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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. – Marc van Leeuwen Jul 19 '12 at 23:05
@MarcvanLeeuwen : Consider e.g. $A = \pmatrix{0 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 1\cr}$, $B = \pmatrix{0 & 5 & 5\cr -1 & -5 & 0\cr 1 & 12 & 7\cr}$ – Robert Israel Jul 20 '12 at 2:38
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. – progressiveforest Jul 21 '12 at 0:56

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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I've never seen $A \oplus B$ used by itself as a condition before. What does it mean? – Eric Stucky Aug 6 '12 at 3:42
@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. – user31373 Aug 6 '12 at 22:18

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. – copper.hat Jul 20 '12 at 1:00

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