Limit of powers of matrix if spectral radius < 1
$begingroup$
I want to show $lim_{krightarrow infty}A^k=0$ iff $sigma(A)<1$.
I have $0leq ||A^k||leq ||A||^k=||Ax||^k=||lambda_{max}x||^k=|lambda_{max}|^k<1$ if we choose $x$ s.t. $||x||<1$ and $Ax=lambda_{max}x$ for the backwards direction and $lim_{krightarrowinfty}||A^kx||=|lambda_{max}|^k ||x||= 0Rightarrow sigma(A)<1$ if we choose $x$ in the same way for the forwards direction. Is this sufficient?
linear-algebra numerical-linear-algebra
asked Jan 24 at 3:24
HBHSUHBHSU
302111
$endgroup$
add a comment |
$begingroup$
I want to show $lim_{krightarrow infty}A^k=0$ iff $sigma(A)<1$.
I have $0leq ||A^k||leq ||A||^k=||Ax||^k=||lambda_{max}x||^k=|lambda_{max}|^k<1$ if we choose $x$ s.t. $||x||<1$ and $Ax=lambda_{max}x$ for the backwards direction and $lim_{krightarrowinfty}||A^kx||=|lambda_{max}|^k ||x||= 0Rightarrow sigma(A)<1$ if we choose $x$ in the same way for the forwards direction. Is this sufficient?
linear-algebra numerical-linear-algebra
asked Jan 24 at 3:24
HBHSUHBHSU
302111
$endgroup$
$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35
add a comment |
$begingroup$
I want to show $lim_{krightarrow infty}A^k=0$ iff $sigma(A)<1$.
I have $0leq ||A^k||leq ||A||^k=||Ax||^k=||lambda_{max}x||^k=|lambda_{max}|^k<1$ if we choose $x$ s.t. $||x||<1$ and $Ax=lambda_{max}x$ for the backwards direction and $lim_{krightarrowinfty}||A^kx||=|lambda_{max}|^k ||x||= 0Rightarrow sigma(A)<1$ if we choose $x$ in the same way for the forwards direction. Is this sufficient?
linear-algebra numerical-linear-algebra
asked Jan 24 at 3:24
HBHSUHBHSU
302111
$endgroup$
I want to show $lim_{krightarrow infty}A^k=0$ iff $sigma(A)<1$.
I have $0leq ||A^k||leq ||A||^k=||Ax||^k=||lambda_{max}x||^k=|lambda_{max}|^k<1$ if we choose $x$ s.t. $||x||<1$ and $Ax=lambda_{max}x$ for the backwards direction and $lim_{krightarrowinfty}||A^kx||=|lambda_{max}|^k ||x||= 0Rightarrow sigma(A)<1$ if we choose $x$ in the same way for the forwards direction. Is this sufficient?
linear-algebra numerical-linear-algebra
linear-algebra numerical-linear-algebra
asked Jan 24 at 3:24
HBHSUHBHSU
302111
asked Jan 24 at 3:24
HBHSUHBHSU
302111
asked Jan 24 at 3:24
HBHSUHBHSU
302111
asked Jan 24 at 3:24
HBHSUHBHSU
302111
asked Jan 24 at 3:24
HBHSUHBHSU
302111
302111
$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35
add a comment |
$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35
$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35
$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $|A| = |Ax| = |lambda_{max}| |x|$. For take
begin{equation}
A = begin{pmatrix}
r & 1 \
0 & r end{pmatrix}.tag{1}
end{equation}
Then the only eigenvalue of $A$ is $r$ but $|A| > r$ since $|Ae_2| = sqrt{1+r^2}$, where $e_2 = begin{pmatrix} 0 & 1 end{pmatrix}^T$. (Here, $|cdot|$ is the vector 2-norm and matrix 2-norm $|A| = max_{|x|le 1} |Ax|$.) However, by direct computations of powers of $A$, we observe
$$
A^k = begin{pmatrix}
r^k & kr^{k-1} \
0 & r^k
end{pmatrix} to 0 textrm{ as } ktoinfty textrm{ provided } |r| < 1.
$$
One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0le|A^k|= |VJ^k V^{-1}| le |V| |J|^k |V^{-1}|$. From here, the analysis reduces to showing that if $|lambda| < 1$, then a Jordan block $J$ with eigenvalue $lambda$ satisfies $J^k to 0$ as $ktoinfty$.
