## When does a polynomial have all pure imaginary roots? [closed]

Let $P(x)=x^{n}+a_{1}x^{n-1}+\cdots+a_{n-1}x+a_{n}$, where $a_1, a_2, \dots, a_n$ are intergers.

Question 1. When does the polynomial $P(x)$ have its zeros all being pure imaginary or zero(here 0 is a root of the given polynomial)?

Question 2. Does there exist a characterization that $P(x)$ have its zeros all being pure imaginary or zero?

-
are your coefficients $a_1,\dots,a_n$ assumed to be real? – Pietro Majer Feb 27 2011 at 9:26
mathoverflow.net/questions/20946/… – J.C. Ottem Feb 27 2011 at 9:53
Here's a hint for the degree 2 case: $P(x) = (x-ia)(x-ib)$. – Franz Lemmermeyer Feb 27 2011 at 10:33
@Pietro Majer, thanks! The coefficients a_i,(i=1,\dots,n) are all intergers – Shunyi Liu Feb 27 2011 at 12:39
@J.C. Ottem, thanks! – Shunyi Liu Feb 27 2011 at 13:37

## closed as too localized by Franz Lemmermeyer, Qiaochu Yuan, Andres Caicedo, Gjergji Zaimi, Dmitri PavlovFeb 27 2011 at 19:23

A necessary and sufficient condition is that $P(x)$ is a power of $x$ times a product of terms $x^2+c$ with $c$ real and positive. Hence $P(x)$ is $x^{n-2m}Q(x^2)$ where $m\ge0$ and $Q(t)$ is a unitary polynomial in $t$ of degree $m$ with integer nonnegative coefficients. Finally, a necessary condition is that $a_{k}=0$ for every odd $k$ and one should further require the underlying polynomial $Q$ to be a product of $t+c_i$.

EDIT It appears that the problem of locating the zeroes of a polynomial resurfaces regularly on MO, see for example this question or that one or that one. The answers to these provide the following facts.

First, there are Newton's inequalities. For any real numbers $(t_i)_{1\le i\le m}$, their elementary symmetric means $S_k$ are such that, for every $k$, $S_k^2\ge S_{k-1}S_{k+1}$. Here $S_k=\sigma_k/{m\choose k}$ where $\sigma_k$ denotes the $k$th elementary function in the real numbers $(t_i)_{1\le i\le m}$, see here.

So, write $Q$ as $Q(t)=t^m+b_1t^{m-1}+\cdots+b_m$ (and recall that $b_k=a_{2k}$). Any $Q$ under consideration is such that $b_k$ must be nonnegative for every $k$ and Newton's inequalities indicate that supplementary necessary condition are $S_k^2\ge S_{k-1}S_{k+1}$ for every $k$, where the $S_k$ are based on the $\sigma_k=b_k$. (At first sight, one should choose $\sigma_k=(-1)^kb_k$ but the $(-1)^k$ disappear.) Hence, a necessary condition is that, for every $k$, $$k(m-k)b_k^2\ge (k+1)(m-k+1)b_{k-1}b_{k+1}.$$ Second, a complete characterization of the polynomials $Q$ with only real negative roots is based on the notion of Hermite forms, see this answer.

Recall that the Hermite form of a polynomial $Q$ of degree $m$ is a symmetric matrix, usually denoted by $H_1(Q)$, of size $m\times m$ with entries $(h_{ij}(Q))$, defined by $$h_{ij}(Q)=s_{i+j-2}(Q),$$ where, for every $k$, $s_k(Q)$ is the sum of the $k$th powers of the roots of $Q$ (and $s_0(Q)=m$), see these lecture notes. Recall that the $s_k(Q)$ are well known functions of the elementary symmetric functions $\sigma_k=(-1)^kb_k$, see here. Recall also that the signature of a symmetric matrix is equal to the number of its positive eigenvalues minus the number of its negative eigenvalues. Then the signature of $H_1(Q)$ is equal to the number of real roots of $Q$.

