Almost every linear map $V\rightarrow V$ (in a certain affine subspace of such maps) is invertible I'm writing up a paper. There is a result that I want to be rigorous about, but I'm not exactly sure how. Here is the set-up:
I have some real Euclidean space $V$ which is isomorphic to $\mathbb{R}^n$. Consider the set of all linear maps $\operatorname{L}(V)$ from $V$ to itself, which is isomorphic to the set of $n\times n$ matrices over $\mathbb{R}$. This is also a real Euclidean space and is isomorphic to $\mathbb{R}^{n^2}$. Finally, let $A\subset\operatorname{L}(V)$ be some affine subspace that does not contain the origin. (In my paper, this is essentially the affine space of all linear maps $f:V\rightarrow V$ satisfying $f^*(v)=v$ for some fixed choice of nonzero vector $v\in V$.)
What I want to say is this: "Almost all maps in $A$ are invertible (in the sense that, with respect to the induced Lebesgue measure on $A$, the set of non-invertible maps has measure zero)."
This is certainly true. But my coauthor is not convinced this is as trivial as I think it is---and would like us to provide 'rigorous' reasoning for this.
My reasoning: We may consider $A$ as an affine subspace of $\mathbb{R}^{n^2}$. The determinant $\operatorname{det}:\mathbb{R}^{n^2}\rightarrow\mathbb{R}$ is a polynomial, and thus $\operatorname{det}$ is either constant on $A$ or the set of zeros on $A$ has measure zero. The desired result follows from the observation that a linear transformation is invertible if and only if its determinant is nonzero.
Is this valid reasoning? Is there something accessible that I could cite here?

As an aside, I wanted to mention where this is coming from. In quantum information theory, a quantum channel is a linear map $\Phi:M_m\rightarrow M_m$ that is completely positive and trace preserving. In particular, every quantum channel is also Hermitian preserving, so we may view it as a linear map on the set of $m\times m$ Hermitian matrices, which is a real Euclidean space. What I want to say is the following: Almost all quantum channels are invertible as linear maps. (Although, the inverse mapping is usually not also a channel.)
 A: Here's one way of doing this in your case.
You are looking at $A_v= \{ T\in L(V) : Tv=v \}$ where $v$ is a non-zero vector. Extend $v$ to a basis. Then with respect to this basis $T\in A_v$ iff it has a matrix of the form $$[T]= \begin{bmatrix} 1 & * \\ 0 &B\end{bmatrix}$$ where $B\in M_{n-1}(\mathbb R)$
So you have identified $A_v \cong\mathbb R^{n^2-n}$ and $T\in A_v$ is invertible iff $\det B \neq 0$. So it is the complement of the zero set of a polynomial in $\mathbb R^{n^2-n}$ and hence has measure $0$.
EDIT: Let's see your problem in a more general framework where $V$ is some vector space over an infinite field $k$ and you ask the same question. $L(V)=M_n(k)$ is equipped with Zariski topology. It is easy to see that $M_n(k)$ is irreducible. So any non empty open set is dense. In particular $GL_n(k)=\{ T \in L(V) : \det(T)\neq 0 \}$ is a dense open subset. Since $A \subset M_n(k)$ is an affine linear subspace, it is also irreducible. So if $A\cap GL_n(k)$ is non-empty, then it is a dense open subset of $A$. The upshot is the existence of one invertible map gives you density of invertible maps in that affine subspace.