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
$endgroup$
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
add a comment |
$begingroup$
This is wrong. An $x$ that maximizes $|Ax|/|x|$ is not necessarily an eigenvector. For example, try $$ A= pmatrix{0 & 2cr 0 & 0cr}$$ where the only eigenvalue is $0$, but $|A| = 2$. Here $x = pmatrix{0cr 1}$ satisfies $|A x| = 2$ and $|x| = 1$. By looking at $|A|$ alone you wouldn't know that $lim_{nto infty} A^n = 0$ (in fact in this case $A^2 = 0$).
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
$endgroup$
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
$begingroup$
If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $|A| = |Ax| = |lambda_{max}| |x|$. For take
begin{equation}
A = begin{pmatrix}
r & 1 \
0 & r end{pmatrix}.tag{1}
end{equation}
Then the only eigenvalue of $A$ is $r$ but $|A| > r$ since $|Ae_2| = sqrt{1+r^2}$, where $e_2 = begin{pmatrix} 0 & 1 end{pmatrix}^T$. (Here, $|cdot|$ is the vector 2-norm and matrix 2-norm $|A| = max_{|x|le 1} |Ax|$.) However, by direct computations of powers of $A$, we observe
$$
A^k = begin{pmatrix}
r^k & kr^{k-1} \
0 & r^k
end{pmatrix} to 0 textrm{ as } ktoinfty textrm{ provided } |r| < 1.
$$
One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0le|A^k|= |VJ^k V^{-1}| le |V| |J|^k |V^{-1}|$. From here, the analysis reduces to showing that if $|lambda| < 1$, then a Jordan block $J$ with eigenvalue $lambda$ satisfies $J^k to 0$ as $ktoinfty$.
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
$endgroup$
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
add a comment |
$begingroup$
If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $|A| = |Ax| = |lambda_{max}| |x|$. For take
begin{equation}
A = begin{pmatrix}
r & 1 \
0 & r end{pmatrix}.tag{1}
end{equation}
Then the only eigenvalue of $A$ is $r$ but $|A| > r$ since $|Ae_2| = sqrt{1+r^2}$, where $e_2 = begin{pmatrix} 0 & 1 end{pmatrix}^T$. (Here, $|cdot|$ is the vector 2-norm and matrix 2-norm $|A| = max_{|x|le 1} |Ax|$.) However, by direct computations of powers of $A$, we observe
$$
A^k = begin{pmatrix}
r^k & kr^{k-1} \
0 & r^k
end{pmatrix} to 0 textrm{ as } ktoinfty textrm{ provided } |r| < 1.
$$
One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0le|A^k|= |VJ^k V^{-1}| le |V| |J|^k |V^{-1}|$. From here, the analysis reduces to showing that if $|lambda| < 1$, then a Jordan block $J$ with eigenvalue $lambda$ satisfies $J^k to 0$ as $ktoinfty$.
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
$endgroup$
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
add a comment |
$begingroup$
If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $|A| = |Ax| = |lambda_{max}| |x|$. For take
begin{equation}
A = begin{pmatrix}
r & 1 \
0 & r end{pmatrix}.tag{1}
end{equation}
Then the only eigenvalue of $A$ is $r$ but $|A| > r$ since $|Ae_2| = sqrt{1+r^2}$, where $e_2 = begin{pmatrix} 0 & 1 end{pmatrix}^T$. (Here, $|cdot|$ is the vector 2-norm and matrix 2-norm $|A| = max_{|x|le 1} |Ax|$.) However, by direct computations of powers of $A$, we observe
$$
A^k = begin{pmatrix}
r^k & kr^{k-1} \
0 & r^k
end{pmatrix} to 0 textrm{ as } ktoinfty textrm{ provided } |r| < 1.
$$
One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0le|A^k|= |VJ^k V^{-1}| le |V| |J|^k |V^{-1}|$. From here, the analysis reduces to showing that if $|lambda| < 1$, then a Jordan block $J$ with eigenvalue $lambda$ satisfies $J^k to 0$ as $ktoinfty$.
