Existence of $lim_{n rightarrow infty}A^n$
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I was studying Markov chain and in it, it is useful to have a transition matrix $M$ that $lim_{n rightarrow infty}M^n$ exist.
So I thought about the existence of $lim_{n rightarrow infty}A^n$ in general($A$ is a square matrix). and it was easy to prove:
for every diagonalizable matrix $A= QDQ^{-1}$, the $lim_{n rightarrow infty}A^n$ exists if and only if elements of $D$ be in the range $(-1,1]$.
but what about not diagonalizable matrices? what is the condition for them?
EXAMPLES:
$E_1$ is not diagonalizable and $lim_{n rightarrow infty}E_1^n$ exists, $E_2$ is not diagonalizable and $lim_{n rightarrow infty}E_2^n$ does not exists.
$$ E_1 = begin{bmatrix}0.1 & 0.1 \0 & 0.1 end{bmatrix}, E_2 = begin{bmatrix}1 & 1 \0 & 1 end{bmatrix}$$
linear-algebra matrices limits
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
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add a comment |
$begingroup$
I was studying Markov chain and in it, it is useful to have a transition matrix $M$ that $lim_{n rightarrow infty}M^n$ exist.
So I thought about the existence of $lim_{n rightarrow infty}A^n$ in general($A$ is a square matrix). and it was easy to prove:
for every diagonalizable matrix $A= QDQ^{-1}$, the $lim_{n rightarrow infty}A^n$ exists if and only if elements of $D$ be in the range $(-1,1]$.
but what about not diagonalizable matrices? what is the condition for them?
EXAMPLES:
$E_1$ is not diagonalizable and $lim_{n rightarrow infty}E_1^n$ exists, $E_2$ is not diagonalizable and $lim_{n rightarrow infty}E_2^n$ does not exists.
$$ E_1 = begin{bmatrix}0.1 & 0.1 \0 & 0.1 end{bmatrix}, E_2 = begin{bmatrix}1 & 1 \0 & 1 end{bmatrix}$$
linear-algebra matrices limits
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
$endgroup$
add a comment |
$begingroup$
I was studying Markov chain and in it, it is useful to have a transition matrix $M$ that $lim_{n rightarrow infty}M^n$ exist.
So I thought about the existence of $lim_{n rightarrow infty}A^n$ in general($A$ is a square matrix). and it was easy to prove:
for every diagonalizable matrix $A= QDQ^{-1}$, the $lim_{n rightarrow infty}A^n$ exists if and only if elements of $D$ be in the range $(-1,1]$.
but what about not diagonalizable matrices? what is the condition for them?
EXAMPLES:
$E_1$ is not diagonalizable and $lim_{n rightarrow infty}E_1^n$ exists, $E_2$ is not diagonalizable and $lim_{n rightarrow infty}E_2^n$ does not exists.
$$ E_1 = begin{bmatrix}0.1 & 0.1 \0 & 0.1 end{bmatrix}, E_2 = begin{bmatrix}1 & 1 \0 & 1 end{bmatrix}$$
linear-algebra matrices limits
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
$endgroup$
I was studying Markov chain and in it, it is useful to have a transition matrix $M$ that $lim_{n rightarrow infty}M^n$ exist.
So I thought about the existence of $lim_{n rightarrow infty}A^n$ in general($A$ is a square matrix). and it was easy to prove:
for every diagonalizable matrix $A= QDQ^{-1}$, the $lim_{n rightarrow infty}A^n$ exists if and only if elements of $D$ be in the range $(-1,1]$.
but what about not diagonalizable matrices? what is the condition for them?
EXAMPLES:
$E_1$ is not diagonalizable and $lim_{n rightarrow infty}E_1^n$ exists, $E_2$ is not diagonalizable and $lim_{n rightarrow infty}E_2^n$ does not exists.
$$ E_1 = begin{bmatrix}0.1 & 0.1 \0 & 0.1 end{bmatrix}, E_2 = begin{bmatrix}1 & 1 \0 & 1 end{bmatrix}$$
linear-algebra matrices limits
linear-algebra matrices limits
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
asked Jan 12 at 6:01
Peyman mohseni kiasariPeyman mohseni kiasari
1209
1209
add a comment |
add a comment |
1 Answer
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You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $1$. Additionally, if $1$ is an eigenvalue, it cannot be defective (defective eigenvalues of $1$ lead to polynomial growth).
answered Jan 12 at 6:03
IanIan
68.7k25390
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $1$. Additionally, if $1$ is an eigenvalue, it cannot be defective (defective eigenvalues of $1$ lead to polynomial growth).
answered Jan 12 at 6:03
IanIan
68.7k25390
$endgroup$
add a comment |
$begingroup$
You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $1$. Additionally, if $1$ is an eigenvalue, it cannot be defective (defective eigenvalues of $1$ lead to polynomial growth).
answered Jan 12 at 6:03
IanIan
68.7k25390
$endgroup$
add a comment |
$begingroup$
You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $1$. Additionally, if $1$ is an eigenvalue, it cannot be defective (defective eigenvalues of $1$ lead to polynomial growth).
answered Jan 12 at 6:03
IanIan
68.7k25390
$endgroup$
You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $1$. Additionally, if $1$ is an eigenvalue, it cannot be defective (defective eigenvalues of $1$ lead to polynomial growth).
answered Jan 12 at 6:03
IanIan
68.7k25390
answered Jan 12 at 6:03
IanIan
68.7k25390
answered Jan 12 at 6:03
IanIan
68.7k25390
answered Jan 12 at 6:03
IanIan
68.7k25390
68.7k25390
add a comment |
add a comment |
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