Eigenvalues & Eigenvectors of the Backward Shift Operator
So I'm trying to find the eigenvalues and eigenvectors of the backwards shift operator $T:F^infty to F^infty$ defined as $Tv=T(z_1,z_2,....)=(z_2,z_3,....)$.
Arriving at the system of equations from $Tv = lambda v$:
$$z_2 = lambda z_1$$
$$z_3 = lambda z_2$$
$$vdots$$
$$z_{n+1} = lambda z_n$$
$$vdots$$
(1) If we fix $z_1$ we arrive at the set of eigenvectors,
$$ { (z_1,lambda z_1, lambda^2 z_1, ....) : lambda epsilon mathbb{F} }$$
(2) If $lambda = 1$, then our eigenvectors are then
$$ { (z,z,...) : z epsilon mathbb{F} }$$
(3) If $lambda = 0$, then clearly $(z,0,0,...)$ describes the eigenvectors.
If $lambda ne 0$, then we have
$$z_1 = frac{z_2}{lambda}$$
$$z_2 = frac{z_3}{lambda}$$
$$vdots$$
$$z_{n} = frac{z_{n+1}}{lambda}$$
$$vdots$$
But this system of equations also implies that the 1st element $z_1$ is a function of $z_2$, which is a function of $z_3$, and so on. So they are all "chained together" to infinity. That is to say,
$$ z_1 = frac{1}{lambda} frac{z_3}{lambda} $$
$$ = frac{1}{lambda^2} frac{z_4}{lambda} $$
$$ vdots $$
$$ = lim_{n to infty} frac{z_n} {lambda^{n-1}} $$
And we can repeat the argument for any $z_i$ since the vector has infinite length. Now if $lambda = 1$, we have the previous case (set 2). If $lambda = -1$, then then the limit is not necessarily convergent, e.g. $z_n = (-1)^n$ could be a constraint, but $z_n=1$, $n=1,2...$ would not give a convergent limit. But if $|lambda| lt 1$, then the limit implies $z_1 to infty$, ... ,$z_n to infty$. Similarly if $|lambda| gt 1$, then we must get $z_1 to 0,z_2 to 0,...)$.
Doesn't this contradict the result that set 1 describes eigenvalues of T?
In comparing to similar posts such as this thread, I see that set 1 should be the correct answer, but I'm not sure where my logic has lead me astray to believe otherwise. I suspect that " we can repeat the argument for any $z_i$" might not be true, but I'm not sure why. Any ideas?
linear-algebra limits
asked Nov 23 at 7:18
chris fritz
63
add a comment |
So I'm trying to find the eigenvalues and eigenvectors of the backwards shift operator $T:F^infty to F^infty$ defined as $Tv=T(z_1,z_2,....)=(z_2,z_3,....)$.
Arriving at the system of equations from $Tv = lambda v$:
$$z_2 = lambda z_1$$
$$z_3 = lambda z_2$$
$$vdots$$
$$z_{n+1} = lambda z_n$$
$$vdots$$
(1) If we fix $z_1$ we arrive at the set of eigenvectors,
$$ { (z_1,lambda z_1, lambda^2 z_1, ....) : lambda epsilon mathbb{F} }$$
(2) If $lambda = 1$, then our eigenvectors are then
$$ { (z,z,...) : z epsilon mathbb{F} }$$
(3) If $lambda = 0$, then clearly $(z,0,0,...)$ describes the eigenvectors.
