$u,v$ two endomorphisms such as $u\circ v = v \circ u $ and $\ker (u) \cap \ker (v) = \{0\}$, Show that $GL(E) \cap Span(u,v) \ne \emptyset$ 
Let E a  finite dimensional complex vector space, $u$ and  $v$ two endomorphisms of E such as $u\circ v = v \circ u $ and $\ker (u)  \cap \ker (v) = \{0\}$ Show that there exist $(\alpha,\beta) \in \mathbb{C}^{2}$ such as $\beta u + \alpha v$ is an automorphism.

Let $n$ denote the dimension of E.
If $u$ or $v$ are automorphisms we can pose $(\alpha,\beta) = (0,1)$ or $(\alpha,\beta) = (1,0) $. Else, the problem can be rephrase : show that there exist $\alpha \in \mathbb{C}$ such as $u + \alpha v$ is an automorphism.
Since $u$ and $v$ commute they are simultaneous trigonalisable (on $\mathbb{C}$),
$(\lambda_1,\dots,\lambda_n)$ the eigenvalues of $u$
$(\mu_1,\dots,\mu_n)$ the eigenvalues of $v$
The problem can be rephrase: show that $\forall i \in \{1,\dots, n \}, \lambda_{i} = 0 \implies \mu_{i} \ne 0$
Do you have any hint ?
 A: Let $A$ and $B$ be the matrices for $u$ and $v$ respectively after triangulation and scaling.
Let $k = \min \{i:\lambda_i = \mu_i = 0\}$ and assume $k > 1$ (otherwise $0 \ne e_1 \in \ker(A \cap B))$. Let $A_k$ be the top left $k\times k$ submatrix of $A$ and similarly for $B_k$. Let $(A_k)_{ii} = a_i$ and $(A_k)_{ij} = a_{ij}$ otherwise, and similarly for $B$. We have
$$(A_kB_k)_{k-1 k} = a_{k-1}b_{k-1k} = b_{k-1}a_{k-1k} = (B_kA_k)_{k-1 k} \tag 1$$
since $A_k$ and $B_k$ commute ($u$ and $v$ commute on the associated invariant subspace).
Now let $v = (v_1,...,v_k)^T$. The idea is to set the $v_i$ so that $A_kv = B_kv = 0$, leading to the desired contradiction.
Unfortunately equation $(1)$ has cases. I will do one here but they are all similar. If $a_{k-1k} = a_{k-1} = 0$ then $b_{k-1} \ne 0$ so we can set $v_{k-1} = -b_{k-1k}/b_{k-1}$. Let $A_k'$ be the matrix obtained by adding $v_{k-1}$ times the second last column of $A_k$ to the last column (this will "kill" $(A_k)_{k-1k}$) and similarly for $B_k'$ ("killing"  $(B_k)_{k-1k}$). Eq. $(1)$ always allows you to find a suitable $v_{k-1}$ to do this.
Now the last row of both $A_k'$ and $B_k'$ and their second last columns are irrelevant so they can be "deleted". The resulting submatrices are of dimension $k-1 \times k-1$ and have the same structure (bottom right corner $0$) as the original $k \times k$ matrices $A_k$ and $B_k$. They also continue to commute on the smaller invariant subspace. So we can repeat this process to set all of the $v_i$. Note that we have taken $v_k=1$.
The resulting $0 \ne (v_1,...,v_k,0,...,0)^T \in \ker(A \cap B)$ gives the contradiction in the larger space.