In the context of this question, one already knows that $Q$ has no positive real root because $b_k$ is nonnegative for every $k$ and one wants $Q$ to have $m$ real roots. Hence the signature of $H_1(Q)$ must be $m$, that is, $H_1(Q)$ must be positive definite.

Finally, necessary and sufficient conditions are that the odd numbered $a_k$ are zero and that the even numbered $a_k$ are nonnegative and define a polynomial $Q$ such that the symmetric matrix $H_1(Q)$ is positive definite.

-
 @Didier Piau, thanks for your reply! Unfortunately I am mainly interested in what conditions of the coefficients $a_{i}$ are satisfied to leave $P(x)$ have its roots pure imaginary or o. – Shunyi Liu Feb 27 2011 at 12:58 @Shunyi Liu: Are you asking for a condition on the coefficients of $Q$ (the $Q$ in my post) equivalent to the fact that all the roots of $Q$ are real negative? – Didier Piau Feb 27 2011 at 15:10 Why is invertibility of $H_1(Q)$ equivalent to saying the signature is $m$? It looks like the condition to check was that $H_1(Q)$ is positive definite. – Douglas Zare Feb 27 2011 at 20:45 @Douglas Let $A$ be symmetric of size $m\times m$. Then: $A$ is positive definite iff every eigenvalue of $A$ is positive iff the number of positive eigenvalues of $A$ is $m$ iff the signature of $A$ is $m$. Or am I missing something? – Didier Piau Feb 28 2011 at 7:42 @Didier Piau, thank you very much for your kind help. It is easy to see that the final conditons you obtained are necessary. But how to show that these conditions are also sufficient? Besides, whether $Q$ has only real negative roots is equivelent to $\det(H_{1}(Q))\ne0$? – Shunyi Liu Feb 28 2011 at 10:17

This should be a comment but I don't have enough reputation to leave one. First of all, $0$ is a root of $P(x)$ if and only if the independent term $a_n$ is null. Then, you can characterize the coefficients in terms of elementary symmetric polynomials and their degree: $a_i = (-1)^{n-i} \sigma_n$, where $\sigma_n = \sum_{1 \le j_1 < j_2 < \cdots < j_n \le n} r_{j_1}r_{j_2}\dotsm r_{j_n}$, and $r_1, \dotsc , r_n$ are the roots of the polynomial. From this you can see that a necessary condition for $r_1, \dotsc , r_n$ to be imaginary is that $a_k$ is of the form $c_ki$ if $k \equiv n \pmod 4$, $c_k$ if $k \equiv n-1 \pmod 4$, $-c_ki$ if $k \equiv n-2 \pmod 4$, and $-c_k$ if $k \equiv n-3 \pmod 4$, where $c_k$ is a real number for all $k \in \mathbb{N}$.

-
 @Abel, thanks very much for your reply! I think it is easy to find some necessary conditions for $r_{1},\dots, r_{n}$ to be imaginary. For example, $a_{k}=0$ for all odd numbers k. However, it is hard to find a necessary and sufficient condition for $r_{1},\dots, r_{n}$ to be imaginary according the relations between the coefficients of the polynomial. – Shunyi Liu Feb 27 2011 at 13:24

The general study of connections between the coefficients of a polynomial, the locations of its roots, the roots of its derivative, et cetera, is called the Geometry of Zeros. There are books, I believe, with this title. Google turned up this survey.

As with most things, your exact question is likely not in print anywhere (I like Didier's answer), but there is an active and beautiful field with powerful techniques that are worth study.

-
 @Kevin O'Bryant, thanks for the interesting reference! – Shunyi Liu Feb 28 2011 at 9:57

There is a method due to Sturm which allows the determination of the number of roots of a polynomial in any real interval, in particular you may apply this to find the number of negative roots. The following Wikipedia article may serve as an initial reference: http://en.wikipedia.org/wiki/Sturm%27s_theorem

-
 @Michael Renardy, thanks for your kind reply. – Shunyi Liu Feb 28 2011 at 9:58