# When are Linear Operator and Identity transversal in a Vector Space?

Let $$A:V \to V$$ a Linear Operator on the Vector Space $$V$$ and $$Id:V \to V$$ the Identity map, when is $$A$$ transverse to $$Id$$ ? My intuition says that $$A$$ may not have $$1$$ as eigenvalue but I didn't find out if it is really true

Since

Two maps $$f: X \to Z$$ and $$g:Y\to Z$$ are transversal if, for every $$x \in X$$ and $$y\in Y$$ with $$f(x) = z = g(y)$$, the diferentials on these points spans the entire tangent space at $$z$$ in sense that $$im(df) + im(dg) \simeq T_zZ$$ (by https://ncatlab.org/nlab/show/transversal+maps)

We have

$$A: V \to V$$ and $$Id:V\to V$$ are transversal if, for every $$x \in V$$ and $$y\in V$$ with $$A(x) = z = Id(y)$$ , the diferentials at these points spans the entire tangent space at $$z$$ in sense that $$im(A) + im(Id) \simeq T_zV \simeq V$$ , i.e.

$$A(x) + Id(y) = z \ \ \forall \ \ x, y \in V \ s.t. \ \ A(x) = z = Id(y) , \ i.e. \\ \iff (A - I)(x) \ = x \ \ * \ for \ x \ = y \ = z$$

Since $$(A-\lambda I)$$ is singular $$\iff det(A-\lambda I)=0$$ $$\iff \lambda$$ is eigenvalue of $$A$$ , if $$1$$ is eigenvalue of $$A$$ then equation $$\ \ * \ \$$ doesn't means isomorphism and then, by if and only if, doesn't means transversality of $$A$$ and $$Id$$.

My first problem is:

I assumed at equation $$\ \ * \ \$$ the equality $$x \ = y\ \ =z$$, which makes sense to me because $$A$$ and $$Id$$ are global maps and then the transversality must be with all points of $$V$$ but I'm feeling that something is missing. $$z \ = y \$$ because $$Id$$, no doubt, but $$x \ = y \$$?

My second problem is:

At https://ncatlab.org/nlab/show/transversal+maps the author says

'In particular, a submersion is transversal to all functions.'

But isn't Identity and non-singular Linear Operator surjective , equals to they derivative , because they are their best linear approximation, and then are submersions ?

Submersions are surjective on tangent spaces, by definition. This means the image of $$df$$ is the entire tangent space. In particular, no matter what other subspace you add to it, you will get the entire tangent space. This means it is transverse to everything.