Proof involving the spectral radius and the Jordan canonical form
$begingroup$
Let $A$ be a square matrix. Show that if $$lim_{n to infty} A^{n} = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.
Hint: Use the Jordan canonical form.
I am 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 matrices jordan-normal-form spectral-radius
$endgroup$
add a comment |
$begingroup$
Let $A$ be a square matrix. Show that if $$lim_{n to infty} A^{n} = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.
Hint: Use the Jordan canonical form.
I am 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 matrices jordan-normal-form spectral-radius
$endgroup$
add a comment |
$begingroup$
Let $A$ be a square matrix. Show that if $$lim_{n to infty} A^{n} = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.
Hint: Use the Jordan canonical form.
I am 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 matrices jordan-normal-form spectral-radius
$endgroup$
Let $A$ be a square matrix. Show that if $$lim_{n to infty} A^{n} = 0$$ then $rho(A) < 1$, where $rho(A)$ denotes the spectral radius of $A$.
Hint: Use the Jordan canonical form.
I am 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 matrices jordan-normal-form spectral-radius
linear-algebra matrices jordan-normal-form spectral-radius
edited 25 mins ago
Rodrigo de Azevedo
13.2k41961
13.2k41961
asked 1 hour ago
mXdXmXdX
1068
1068
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2 Answers
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$begingroup$
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 $A^n to 0$.
$endgroup$
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$begingroup$
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.
$endgroup$
$begingroup$
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
1 hour 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.
1 hour 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
1 hour ago
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
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 $A^n to 0$.
$endgroup$
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 $A^n to 0$.
$endgroup$
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 $A^n to 0$.
$endgroup$
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 $A^n to 0$.
answered 1 hour ago
Robert IsraelRobert Israel
332k23221478
332k23221478
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$begingroup$
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.
$endgroup$
$begingroup$
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
1 hour 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.
1 hour 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
1 hour 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.
$endgroup$
$begingroup$
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
1 hour 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.
1 hour 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
1 hour 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.
$endgroup$
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.
answered 1 hour ago
N. S.N. S.
105k7115210
105k7115210
$begingroup$
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
1 hour 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.
1 hour 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
1 hour ago
add a comment |
$begingroup$
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
1 hour 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.
1 hour 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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour 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.
1 hour 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.
1 hour 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
1 hour 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
1 hour ago
add a comment |
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