I have a subvariety $V \subset X$, and I want to compute the dimension of the connected component of $\textrm{Hilb}(X)$ containing $[V]$.
I can give explicit deformations of $V$ showing that the dimension of this component is at least $m$. I can also show that $H^0(V, N\_{V /X}) \leq m$. Do these facts together imply that the dimension of this component is exactly $m$, and that $\textrm{Hilb}(X)$ is smooth at $[V]$? I am a little worried about how the tangent sheaf works in the singular case.



Dimension of Hilbert scheme?
In general, the Zariski tangent space of $\textrm{Hilb}(X)$ at $[V]$ is naturally isomorphic to $\textrm{Hom}\_V(I\_V/I\_V^2, \, \mathcal{O}\_V)$. 


When $X$ and $V$ are both smooth and projective, this group equals $H^0(V, \, N\_{V/X})$. Therefore in your case we can write $$m \leq \dim \_{[V]}\textrm{Hilb}(X) \leq \dim T\_{[V]} \textrm{Hilb}(X) = H^0(V, \, N\_{V/X}) \leq m.$$


This means $$\dim\_{[V]}\textrm{Hilb}(X) = \dim T\_{[V]} \textrm{Hilb}(X)=m,$$
that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.


