Question about linear algebra in Benson's book: intersections of images or sum of kernels I am not sure if this question is suitable in here. I asked this question in Mathematics some days ago.
The following proposition is in Benson's book “Representation theory of elementary abelian groups and vector bundles”:

Proposition 4.1.2 Suppose $k$ is algebraically closed. Let $A_1,A_2\in M_{n\times m}(k)$, regarded as maps from $k^m$ to $k^n$. Then:
  (1) Suppose that forall $\lambda,\mu\in k$, not both zero, $\lambda A_1+\mu A_2$ is injective. Then $$\bigcap_{(\lambda,\mu)\neq (0,0)}Im(\lambda A_1+\mu A_2)=0$$
  (2) Suppose that forall $\lambda,\mu\in k$, not both zero, $\lambda A_1+\mu A_2$ is surjective. Then $$\sum_{(\lambda,\mu)\neq (0,0)}Ker(\lambda A_1+\mu A_2)=k^m$$

This is very interesting proposition and the proof in the book is very beautiful. The proof in the book used field extension. I am wondering if this can be proved by methods of linear algebra. Any helps will be appreciated.
The following is proof in the book:

Th4.1.1 is that the rank is preserved when consider field extension and take algebraic independent variables.
 A: I don't have a proof which only uses linear algebra, but I have one in the context projective geometry.
I will prove the second statement you mention. Let $m \geq n$ and let $M$ be the generic matrix of size $m \times n$ with linear coefficients. Let $V = \mathbb{k}^m$, $W = \mathbb{k}^n$ and $X = \mathbb{P}(\mathrm{Hom(V,W)})$. The matrix $M$ induces a map:
$$ V \otimes \mathcal{O}_X(-1) \longrightarrow W \otimes \mathcal{O}_X.$$
Assume there exists $A_1,A_2 \in X$ such that for all $(\lambda,\mu) \in \mathbb{k}^2 -\{0,0\}$, the matrix $\lambda A_1 + \mu A_2$ is surjective. This is equivalent to saying that the restriction of $M$ to the $\mathbb{P}^1$ joining $A_1$ to $A_2$ in $X$ is surjective. We thus have an exact sequence:
$$0 \longrightarrow F \longrightarrow V\otimes \mathcal{O}_{\mathbb{P}^1_{A_1,A_2}}(-1) \longrightarrow W \otimes \mathcal{O}_{\mathbb{P}^1_{A_1,A_2}} \longrightarrow 0,$$
where $F$ is the kernel of $M$ restricted to this $\mathbb{P}^1_{A_1,A_2}$.
Assume by absurd that $\sum_{A \in \mathbb{P}^1_{A_1,A_2}} Ker A \neq \mathbb{k}^m.$ Dualizing, this is equivalent to $\bigcap_{A \in \mathbb{P}^1_{A_1,A_2}} (Ker{A})^{\perp} \neq 0$, where $(Ker A)^{\perp} \subset V^{*}$ is the orthogonal to $Ker A$ in $V^*$. 
From the above exact sequence, we see $F^{\perp} \simeq W^* \otimes \mathcal{O}_{\mathbb{P}^1_{A_1,A_2}}(-1)$. Now the hypothesis $\bigcap_{A \in \mathbb{P}^1_{A_1,A_2}} (Ker{A})^{\perp} \neq 0$ exactly translates as $F^{\perp}$ has a non-zero constant section over $\mathbb{P}^1_{A_1,A_2}$, namely the map:
$$ x  \longrightarrow (x,p),$$
where $p$ is a non-zero vector in $\bigcap_{A \in \mathbb{P}^1_{A_1,A_2}} (Ker{A})^{\perp}$.
This is impossible as $H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(-1))=0$.
