Kernel of a matrix pencil Let $A,B$ be $n\times n$ singular real matrices such $ker A\cap ker B=\{0\}$, how could I show that there exists $x\in \mathbb R$ such that $ker (A+xB)=\{0\}$?

Let $A = \left( \begin{array}{cc}
1 & 1  \\
1 & 1  \\ \end{array} \right)$ and $B = \left( \begin{array}{cc}
0 & 1  \\
0 & 1  \\ \end{array} \right)$. Then $ker(A) \cap ker(B) = \{ 0 \}$ and 
$$ \det(A + xB) = \det\left( \begin{array}{cc}
1 & 1 + x  \\
1 & 1 + x  \\ \end{array} \right) = 0$$ for all $x \in \mathbb{R}$. So such $x$ does not exists.
