I am trying to prove an Exercise in Bhatia's Matrix Analysis, and I'm unsure how to approach the problem.

Let $f(z)=z^n+a_1z^{n-1}+\cdots+a_n=(z-\lambda_1)\cdots(z-\lambda_n)$ be a given monic polynomial. Let $\mu_1,\ldots,\mu_n$ be the numbers $|a_k|^{1/k}$ for $k=1,\ldots,n$, rearranged in decreasing order. Show that the roots satisfy $$|\lambda_i|\leq\mu_1+\mu_2$$ for all $i=1,\ldots,n$.

Any help is greatly appreciated.


I will sketch an answer and be happy to add details if needed:

1: if all $a_k$ are zero, roots are zero, problem clear, so assume at least one of them is non-zero; also we can assume $n \geq 2$ as the problem is clear by inspection

2: let $0 \leq a \leq b$ real numbers, and $n \geq 2$; we can prove an inequality: $c_n^n+c_{n-1}^{n-1}(a+b)+c_{n-2}^{n-2}(a+b)^2+..c_1(a+b)^{n-1} \leq (a+b)^n$, where $c_k$ are all $a$ except one that is $b$; the proof for example works by showing that LHS increases if the place of the one $b$ is lower (as index) and then considering the case $c_1=b$ which is fairly straightforward.

3: consider the Cauchy polynomial associated to $f$, namely $g(z) = f(z)=z^n- |a_1|z^{n-1}-|a_2|z^{n-2}-\cdots-|a_n|$

It is easy to see that (assuming not all $a$ zero) $g$ has an unique positive roof $c(f)$ (divide by $z^n$ and show the coefficient part is decreasing in $r>0$) and then using the triangle inequality $|\lambda_i|\leq c(f)$ when $\lambda_i$ is any of the roots of $f$; in particular if $R>0$ satisfies $g(R) \geq 0$, then $|\lambda_i|\leq R$ for any root of $f$

4: using the given coefficient inequality with $a=\mu_2, b=\mu_1$, and the definition of $\mu_1, \mu_2$ it follows that $|a_k| \leq a^k$ for all but one $k$, where $|a_k| =b^k$

5: using the inequality at point 2: it then follows $g(a+b) \geq 0$, so using point 3: we are done

  • $\begingroup$ Hi @Conrad! First up, thank you for the response. I am currently studying your sketch, and I seem to be stuck on step 2 because I can't prove the inequality. Is there an easier way to see it? Again, thank you! $\endgroup$ – chhro Mar 10 at 15:57
  • $\begingroup$ Just checking in to say that I now fully understand all steps except 2 which I will think about more. Highly appreciate the help! $\endgroup$ – chhro Mar 10 at 17:22
  • $\begingroup$ Show 2 with $b=c_1$ and all others $a$ - this is straightforward by passing the $b$ term and subtracting and the induction or just step by step subtracting from the highest term and then show that that's best you can do as position b goes $\endgroup$ – Conrad Mar 10 at 17:44
  • $\begingroup$ Thanks for the clarification @Conrad! That settles it. $\endgroup$ – chhro Mar 11 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.