Consider two matrices $A, B\in\mathcal{M}\_n(\mathbb{C})$, such that $A, B$ has no common eigenvectors. Is it true that for some nonzero $t\in\mathbb{C}$, matrix $A+tB$ is similar to diagonal matrix $\text{diag}(a\_1, a\_2, \ldots, a\_n)$, where $a\_i\not = a\_j, \forall i\not= j$?



Pairs of matrices
It's false for every $n\ge 3$.


Consider $A\_n,B\_n$ $n\times n$ square matrices with no common eigenvalue (these exist for $n=0$ and $n\ge 2$). Then
$$A=\begin{pmatrix} A\_2 & 0 & 0\\ 0 & A\_2 & 0\\ 0 & 0 & A\_{n-4}\end{pmatrix},\qquad B=\begin{pmatrix} B\_2 & 0 & 0\\ 0 & B\_2 & 0\\ 0 & 0 & B\_{n-4}\end{pmatrix}$$ 
work for $n=4$ and $n\ge 6$.


It remains to do $n=3,5$. Define 
$$N=\begin{pmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 &1& 0\end{pmatrix},\qquad N'=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & -1\\ 0 &0& 0\end{pmatrix}$$
Then $N$ and $N'$ have no common eigenvector but the plane spanned by $N,N'$ consists of nilpotent matrices.
Then the matrices
$$A=\begin{pmatrix} N & 0\\ 0 & A\_{n-3}\end{pmatrix},\qquad B=\begin{pmatrix} N' & 0\\ 0 & B\_{n-3}\end{pmatrix}$$ 
work for all $n=3$ and $n\ge 5$. 


