I was primarily interested in the following question. Let $n\geq 3$, and let $X\subset \mathbb{P}^n$ be a degree $d$ hypersurface. Assume that its singularity locus $S$ (with reduced structure) is irreducible and smooth of dimension $k$. (As pointed out by @abx , this should not be a strict inclusion)
Is it true that
$$
\dim S^\vee \geq \dim X^\vee ?
$$
The statement is true for quadratic hypersurfaces and hypersurfaces with isolated singularities. I am wondering if in general this holds.


Also, is there any criteria on when this inequality is strict ?


I was reading the book Discriminants, Resultants, and Multidimensional Determinants by Gelfand-Kapranov-Zelevinksy. In the first chapter they introduce the Katz dimension theorem, which computes the dimension of the dual varieties via local coordinates. But I found it difficult to use, since it's quite hard to write down the local coordinates. Is there any comments on the Katz dimension theorem, especially on how to use it ?



On the dimension of the dual variety of a singular hypersurface
No. Consider the cubic surface $X$ defined by the equation $$x\_0^2x\_2+x\_1^2x\_3=0.$$


Then $X$ is singular along the line $x\_0=x\_1=0.$ Then $X^{\vee}$ is a hypersurface (actually isomorphic to $X$) and $S^{\vee}\cong \mathbb P^1$ (consisting of the hyperplanes containing the line). So


$$\dim S^{\vee}=1<2=\dim X^{\vee}.$$


