Mixing
Spectral Radius proof
-1
$begingroup$
Let A be a symmetric matrix with positive Eigenvalues.Proof that ρ(Α^4)=(ρ(Α))^4 where ρ is the spectral radius.
How should i go about that?
I am kinda confused on how should i use the information i am given
matrices numerical-methods eigenvalues-eigenvectors
asked Jan 27 at 11:04
EdwardEdward
1
$endgroup$
add a comment |
-1
$begingroup$
Let A be a symmetric matrix with positive Eigenvalues.Proof that ρ(Α^4)=(ρ(Α))^4 where ρ is the spectral radius.
How should i go about that?
I am kinda confused on how should i use the information i am given
matrices numerical-methods eigenvalues-eigenvectors
asked Jan 27 at 11:04
EdwardEdward
1
$endgroup$
$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57
add a comment |
-1
-1
-1
$begingroup$
Let A be a symmetric matrix with positive Eigenvalues.Proof that ρ(Α^4)=(ρ(Α))^4 where ρ is the spectral radius.
How should i go about that?
I am kinda confused on how should i use the information i am given
matrices numerical-methods eigenvalues-eigenvectors
asked Jan 27 at 11:04
EdwardEdward
1
$endgroup$
Let A be a symmetric matrix with positive Eigenvalues.Proof that ρ(Α^4)=(ρ(Α))^4 where ρ is the spectral radius.
How should i go about that?
I am kinda confused on how should i use the information i am given
matrices numerical-methods eigenvalues-eigenvectors
matrices numerical-methods eigenvalues-eigenvectors
asked Jan 27 at 11:04
EdwardEdward
1
asked Jan 27 at 11:04
EdwardEdward
1
asked Jan 27 at 11:04
EdwardEdward
1
asked Jan 27 at 11:04
EdwardEdward
1
asked Jan 27 at 11:04
EdwardEdward
1
1
$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57
add a comment |
$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57
$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
The spectral radius formula is $rho (A)=lim |A^{n}|^{1/n}$. From this $rho (A^{4})=lim |A^{4n}|^{1/n}=(lim |A^{4n}|^{1/4n})^{4}$ which is $rho (A) ^{4}$. See Gelfand Formula in https://en.wikipedia.org/wiki/Spectral_radius
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
add a comment |
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1 Answer
1
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1 Answer
1
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oldest
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1
$begingroup$
The spectral radius formula is $rho (A)=lim |A^{n}|^{1/n}$. From this $rho (A^{4})=lim |A^{4n}|^{1/n}=(lim |A^{4n}|^{1/4n})^{4}$ which is $rho (A) ^{4}$. See Gelfand Formula in https://en.wikipedia.org/wiki/Spectral_radius
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
add a comment |
1
$begingroup$
The spectral radius formula is $rho (A)=lim |A^{n}|^{1/n}$. From this $rho (A^{4})=lim |A^{4n}|^{1/n}=(lim |A^{4n}|^{1/4n})^{4}$ which is $rho (A) ^{4}$. See Gelfand Formula in https://en.wikipedia.org/wiki/Spectral_radius
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
add a comment |
1
1
1
$begingroup$
The spectral radius formula is $rho (A)=lim |A^{n}|^{1/n}$. From this $rho (A^{4})=lim |A^{4n}|^{1/n}=(lim |A^{4n}|^{1/4n})^{4}$ which is $rho (A) ^{4}$. See Gelfand Formula in https://en.wikipedia.org/wiki/Spectral_radius
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
$endgroup$
The spectral radius formula is $rho (A)=lim |A^{n}|^{1/n}$. From this $rho (A^{4})=lim |A^{4n}|^{1/n}=(lim |A^{4n}|^{1/4n})^{4}$ which is $rho (A) ^{4}$. See Gelfand Formula in https://en.wikipedia.org/wiki/Spectral_radius
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
answered Jan 27 at 11:49
Kavi Rama MurthyKavi Rama Murthy
69.5k53170
69.5k53170
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
add a comment |
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
$begingroup$
I wasn't aware of this formula??We never talked about it in Uni.Thank you this worked.
$endgroup$
– Edward
Jan 27 at 11:56
add a comment |
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$begingroup$
Are you aware (or, can you prove) that if $lambda$ is an eigenvalue of a matrix $A$, then $lambda^k$ is an eigenvalue of $A^k$ for all positive integers $k$?
$endgroup$
– Gerry Myerson
Jan 27 at 11:34
$begingroup$
How would I go about that?And how could i use it?
$endgroup$
– Edward
Jan 27 at 11:39
$begingroup$
Start with the definition of eigenvalue.
$endgroup$
– Gerry Myerson
Jan 27 at 11:43
$begingroup$
@GerryMyerson It is easy to verify that $lambda^{k}$ is an eigen value of $A^{k}$ if $lambda $ is an eigen value of $A$. But this is not enough to find the spectral radius.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 11:52
$begingroup$
The spectral radius is the largest absolute value of an eigenvalue, @Kavi. So I think what I've hinted at is all that's needed.
$endgroup$
– Gerry Myerson
Jan 27 at 11:57