If $lambda ne 0$, then we have
$$z_1 = frac{z_2}{lambda}$$
$$z_2 = frac{z_3}{lambda}$$
$$vdots$$
$$z_{n} = frac{z_{n+1}}{lambda}$$
$$vdots$$
But this system of equations also implies that the 1st element $z_1$ is a function of $z_2$, which is a function of $z_3$, and so on. So they are all "chained together" to infinity. That is to say,
$$ z_1 = frac{1}{lambda} frac{z_3}{lambda} $$
$$ = frac{1}{lambda^2} frac{z_4}{lambda} $$
$$ vdots $$
$$ = lim_{n to infty} frac{z_n} {lambda^{n-1}} $$
And we can repeat the argument for any $z_i$ since the vector has infinite length. Now if $lambda = 1$, we have the previous case (set 2). If $lambda = -1$, then then the limit is not necessarily convergent, e.g. $z_n = (-1)^n$ could be a constraint, but $z_n=1$, $n=1,2...$ would not give a convergent limit. But if $|lambda| lt 1$, then the limit implies $z_1 to infty$, ... ,$z_n to infty$. Similarly if $|lambda| gt 1$, then we must get $z_1 to 0,z_2 to 0,...)$.
Doesn't this contradict the result that set 1 describes eigenvalues of T?
In comparing to similar posts such as this thread, I see that set 1 should be the correct answer, but I'm not sure where my logic has lead me astray to believe otherwise. I suspect that " we can repeat the argument for any $z_i$" might not be true, but I'm not sure why. Any ideas?
linear-algebra limits
asked Nov 23 at 7:18
chris fritz
63
1
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
1
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00
add a comment |
So I'm trying to find the eigenvalues and eigenvectors of the backwards shift operator $T:F^infty to F^infty$ defined as $Tv=T(z_1,z_2,....)=(z_2,z_3,....)$.
Arriving at the system of equations from $Tv = lambda v$:
$$z_2 = lambda z_1$$
$$z_3 = lambda z_2$$
$$vdots$$
$$z_{n+1} = lambda z_n$$
$$vdots$$
(1) If we fix $z_1$ we arrive at the set of eigenvectors,
$$ { (z_1,lambda z_1, lambda^2 z_1, ....) : lambda epsilon mathbb{F} }$$
(2) If $lambda = 1$, then our eigenvectors are then
$$ { (z,z,...) : z epsilon mathbb{F} }$$
(3) If $lambda = 0$, then clearly $(z,0,0,...)$ describes the eigenvectors.
If $lambda ne 0$, then we have
$$z_1 = frac{z_2}{lambda}$$
$$z_2 = frac{z_3}{lambda}$$
$$vdots$$
$$z_{n} = frac{z_{n+1}}{lambda}$$
$$vdots$$
But this system of equations also implies that the 1st element $z_1$ is a function of $z_2$, which is a function of $z_3$, and so on. So they are all "chained together" to infinity. That is to say,
$$ z_1 = frac{1}{lambda} frac{z_3}{lambda} $$
$$ = frac{1}{lambda^2} frac{z_4}{lambda} $$
$$ vdots $$
$$ = lim_{n to infty} frac{z_n} {lambda^{n-1}} $$
And we can repeat the argument for any $z_i$ since the vector has infinite length. Now if $lambda = 1$, we have the previous case (set 2). If $lambda = -1$, then then the limit is not necessarily convergent, e.g. $z_n = (-1)^n$ could be a constraint, but $z_n=1$, $n=1,2...$ would not give a convergent limit. But if $|lambda| lt 1$, then the limit implies $z_1 to infty$, ... ,$z_n to infty$. Similarly if $|lambda| gt 1$, then we must get $z_1 to 0,z_2 to 0,...)$.
Doesn't this contradict the result that set 1 describes eigenvalues of T?
In comparing to similar posts such as this thread, I see that set 1 should be the correct answer, but I'm not sure where my logic has lead me astray to believe otherwise. I suspect that " we can repeat the argument for any $z_i$" might not be true, but I'm not sure why. Any ideas?
linear-algebra limits
asked Nov 23 at 7:18
chris fritz
63
So I'm trying to find the eigenvalues and eigenvectors of the backwards shift operator $T:F^infty to F^infty$ defined as $Tv=T(z_1,z_2,....)=(z_2,z_3,....)$.
