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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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    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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This is not a complete answer, just some sufficient and some necessary conditions.

Let's look at the opposite condition. Fix $u,v\in\mathrm{End}_k(V)$ two singular (non proportional) endomorphisms of the finite dimensional $k$-vector space $V$: what conditions are necessary or sufficient for there to be a scalar $\lambda$ such that $u+\lambda v\in\mathrm{GL}(V)$?

The following conditions are necessary:

  1. $\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)$;
  2. $\mathrm{Im}(u)+\mathrm{Im}(u)=V$;
  3. $v\big(\mathrm{Ker}(u)\big)\oplus u\big(\mathrm{Ker}(v)\big)$.

The first is @chaochuang's answer, and the second is @Marc van Leeuwen' comment. To see the necessity of the third condition, write down matrices in a basis adapted to $V=\mathrm{Ker}(u)\oplus\mathrm{Ker}(v)\oplus S$. The following conditions are sufficient (at least when $k$ is big enough, for instance if $\mathbb Q\subset k$):

  1. The induced morphism $\tilde{v}:\mathrm{Ker}(u)\rightarrow V\rightarrow V/\mathrm{Im}(u)$ is an isomorphism i.e. $v(\mathrm{Ker}(u))\oplus\mathrm{Im}(u)$;
  2. The induced morphism $\tilde{u}:\mathrm{Ker}(v)\rightarrow V\rightarrow V/\mathrm{Im}(v)$ is an isomorphism i.e. $u(\mathrm{Ker}(v))\oplus\mathrm{Im}(v)$.

To see the sufficiency of the first point, write down the matrix for $u+\lambda v$ where the initial base of $V$ is adapted to $S\oplus \mathrm{Ker}(u)=V$, and the final base is the images under $u$ of the basis vectors chosen in $S$ and the images under $v$ of the basis vectors in $\mathrm{Ker}(u)$, and let $\lambda\rightarrow 0$: the determinant of this matrix is a polynomial in $\lambda$ of the form $\lambda^d(1+\cdots)$ where $d=\dim~\mathrm{Ker}(u)$.

Going back to the initial problem, we get $3$ sufficient conditions for the span of $u$ and $v$ to be completely singular, and $2$ necessary conditions.

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    @EricStucky It's a shorthand for "$A\oplus B$ makes sense", which is a longhand for $A\cap B=\{0\}$. It makes sense if you think of a programming language in which the procedure "Take Direct Sum" returns *False* if it encountered an error, and returns a truth-y value otherwise.2012-08-06
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A sufficient condition is $P$ and $Q$ share at least one eigenvector for 0 eigenvalue:

$P\mathbf{v}= 0 \mathbf{v}$, $Q\mathbf{v}= 0 \mathbf{v}$

so for any $\lambda,\mu\in\mathbb{R}$, we have $(\lambda P+\mu Q) \mathbf{v} = 0 \mathbf{v}$, which means 0 is still an eigenvalue of $\lambda P+\mu Q$, i.e. $\det(\lambda P+\mu Q)=0$.

I doubt it's also a necessary condition.

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    It is not necessary, take P = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, Q = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}. The kernels are orthogonal, but the above property holds.2012-07-20