Using this theorem we can say that if $\mathcal{M}(T)$ diagonalizable then $$\begin{align*} \mathcal{M}(T)&=PDP^{-1}\\ &=\begin{bmatrix}v_{1}^T & \cdots & v_n^T \\ \end{bmatrix}\begin{bmatrix}\lambda_{1} & \cdots & 0 \\ 0 & \ddots & 0 \\ 0 & \cdots & \lambda_{n}\end{bmatrix} \begin{bmatrix}v_{1}^T & \cdots & v_n^T \\ \end{bmatrix}^{-1} \end{align*}$$where $\lambda_1,\cdots,\lambda_{n}$ are eigenvalues of $T$ and $v_1,\cdots,v_n$ are the corresponding eigenvectors.