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I'd like to know if the following problem of elementary linear algebra is already solved / solvable.

For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?

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    This is obviously the case if either all columns of the two matrices taken together are contained in some hyperplane, or the corresponding situation holds for all rows taken together. I'm not sure whether there are other possiblilties.2012-07-19
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    @MarcvanLeeuwen : Consider e.g. $A = \pmatrix{0 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr}$, $B = \pmatrix{0 & 5 & 5\cr -1 & -5 & 0\cr 1 & 12 & 7\cr}$2012-07-20
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    I think that a possible strategy in tackling this problem is to take certain $\lambda_1,\mu_1,\lambda_2,\mu_2$ satisfying $\begin{vmatrix}\lambda_1&\mu_1\\ \lambda_2&\mu_2\end{vmatrix}\neq0$ so that we can take $(\lambda_1P+\mu_1Q,\lambda_2P+\mu_2Q)$ as the new base instead of $(P,Q)$. By doing so we may assume certain special properties on $P$ and $Q$. As a (rather weak) example, we can make the top left corner of $P$ and bottom right corner of $Q$ become zero.2012-07-21

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