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2

$begingroup$

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
$$|lambda_i|leqmu_1+mu_2$$
for all $i=1,ldots,n$.

Any help is greatly appreciated.

polynomials

share|cite|improve this question

asked Mar 10 at 3:37

chhrochhro

1,310311

$endgroup$

add a comment | 

2

$begingroup$

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
$$|lambda_i|leqmu_1+mu_2$$
for all $i=1,ldots,n$.

Any help is greatly appreciated.

polynomials

share|cite|improve this question

asked Mar 10 at 3:37

chhrochhro

1,310311

$endgroup$

add a comment | 

$begingroup$

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
$$|lambda_i|leqmu_1+mu_2$$
for all $i=1,ldots,n$.

Any help is greatly appreciated.

polynomials

share|cite|improve this question

asked Mar 10 at 3:37

chhrochhro

1,310311

$endgroup$

share|cite|improve this question

asked Mar 10 at 3:37

chhrochhro

1,310311

share|cite|improve this question

asked Mar 10 at 3:37

chhrochhro

1,310311

asked Mar 10 at 3:37

chhrochhro

1,310311

asked Mar 10 at 3:37

chhrochhro

1,310311

1 Answer
1

active

oldest

votes

1

$begingroup$

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

share|cite|improve this answer

edited Mar 10 at 12:33

answered Mar 10 at 11:59

ConradConrad

88835

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

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

share|cite|improve this answer

edited Mar 10 at 12:33

answered Mar 10 at 11:59

ConradConrad

88835

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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

share|cite|improve this answer

edited Mar 10 at 12:33

answered Mar 10 at 11:59

ConradConrad

88835

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

edited Mar 10 at 12:33

answered Mar 10 at 11:59

ConradConrad

88835

$endgroup$

share|cite|improve this answer

edited Mar 10 at 12:33

answered Mar 10 at 11:59

ConradConrad

88835

answered Mar 10 at 11:59

ConradConrad

88835

answered Mar 10 at 11:59

ConradConrad

88835