Define the <smooth map between manifolds>[smooth map]
$$
\Phi:X_{m,n}\times\mathbb R^n\longrightarrow\mathbb R^m,
\qquad \Phi(M,v)=Mv.
$$
At $(M,v)$ its <derivative> in the direction $(A,w)$ is
$$
D\Phi_{(M,v)}(A,w)=Av+Mw.
$$
The restriction to variations $(0,w)$ is the <surjective linear map> $M:\mathbb R^n\to\mathbb R^m$, since $M$ has full row rank. Thus $0$ is a <regular value> of $\Phi$. The <regular level set theorem> now shows that
$$
E_{m,n}=\Phi^{-1}(0)
$$
is an <embedded submanifold> of $X_{m,n}\times\mathbb R^n$, of dimension $mn+n-m$.

Solution

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

A linear map $T:V\to W$ is surjective when its <image and preimage of a function>[image] is all of $W$. For finite-dimensional spaces this is equivalent to $\operatorname{rank}T=\dim W$.

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