Necessary conditions for existence of linear combination of these matrices to be singular

Question

I'm doing some research in Control theory, and a stumbled with this problem. Any help is appreciated.

QUESTION Let $P_1,\dots,P_m$ be $m$ symmetric positive definite $n\times n$ matrices with $m<n$ and real entries. I'm looking for necessary conditions for the existence of nontrivial real coefficients $\alpha_1,\dots,\alpha_m$ (not all $\alpha_i$ equal to 0) such that $$ H(\alpha) = \sum_{i=1}^m \alpha_iP_i $$ is singular (det$H(\alpha)=0$) and $\sum_{i=1}^m\alpha_m=0$.

More details

Of course there are many ways to come up with some sufficient conditions. For instance, if $m>3$ one can set $\alpha_0 = 1, \alpha_2 = s, \alpha_3 = (-1-s)$ and $\alpha_i=0, i>3$ for some $s$, and then solve the polynomial $\det(P_1 + sP_2 + (-1-s)P_3) = 0$ for $s$ (then check one of those values for $s$ is real). There are many other similar ways in which one can come up with other values for $\alpha_1,\dots,\alpha_m$. But I'm interested for a necessary condition which one can check by looking at the matrices $P_1,\dots,P_m$ and see if there exists such $\alpha_1,\dots,\alpha_m$. Such conditions are easy to derive when $m\leq 3$ (using a similar reasoning as the example before), but I'm interested in the general case.

Of course I know that this may be hard in general, but any help, references and suggestions are appreciated.

Answer 1

Eigenvalues are continuous functions of the matrix entries if this is expressed carefully. Consider $H(c) = cP_1+P_2+\cdots+P_m$. When $c$ is large and negative, the eigenvalues of $H(c)$ are all negative. When $c$ is positive, $H(c)$ is psd and the eigenvalues are all positive. So there is some $c$ such that $H(c)$ has a zero eigenvalue.

EDIT. Nathaniel noted that I had missed the condition $\sum_i \alpha_i=0$. Oops. I'll try again.

Assume $m=3$ (as we can take $\alpha_i=0$ for $i\ge 4$). Choose arbitrary non-zero $\beta_1,\beta_2,\beta_3$ summing to 0. Choose continuous functions $f_1,f_2,f_3:[0,1]\to\mathbb R$ such that, for each $i$, $f_i(0)=-\beta_i$ and $f_i(1)=\beta_i$, and also that $f_1(t),f_2(t),f_3(t)$ always sum to 0 but are never all 0 at the same time. (Linear functions won't do but almost anything else will do.)

Define $H_t=f_1(t)P_1 + f_2(t)P_2 + f_3(t)P_3$.

First consider matrices of odd order. If $H_0$ and $H_1$ are nonsingular, their determinants have opposite sign, so $H_t$ is singular for some $t$ by continuity.

For matrices of even order, the same is true if the numbers of positive and negative eigenvalues of $H_0$ are different. But if those numbers are the same I'm not sure what to do. Can we always choose $\beta_1,\beta_2,\beta_3$ so that $H_0$ has different numbers of positive and negative eigenvalues?