Converting my comments to an answer.
It is known from work of Björner in the late 80s/early 90s that a matroid (independence) complex is shellable: see e.g. his "Homology and shellability of matroids and geometric lattices" available online at https://www.cefns.nau.edu/~falk/classes/690/extras/bjorner.pdf.
Hence, such a complex has the homotopy type of a wedge of spheres. More precisely, since a the complex of matroid $M$ is a pure $(r(M)-1)$-dimensional complex (where $r(M)$ is the rank), it has the homotopy type of a wedge of $\widetilde{\chi}(M)$ $(r(M)-1)$-spheres. Here $\widetilde{\chi}(M)$ is the reduced Euler characteristic of the complex, which is equal to a Tutte polynomial evaluation $\widetilde{\chi}(M)=T_M(0,1)$.
So if this number $T_M(0,1)$ is nonzero, then we will have an exact equality $\eta(M) = r(M)$. Otherwise, the matroid complex of $M$ is contractible so $\eta(M)=\infty$ (this happens e.g. for a simplex, but shouldn't happen for most matroids...). Either way your bound holds.