True or False, diagonalization problem 
Let $B_c= \left\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\right\}$, $T:\mathbb{R}^4\rightarrow \mathbb{R}^4$ a linear operator such that \begin{equation} \det\left[T-I\lambda\right]_{B_{c}}=(2-\lambda)^4\qquad \text{and}\qquad T((0,0,0,1)) = (1,0,0,1)\end{equation}
  Is T diagonalizable?

I've tried to use the following reasoning: Since the algebraic multiplicity of the eigenvalue $2$ is $4$, we know that $T$ is diagonalizable if (and only if) the dimension of the eigenspace associated with the eigenvalue $2$ is $4$. But we also know that there's a vector in $\mathbb{R}^4$ that don't belong to  $S_{2}$ (because $(0,0,0,1)$ is not an eigenvector). Can I use this argument to show that $T$ is not diagonalizable? Thanks in advance

Since $T((0,0,0,1)) \neq 2 \cdot (0,0,0,1)$ we know that $(0,0,0,1)$ isn't in the eigenspace $\operatorname{Ker}(T-2I)$. The dimension of the eigenspace is therefore strictly less than the dimension of $\mathbb{R}^4$.