Arriving at the system of equations from $Tv = lambda v$:
$$z_2 = lambda z_1$$
$$z_3 = lambda z_2$$
$$vdots$$
$$z_{n+1} = lambda z_n$$
$$vdots$$
(1) If we fix $z_1$ we arrive at the set of eigenvectors,
$$ { (z_1,lambda z_1, lambda^2 z_1, ....) : lambda epsilon mathbb{F} }$$
(2) If $lambda = 1$, then our eigenvectors are then
$$ { (z,z,...) : z epsilon mathbb{F} }$$
(3) If $lambda = 0$, then clearly $(z,0,0,...)$ describes the eigenvectors.
If $lambda ne 0$, then we have
$$z_1 = frac{z_2}{lambda}$$
$$z_2 = frac{z_3}{lambda}$$
$$vdots$$
$$z_{n} = frac{z_{n+1}}{lambda}$$
$$vdots$$
But this system of equations also implies that the 1st element $z_1$ is a function of $z_2$, which is a function of $z_3$, and so on. So they are all "chained together" to infinity. That is to say,
$$ z_1 = frac{1}{lambda} frac{z_3}{lambda} $$
$$ = frac{1}{lambda^2} frac{z_4}{lambda} $$
$$ vdots $$
$$ = lim_{n to infty} frac{z_n} {lambda^{n-1}} $$
And we can repeat the argument for any $z_i$ since the vector has infinite length. Now if $lambda = 1$, we have the previous case (set 2). If $lambda = -1$, then then the limit is not necessarily convergent, e.g. $z_n = (-1)^n$ could be a constraint, but $z_n=1$, $n=1,2...$ would not give a convergent limit. But if $|lambda| lt 1$, then the limit implies $z_1 to infty$, ... ,$z_n to infty$. Similarly if $|lambda| gt 1$, then we must get $z_1 to 0,z_2 to 0,...)$.
Doesn't this contradict the result that set 1 describes eigenvalues of T?
In comparing to similar posts such as this thread, I see that set 1 should be the correct answer, but I'm not sure where my logic has lead me astray to believe otherwise. I suspect that " we can repeat the argument for any $z_i$" might not be true, but I'm not sure why. Any ideas?
linear-algebra limits
linear-algebra limits
asked Nov 23 at 7:18
chris fritz
63
asked Nov 23 at 7:18
chris fritz
63
edited Nov 24 at 8:01
asked Nov 23 at 7:18
chris fritz
63
asked Nov 23 at 7:18
chris fritz
63
asked Nov 23 at 7:18
chris fritz
63
63
1
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
1
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00
add a comment |
1
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
1
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00
1
1
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
1
1
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00
add a comment |
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1
What is the definition of $F^{infty}$?
– Kavi Rama Murthy
Nov 23 at 7:21
"if $lambda = -1$, then the limit is not defined" - how do you conclude this? What if $z_n = (-1)^n$?
– Bungo
Nov 23 at 7:27
1
Your limit argument in general seems to be assuming that $z_n$ is constant as $n to infty$. I'm not sure why you need a limit argument at all. Plainly for any nonzero $z$, the definition shows that $(z, lambda z, lambda^2 z, lambda^3 z , ldots)$ works as an eigenvector corresponding to the eigenvalue $lambda$. This is true for any real or even complex $lambda$, or indeed in any field.
– Bungo
Nov 23 at 7:31
I'm using the limit argument because the system of equations imply that each entry depends on the next (i.e $z_n = f(z_{n+1}))$. Since the list has infinite length, this should mean that the limiting behavior of the entries is relevant to at least the first entry. The only instance I could think of where an entry doesn't depend on the next ad infinitum is if the next entry is zero, but this would imply all entries are zero (for nonzero $lambda$).
– chris fritz
Nov 24 at 7:59
Still unclear why $(z,lambda z, lambda^2 z, ldots)in F^{infty}$ does not (?) hold for any $z$ and $lambda$. Just rephrasing things asked already: what is your $F^{infty}$ ($implies$ why does the limit behavior bother you)?
– metamorphy
Dec 1 at 23:00