Is every $T$ s.t. $T:\mathbb{R^3} \to \mathbb{R^3}$ s.t. $T(w)=2w, \forall w \in W$ and $T(z) \notin W$ if $z \notin W$ necessarily triangulable?

Question

Let $W \subset \mathbb{R^3}$ s.t. $W=span \{v_1: \begin{pmatrix} 1 \\ -3 \\ 2 \\ \end{pmatrix}, v_2: \begin{pmatrix} 2 \\ 0 \\ 1 \\ \end{pmatrix} \}$

Find a linear application $T:\mathbb{R^3} \to \mathbb{R^3}$ s.t. $T(w)=2w, \forall w \in W$ and $T(z) \notin W$ if $z \notin W$: is every $T$, with these properties, necessarily triangulable? And diagonalizable?

Find $T$ is quite simple: completing the basis of $W$ to a basis of $\mathbb{R^3}$, for example $e_2 = (0, 1, 0)$, it works if $T(v_1)=2v_1$, $T(v_2)=2v_2$ and $T(e_2)=e_2$. The Matrix which represents $T$ is:

$A_T = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 1 & -2 \\ 0 & 0 & 2 \\ \end{bmatrix} $ , which is diagonalizable: $ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $

How to answer to the first question? The second one is given because of the definition of eigenvalues. Thank you.

Answer 1

Given any $v_3\notin\langle v_1,v_2\rangle$, $(v_1,v_2,v_3)$ is a basis of $\mathbb{R}^3$ and the matrix of $T$ with respect to this basis is of the type$$\begin{bmatrix}2&0&a\\0&2&b\\0&0&c\end{bmatrix},$$with $c\neq0$. Such a matrix is already an upper triangular matrix. But it doesn't have to be diagonalizable. Take $a=0$, $b=1$, and $c=2$, for instance.

Answer 2

Hint The characteristic polynomial of $T$ splits on $\mathbb R$. Hence $T$ is triangularizable.