Proof involving the spectral radius and Jordan Canonical form Announcing the arrival of Valued...
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Proof involving the spectral radius and Jordan Canonical form
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Spectral radius of the Volterra operatorExample that the Jordan canonical form is not “robust.”The unit vector in the direction of uWhat is the purpose of Jordan Canonical Form?Confusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixSpectral radius of perturbed bipartite graphsA proof involving invertible $ntimes n$ matricesProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Computing Canonical Jordan Form over a field $mathbb{Q}$
$begingroup$
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
asked 1 hour ago
mXdXmXdX
1068
$endgroup$
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
asked 1 hour ago
mXdXmXdX
1068
$endgroup$
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
asked 1 hour ago
mXdXmXdX
1068
$endgroup$
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
linear-algebra spectral-radius
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
asked 1 hour ago
mXdXmXdX
1068
1068
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $A^n to 0$.
answered 47 mins ago
Robert IsraelRobert Israel
332k23221478
$endgroup$
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.
answered 1 hour 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
41 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.
36 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
30 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
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 $A^n to 0$.
answered 47 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 $A^n to 0$.
answered 47 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 $A^n to 0$.
answered 47 mins ago
Robert IsraelRobert Israel
332k23221478
$endgroup$
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 47 mins ago
Robert IsraelRobert Israel
332k23221478
answered 47 mins ago
Robert IsraelRobert Israel
332k23221478
answered 47 mins ago
Robert IsraelRobert Israel
332k23221478
answered 47 mins ago
Robert IsraelRobert Israel
332k23221478
332k23221478
add a comment |
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.
answered 1 hour 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
41 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.
36 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
30 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.
answered 1 hour 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
41 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.
36 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
30 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.
answered 1 hour ago
N. S.N. S.
105k7115210
$endgroup$
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
answered 1 hour ago
N. S.N. S.
105k7115210
answered 1 hour ago
N. S.N. S.
105k7115210
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
41 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.
36 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
30 mins 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
41 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.
36 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
30 mins 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
41 mins 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
41 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.
36 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.
36 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
30 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
30 mins ago
add a comment |
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