Define $\vec{F}$ to be a mapfrom the state space to the tangent space
Without formally defining what the linearisation theorem is, the idea behind this theorem is that for any fixed point $x=x^{*}$, the behaviour of the neighbourhood about this fixed point is equivalent to the linear system $\dot{y}=Ay$ for a square matrix A and vector y.
The behaviour about the neighbourhood of a fixed point $x=x^{*}$ can be determined by evaluating the Jacobian $d\vec{F}\left ( x^{*} \right )=\frac{\partial f_{i}}{\partial x_{j}} \mid_{x=x^{*}}$ of the vector field $\vec{F}\left ( x \right )$ at the fixed point $x=x^{*}$. However, a change of coordinate z=Py is required so that an equivalent system $\dot{z}=Jz$ where J is the block diagonal matrix, P is a matrix and z is a vector.
It is not explained in my notes why this change of coordinates is required. Under what circumstances would $\dot{z}=Jz$ be more appropriate than the linear system $\dot{y}=Ay$? Good explanation is needed and will be appreciated!
Thanks in advance.
I am not sure I understand what your question is.
The change of coordinates $z=Py$ is only needed if you want to put the linear system $\dot{y}=Ay$ into the form $\dot{z}=J{z}$ where $J$ is the Jordan normal form of $A$. If $A$ is diagonalizable, then $J$ will be a diagonal matrix.
This is not required but it is useful as it allows for the solution $y(t)$ to be written in the form
$$y(t)=Pe^{Jt}P^{-1} y \ (t_0)$$
where $e^{\cdot}$ is the matrix exponential.