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Given two square matrices $A, B$, when is $$\det(A+tB) = 0$$ for all $t\in \mathbb{R}$?

An easy sufficient condition is that $A$ and $B$'s kernels have nontrivial intersection. Per Henning's comment below, this is not also necessary. Does there exist a nice necessary and sufficient characterization?

user7530
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    No, consider $A=\begin{pmatrix}1&0\0&0\end{pmatrix}$ and $B=\begin{pmatrix}0&1\0&0\end{pmatrix}$. – hmakholm left over Monica Apr 27 '14 at 16:34
  • @HenningMakholm Good example, thanks. – user7530 Apr 27 '14 at 16:39
  • Related: It's not very hard to show that $A+tB$ is invertible for every real $t$ if and only if $A$ is invertible and $A^{-1}B$ doesn't have any nonzero, real eigenvalues. – Mike F Apr 30 '14 at 01:51
  • Also, Henning Makholm's example suggests another easy sufficient condition. Namely: if the column spaces $C(A)$ and $C(B)$ do not together span the whole space, then $A+tB$ is always singular. – Mike F Apr 30 '14 at 02:04
  • Duplicate of http://math.stackexchange.com/q/173026 –  May 01 '14 at 00:43

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