Jordan normal form and spectral decomposition In this post by Terence Tao (exercise 28, point vi), he proves the following theorem (called spectral decomposition):

Theorem:
Let $A$ be a complex square matrix. Then $A$ can be written as

$A=\sum P_j(\lambda_j I+N_j)P_j$

Where $P_j$ are matrices such that $P_j P_l=0$, $P_j^2=P_j$ and $P_1+\dots +P_k=I$
and $N_j$ is a nilpoten matrix of order $\text{mult}(\lambda_j)$ ($\lambda_j$ is the $j$-th eigenvalue)

This theorem seems really connected with the Jordan form of a matrix, although I am not able to prove that: It is clear that the J.F. implies the spectral decomposition, while the opposite is not so easy: I tried to prove it on a canonical basis in order to generalize it after but I didn't get anywhere. 

Question:

Is the J.F. actually a consequence of the spectral decomposition? If so, how to prove it?
This almost implies the Jordan canonical form. What is missing is the normal form of nilpotent matrices. That is, $N_j$ is similar to a block diagonal matrix with blocks of the form
$$
\begin{bmatrix}0&amp;1&amp;0&amp;\ldots &amp;0\\0&amp;0&amp;1&amp;\ldots &amp;0\\\vdots &amp;\vdots &amp;\vdots &amp;\ddots &amp;\vdots \\0&amp;0&amp;0&amp;\ldots &amp;1\\0&amp;0&amp;0&amp;\ldots &amp;0\end{bmatrix}.
$$
This change of basis gives a Jordan basis on each generalized eigenspace, which are the images of the $P_j$.
This almost implies the Jordan canonical form. What is missing is the normal form of nilpotent matrices. That is, $N_j$ is similar to a block diagonal matrix with blocks of the form
$$
\begin{bmatrix}0&amp;1&amp;0&amp;\ldots &amp;0\\0&amp;0&amp;1&amp;\ldots &amp;0\\\vdots &amp;\vdots &amp;\vdots &amp;\ddots &amp;\vdots \\0&amp;0&amp;0&amp;\ldots &amp;1\\0&amp;0&amp;0&amp;\ldots &amp;0\end{bmatrix}.
$$
This change of basis gives a Jordan basis on each generalized eigenspace, which are the images of the $P_j$.