Proof involving the spectral radius and the 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 operatorSpectral radius, second induced normContinuity of the spectral radiusSpectral radius and matrix norm inequality as its consequenceConfusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Spectral radius of a matrixLower bound spectral radius of matrixFinding the Jordan Form of a matrix…
Output the ŋarâþ crîþ alphabet song without using (m)any letters
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Proof involving the spectral radius and the 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 operatorSpectral radius, second induced normContinuity of the spectral radiusSpectral radius and matrix norm inequality as its consequenceConfusion between spectral radius of matrix and spectral radius of the operatorComputing the Jordan Form of a MatrixProof of Gelfand's formula without using $rho(A) < 1$ iff $lim A^n = 0$Spectral radius of a matrixLower bound spectral radius of matrixFinding the Jordan Form of a matrix…
$begingroup$
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
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
$endgroup$
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
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
$endgroup$
$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago
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
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
$endgroup$
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 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
edited 1 hour ago
Rodrigo de Azevedo
13.2k41961
13.2k41961
asked 2 hours ago
mXdXmXdX
1118
asked 2 hours ago
mXdXmXdX
1118
asked 2 hours ago
mXdXmXdX
1118
1118
$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago
add a comment |
$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago
$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago
$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago
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 1 hour ago
Robert IsraelRobert Israel
332k23221478
$endgroup$
add a comment |
$begingroup$
Hint
$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.
Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
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 2 hours ago
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
So, $A^m = PJ^mP^-1$. 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
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Robert IsraelRobert Israel
332k23221478
answered 1 hour ago
Robert IsraelRobert Israel
332k23221478
answered 1 hour ago
Robert IsraelRobert Israel
332k23221478
answered 1 hour ago
Robert IsraelRobert Israel
332k23221478
332k23221478
add a comment |
add a comment |
$begingroup$
Hint
$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.
Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
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 2 hours ago
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
So, $A^m = PJ^mP^-1$. 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
$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.
Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
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 2 hours ago
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
So, $A^m = PJ^mP^-1$. 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
$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.
Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
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 2 hours ago
N. S.N. S.
105k7115210
$endgroup$
Hint
$$A=PJP^-1 \
J=beginbmatrix
lambda_1 & * & 0 & 0 & 0 & ... & 0 \
0& lambda_2 & * & 0 & 0 & ... & 0 \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n \
endbmatrix$$
where each $*$ is either $0$ or $1$.
Prove by induction that
$$J^m=beginbmatrix
lambda_1^m & star & star & star & star & ... & star \
0& lambda_2^m & star & star & star & ... & star \
...&...&...&...&....&....&....\
0 & 0 & 0 & 0&0&...&lambda_n^m \
endbmatrix$$
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 2 hours ago
N. S.N. S.
105k7115210
answered 2 hours ago
N. S.N. S.
105k7115210
answered 2 hours ago
N. S.N. S.
105k7115210
answered 2 hours ago
N. S.N. S.
105k7115210
105k7115210
$begingroup$
So, $A^m = PJ^mP^-1$. 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, $A^m = PJ^mP^-1$. 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, $A^m = PJ^mP^-1$. 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, $A^m = PJ^mP^-1$. 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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$begingroup$
This follows from $A^n v = lambda^n v$. The other direction is straighforward using the Jordan form.
$endgroup$
– copper.hat
21 mins ago