# 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.