Suppose $A\in M_n(\mathbb R)$ with \begin{align} \big|A-\lambda I_n\big|\,&=\,\pm(\lambda-\lambda_1)^{m_1}\cdots(\lambda-\lambda_p)^{m_p} \newline minpoly_A(\lambda)\,&=\,(\lambda-\lambda_1)^{n_1}\cdots(\lambda-\lambda_p)^{n_p} \end{align} admits the Jordan normal form \begin{align} J_A\,=\,\widetilde J_{m_1}(\lambda_1)\oplus\cdots\oplus \widetilde J_{m_p}(\lambda_p) \end{align} where $\widetilde J_{m_i}(\lambda_i)$ is Jordan matrix of order $m_i$ corresponding to eigenvalue $\lambda_i,\ 1\leq i\leq p$.
Let $\dim E(\lambda_i)=r_i$. By Jordan Theorem, we get \begin{align} \widetilde J_{m_i}(\lambda_i)\,=\,J^1(\lambda_i)\oplus\cdots\oplus J^{r_i}(\lambda_i) \end{align} where $\ J^{k_i}(\lambda_i),\ 1\leq k_i\leq r_i$ is the Jordan block corresponding to $\lambda_i$.
So we have \begin{align} J_A\,=\,\Big[J^1(\lambda_1)\oplus\cdots\oplus J^{r_1}(\lambda_1)\Big]\oplus\cdots\oplus\Big[J^1(\lambda_p)\oplus\cdots\oplus J^{r_p}(\lambda_p)\Big] \end{align} Suppose the block $J^{k_1}(\lambda_i)$ is of order $d_{ik_i}$. Let this matrix \begin{align} P\,=\,\begin{bmatrix} \begin{pmatrix} P_{11} & \cdots & P_{1r_1} \end{pmatrix} & \cdots & \begin{pmatrix} P_{p1} & \cdots & P_{pr_p} \end{pmatrix} \end{bmatrix} \end{align} where \begin{align} P_{ik_i}\,=\,\begin{bmatrix} \big(A-\lambda_iI_n\big)^{d_{ik_i}-1}u_{ik_i}^\top & \cdots & \big(A-\lambda_iI_n\big)u_{ik_i}^\top & u_{ik_i}^\top \end{bmatrix} \end{align} with some vector $\ u_{ik_i}\in\ker\big(A-\lambda_i I_n\big)^{n_i}\setminus\ker\big(A-\lambda_i I_n\big)^{n_i-1}\ $ such that all these vectors are linearly independent.
Then, I want to show that \begin{align} P^{-1}AP\,=\,J_A. \end{align}
I have no clue how to prove it. I hope someone will blow my mind with some interesting guidances. Thank you.