorem.14$ - pag 476 that exist a basis for $\mathbb{R}^3$ such that $A_T$ is in ratinal canonical form, but I don't know a method for find this basis.

Answer 1

The vectors $(1,1,1)^T$, $(1,-1,0)^T$, $(0,1,-1)^T$ are eigenvectors and form a basis, so the matrix of $T$ will be diagonal in this basis.