Transversality condition From my textbook:
"If we examine linearizations of $f_\mu (x)$, (where $\dot{x}=f_\mu(x) \ x\in \mathbb{R}^n, \mu\in \mathbb{R}^2$) at the equilibria $f_\mu(x)=0$, then we can formulate a transversality condition which guarantees, for example, that no linearization of $f$, has a zero eigenvalue of multiplicity greater than two and that any equilibrium which does have a zero eigenvalue of multiplicity two has a Jordan normal form with the block
$$\left(
\begin{array}{cc}
  0 & 1 \\
  0 & 0 
\end{array} \right)$$"
What is this transversality condition? I understand that this condition should include the Jacobian of $f$, even though I do not know how to proceed. I have already asked a question kept on transversality: Transversal intersection.
Thank you very much
 A: First, the authors introduce the set $\mathcal M=\{(x,\mu)\in\mathbb R^{n+2}: f_\mu(x)=0\}$. "Using transversality, we expect that $\mathcal M$ is a smooth two-dimensional surface" near $(\mu_0,x_0)$, they say. I think that the appropriate transversality condition for this part is $$\mathrm{rank}\, \frac{\partial f_\mu}{\partial (x,\mu)} (\mu_0,x_0)=n\tag{1}$$ In words, the total derivative of $f_\mu$ with respect to both $x$ and $\mu$, represented by an $n\times (n+2)$ matrix, has maximal rank $n$, and therefore the Implicit Function theorem applies. 
Then we consider the set $$\mathcal J=\left\{ \frac{\partial f_\mu}{\partial x} : (x,\mu)\in\mathcal M\right\}\tag{2}$$
of all Jacobians (derivatives with respect to $x$ only) evaluated at the points of $\mathcal M$. 
This is a parametric two-dimensional surface in the $n^2$-dimensional space of square matrices of size $n$. 
Next, introduce a set $\mathcal B$ of bad square matrices. Specifically, $\mathcal B$ comprises of all matrices which have eigenvalue zero with multiplicity at least three, and also all matrices that have the block $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ in their Jordan normal form. The authors leave it to the reader to verify that $\mathcal B$ has codimension greater than two. Informally, this is true because in order to qualify to be a member of $\mathcal B$, a matrix has to satisfy at least three different equations. 
The idea of transversality now takes the following form: being two-dimensional, $\mathcal J$ will in general be disjoint from the set $\mathcal B$, because the latter set has codimension greater than two. I guess our new "transversality condition" is now $$\mathcal J\cap \mathcal B=\varnothing\tag{3}$$ 
As the authors say in the introduction, they are writing for mixed audience and are trying to avoid "mathematical pedantry". Thus, the lack of precise statements on these pages is supplanted by vigorous hand-waving and repeated mention of the words "transversality". 