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
$endgroup$
If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $|A| = |Ax| = |lambda_{max}| |x|$. For take
begin{equation}
A = begin{pmatrix}
r & 1 \
0 & r end{pmatrix}.tag{1}
end{equation}
Then the only eigenvalue of $A$ is $r$ but $|A| > r$ since $|Ae_2| = sqrt{1+r^2}$, where $e_2 = begin{pmatrix} 0 & 1 end{pmatrix}^T$. (Here, $|cdot|$ is the vector 2-norm and matrix 2-norm $|A| = max_{|x|le 1} |Ax|$.) However, by direct computations of powers of $A$, we observe
$$
A^k = begin{pmatrix}
r^k & kr^{k-1} \
0 & r^k
end{pmatrix} to 0 textrm{ as } ktoinfty textrm{ provided } |r| < 1.
$$
One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0le|A^k|= |VJ^k V^{-1}| le |V| |J|^k |V^{-1}|$. From here, the analysis reduces to showing that if $|lambda| < 1$, then a Jordan block $J$ with eigenvalue $lambda$ satisfies $J^k to 0$ as $ktoinfty$.
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
answered Jan 24 at 4:07
eepperly16eepperly16
3,02611125
3,02611125
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
add a comment |
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
$begingroup$
Is there any way to prove this without using a Jordan normal form? I haven't learned about it yet.
$endgroup$
– HBHSU
Jan 24 at 4:19
add a comment |
$begingroup$
This is wrong. An $x$ that maximizes $|Ax|/|x|$ is not necessarily an eigenvector. For example, try $$ A= pmatrix{0 & 2cr 0 & 0cr}$$ where the only eigenvalue is $0$, but $|A| = 2$. Here $x = pmatrix{0cr 1}$ satisfies $|A x| = 2$ and $|x| = 1$. By looking at $|A|$ alone you wouldn't know that $lim_{nto infty} A^n = 0$ (in fact in this case $A^2 = 0$).
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
This is wrong. An $x$ that maximizes $|Ax|/|x|$ is not necessarily an eigenvector. For example, try $$ A= pmatrix{0 & 2cr 0 & 0cr}$$ where the only eigenvalue is $0$, but $|A| = 2$. Here $x = pmatrix{0cr 1}$ satisfies $|A x| = 2$ and $|x| = 1$. By looking at $|A|$ alone you wouldn't know that $lim_{nto infty} A^n = 0$ (in fact in this case $A^2 = 0$).
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
$endgroup$
add a comment |
$begingroup$
This is wrong. An $x$ that maximizes $|Ax|/|x|$ is not necessarily an eigenvector. For example, try $$ A= pmatrix{0 & 2cr 0 & 0cr}$$ where the only eigenvalue is $0$, but $|A| = 2$. Here $x = pmatrix{0cr 1}$ satisfies $|A x| = 2$ and $|x| = 1$. By looking at $|A|$ alone you wouldn't know that $lim_{nto infty} A^n = 0$ (in fact in this case $A^2 = 0$).
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
$endgroup$
This is wrong. An $x$ that maximizes $|Ax|/|x|$ is not necessarily an eigenvector. For example, try $$ A= pmatrix{0 & 2cr 0 & 0cr}$$ where the only eigenvalue is $0$, but $|A| = 2$. Here $x = pmatrix{0cr 1}$ satisfies $|A x| = 2$ and $|x| = 1$. By looking at $|A|$ alone you wouldn't know that $lim_{nto infty} A^n = 0$ (in fact in this case $A^2 = 0$).
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
answered Jan 24 at 4:08
Robert IsraelRobert Israel
326k23215469
326k23215469
add a comment |
add a comment |
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$begingroup$
If you learned Gelfand's formula, you can prove $lim_{krightarrowinfty}|A|^k =0$.
$endgroup$
– i707107
Jan 26 at 2:35