Yet another solution of an old problem

Consider a problem we already solved (in a weaker setting) in a previous post.
Let $x_1,\ldots, x_n$ be complex numbers, such that $$x_1^k+\cdots +x_n^k=0$$ for all $k\in\{1,\ldots,n\}$. Prove that all of the numbers are zeros.

Here we would investigate another approach. The idea is simple. First we construct a matrix $A$ whose characteristic polynomial has as roots $x_1,\ldots, x_n$. Then we use the fact that $\text{Tr}(A^k)=x_1^k+\cdots+x_n^k$ to successively eliminate all the coefficients, hence showing that $x_1,\ldots,x_n$ are roots of $x^n=0$.
Assume $x^n-a_1 x^{n-1}-\cdots-a_{n-1}x-a_n$ is a polynomial whose roots are precisely $x_1,\ldots,x_n$. Then the matrix
$$  A= \begin{pmatrix}
        a_1       & a_2 & a_3 & \dots & a_{n-1} & a_n \\
        1       & 0 & 0 & \dots  &0 & 0 \\
         0     & 1 & 0  & \dots &0 & 0 \\
         \vdots          &\vdots & \vdots &\vdots   &\ddots   &\vdots\\
        0      & 0 & 0 & \dots &1  & 0
    \end{pmatrix}
 $$ has this polynomial as characteristic.
Since $\text{Tr}(A)=0$ we have $a_1=0$. Now assume we have deduced that $a_1=\cdots=a_k=0$ for some $k\le n-1$.  It is easy to see by induction that $$A^{k+1}=\begin{pmatrix}
        a_{k+1}      & a_{k+2}  & a_{k+3} & \dots & \dots & a_{n}     & 0 & \dots & 0\\
        0    & a_{k+1} & a_{k+2}        & \dots  & \dots & a_{n-1}              & a_n &\dots       & 0\\   
    \vdots   &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots\\
       0       & 0 & 0        & \dots  &a_{k+1}              & \dots     & \dots & \dots & a_n  \\
1      & 0 & 0        & \dots  &\dots            &\dots &\dots &\dots &0  \\
0       & 1 & 0        & \dots  &\dots &\dots &\dots &\dots  &0\\

    \vdots   &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots&\vdots&\vdots\\

          0      & 0 & \dots & 1 & 0 & \dots & \dots &\dots & 0
    \end{pmatrix}$$
 (the matrix may not be perfectly written, but you got the feeling)
From here is evident that $\text {Tr}(A^{k+1})=(k+1) a_{k+1}$. Hence $(k+1)a_{k+1}=0$ thus $a_{k+1}$ is zero. Thus we obtain that all the coefficients are zero. Hence the roots.

Comment. Does this solution generalizes the problem into a more abstract setting?

Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$ Proof. The function $$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$ is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that $$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$ If we make change of variables in the latter integral $t\to 1-t$ we obtain $$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$ On the other hand, using integration by parts, we obtain $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$ Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$ Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$ On the ot...
Read more

A problem proposed by prof. Babev

Consider the sequence $$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$ It could be proven that $$\lim_{n\to \infty}n a_n=\frac{1}{2}$$  Now we are going to prove that $$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$ and show how to derive all such limits. $\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as $$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$ $$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$ Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to $$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$ Since for $s\in[0,1]$, $\displaystyle \...
Read more

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_...
Read more