I am given an operator $T: \mathbb{C}^3 \rightarrow \mathbb{C}^3$ that is represented by the matrix $$ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ Now I am supposed to find the spectral family and the resulting spectral representation of $T$. We learned how to construct the spectral family by defining $$ T_{\lambda} = T -\lambda I $$ for a fixed $\lambda \in \mathbb{R}$. We then define $$ T_{\lambda}^+ = \frac{1}{2}(B_{\lambda} + T_{\lambda}) ~~~~~~~\text{with }B_{\lambda} := (T_{\lambda}^2)^{1/2} $$ Finally we define $E_{\lambda}$ as the projection of $H$ onto $\ker T_{\lambda}^+$.
Now in this specific case I calculated $T_{\lambda}$ to be $$ T_{\lambda} = \begin{pmatrix} -\lambda & 1 & 0 \\ 1 & -\lambda & 0 \\ 0 & 0 & 1 - \lambda \end{pmatrix} $$ The square of this matrix is $$ T_{\lambda}^2 = \begin{pmatrix} \lambda^2 + 1 & -2\lambda & 0 \\ -2\lambda & \lambda^2 + 1 & 0 \\ 0 & 0 & (1 - \lambda)^2 \end{pmatrix} $$ So therefore $$ B_{\lambda} = 2(\lambda - 1) I $$ And finally $$ T_{\lambda}^+ = \frac{1}{2}\begin{pmatrix} \lambda - 2 & 1 & 0 \\ 1 & \lambda - 2 & 0 \\ 0 & 0 & \lambda - 1 \end{pmatrix} ~~~~~~~~~\text{with } a_{\lambda} = 2(\lambda - 1) - \lambda $$ Now to me it seems like $\ker T_{\lambda}^+ = \{0\}$ (with $0$ the null-element of $\mathbb{C}^3$).
My question is how do I get a spectral family from here? And how can I use the spectral theorem to get the spectral representation of this operator with this family?