Fix $I$ and write $I'$ for the complementary column indices. Over $U_I$, decompose $M=(M_I,M_{I'})$ and $v=(v_I,v_{I'})$. The equation $Mv=0$ is equivalent to
$$
v_I=-M_I^{-1}M_{I'}v_{I'}.
$$
Matrix inversion is <smooth function>[smooth] on the <invertible matrix>[invertible matrices], so
$$
(M,u)\longmapsto\left(M,\bigl(-M_I^{-1}M_{I'}u,u\bigr)\right)
$$
is a smooth, fiberwise-linear trivialization $U_I\times\mathbb R^{n-m}\to\pi^{-1}(U_I)$. The sets $U_I$ cover the base, proving that $\pi:E_{m,n}\to X_{m,n}$ is a <vector bundle> of rank $n-m$. It is the <kernel bundle of a constant-rank family of linear maps>, and its rank also follows from the <rank-nullity theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 1 / c / Solution

Let $A_b:V\to W$ vary smoothly with $b$ and have constant rank $r$. The spaces $\ker A_b$ form a <vector bundle> of rank $\dim V-r$. On a neighborhood where one $r$ by $r$ minor is invertible, the corresponding coordinates are smooth linear functions of the remaining coordinates, giving a local trivialization.

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