$A^tto 0$ when its row sum is strictly less than one?
$begingroup$
$A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.
Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.
matrices limits eigenvalues-eigenvectors matrix-norms
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
$endgroup$
add a comment |
$begingroup$
$A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.
Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.
matrices limits eigenvalues-eigenvectors matrix-norms
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
$endgroup$
add a comment |
$begingroup$
$A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.
Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.
matrices limits eigenvalues-eigenvectors matrix-norms
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
$endgroup$
$A_{ntimes n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^tto 0,text{ i.e } a^t_{ij}to 0forall i,jtext{ as } ttoinfty$ given that $0<a_{ij}<1$.
Well, suppose $0<lambda<1$ be the largest eigenvalue of $A$ then $A^tx=lambda^txto 0text{ as } ttoinftyRightarrow A^tto0because xne0$, is my argument works? Well $lambda$ may be negative? Thanks.
matrices limits eigenvalues-eigenvectors matrix-norms
matrices limits eigenvalues-eigenvectors matrix-norms
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
edited Dec 6 '18 at 19:03
Martin Sleziak
44.7k9117272
44.7k9117272
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
asked Dec 6 '18 at 16:17
MarkovMarkov
17.2k957178
17.2k957178
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.
Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.
This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
$endgroup$
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.
Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.
This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
$endgroup$
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
add a comment |
$begingroup$
The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.
Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.
This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
$endgroup$
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
add a comment |
$begingroup$
The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.
Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.
This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
$endgroup$
The maximum absolute row sum of a matrix is a matrix norm, the $infty$-norm.
Therefore, if $|A|_{infty}<1$, then $A^n to 0$ since $|A^n|_{infty} le |A|_{infty}^nto 0$.
This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
edited Dec 6 '18 at 16:39
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
answered Dec 6 '18 at 16:25
lhflhf
164k10170395
164k10170395
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
add a comment |
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact?
$endgroup$
– Markov
Dec 6 '18 at 16:31
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
$begingroup$
@Wow, I think the extra information is redundant.
$endgroup$
– lhf
Dec 6 '18 at 16:37
add a comment |
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