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Proof involving the spectral radius and Jordan Canonical form

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Proof involving the spectral radius and Jordan Canonical form

1

$begingroup$

Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.

I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.

linear-algebra spectral-radius

share|cite|improve this question

asked 58 mins ago

mXdXmXdX

1018

$endgroup$

add a comment | 

1

$begingroup$

Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.

I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.

linear-algebra spectral-radius

share|cite|improve this question

asked 58 mins ago

mXdXmXdX

1018

$endgroup$

add a comment | 

$begingroup$

Let $A$ be a square matrix. Show that if $lim_{n to infty} A^{n} = 0$, then $rho(A) < 1$. Hint: Use the Jordan Canonical form. Here, $rho(A)$ denotes the spectral radius of $A$.

I'm self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem -- I don't know where to start. Any help appreciated.

linear-algebra spectral-radius

share|cite|improve this question

asked 58 mins ago

mXdXmXdX

1018

$endgroup$

share|cite|improve this question

asked 58 mins ago

mXdXmXdX

1018

share|cite|improve this question

asked 58 mins ago

mXdXmXdX

1018

asked 58 mins ago

mXdXmXdX

1018

asked 58 mins ago

mXdXmXdX

1018

2 Answers
2

active

oldest

votes

4

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

1

$begingroup$

Hint

$$A=PJP^{-1} \
J=begin{bmatrix}
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
end{bmatrix}$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=begin{bmatrix}
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
end{bmatrix}$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 43 mins ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  10 mins ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

4

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

add a comment | 

$begingroup$

You don't really need Jordan canonical form. If $rho(A) ge 1$, $A$ has an eigenvalue $lambda$ with $|lambda| ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = lambda^n v$, so $|A^n v| = |lambda|^n |v| ge |v|$ does not go to $0$ as $n to infty$, which is impossible if $A^n to 0$.

share|cite|improve this answer

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

$endgroup$

share|cite|improve this answer

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

answered 27 mins ago

Robert IsraelRobert Israel

332k23221478

1

$begingroup$

Hint

$$A=PJP^{-1} \
J=begin{bmatrix}
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
end{bmatrix}$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=begin{bmatrix}
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
end{bmatrix}$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 43 mins ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  10 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint

$$A=PJP^{-1} \
J=begin{bmatrix}
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
end{bmatrix}$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=begin{bmatrix}
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
end{bmatrix}$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 43 mins ago

N. S.N. S.

105k7115210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right?
  $endgroup$
  – mXdX
  21 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mXdX Well, that is the point. First $$lim_m J^m= lim_m P^{-1} A^m P =0$$ Now, since $lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $lambda_j^m to 0$. This implies that $|lambda_j |<1$
  $endgroup$
  – N. S.
  16 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues.
  $endgroup$
  – mXdX
  10 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint

$$A=PJP^{-1} \
J=begin{bmatrix}
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
end{bmatrix}$$
where each $*$ is either $0$ or $1$.

Prove by induction that
$$J^m=begin{bmatrix}
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
end{bmatrix}$$
where the $star$s represent numbers, that is $J^m$ is an upper triangular matrix
with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

share|cite|improve this answer

answered 43 mins ago

N. S.N. S.

105k7115210

$endgroup$

share|cite|improve this answer

answered 43 mins ago

N. S.N. S.

105k7115210

answered 43 mins ago

N. S.N. S.

105k7115210

answered 43 mins ago

N. S.N. S.

105k7115210