Are eigenvector continuous?
$begingroup$
Lets consider a matrix $Ain mathbb R^{dtimes d}$, where $d=2$ or $3$,
$ A$ has entries $a_{ij}(x)$, where $a_{ij}in C^1(mathbb{R})$. Also, $A$ is positive definite. Given the eigenvalues $lambda_i$ of $A$ are the values that satisfy
$$Av_i = lambda_i v_i,$$
where $v_i$ is the corresponding eigenvector. I think the derivative
$$frac{mathrm dlambda_i}{mathrm dx}$$
exists, since $lambda_i$ are the roots of a polynomial with coefficients that are again multivariate polynomials in $a_{ij}$, therefore being continuous in $lambda_i$.
What I was wondering was if the eigenvectors $v_i$ are as well. Usually the eigenvectors are all vectors which satisfy
$$ (A-lambda_iI)v_i=0,$$
but the matrix $(A-lambda_iI)$ is singular, so I don't know whether the eigenvectors are still continuous.
calculus matrices eigenvalues-eigenvectors
asked Jan 3 at 11:16
User123456789User123456789
457314
$endgroup$
|
show 2 more comments
$begingroup$
Lets consider a matrix $Ain mathbb R^{dtimes d}$, where $d=2$ or $3$,
$ A$ has entries $a_{ij}(x)$, where $a_{ij}in C^1(mathbb{R})$. Also, $A$ is positive definite. Given the eigenvalues $lambda_i$ of $A$ are the values that satisfy
$$Av_i = lambda_i v_i,$$
where $v_i$ is the corresponding eigenvector. I think the derivative
$$frac{mathrm dlambda_i}{mathrm dx}$$
exists, since $lambda_i$ are the roots of a polynomial with coefficients that are again multivariate polynomials in $a_{ij}$, therefore being continuous in $lambda_i$.
What I was wondering was if the eigenvectors $v_i$ are as well. Usually the eigenvectors are all vectors which satisfy
$$ (A-lambda_iI)v_i=0,$$
but the matrix $(A-lambda_iI)$ is singular, so I don't know whether the eigenvectors are still continuous.
calculus matrices eigenvalues-eigenvectors
asked Jan 3 at 11:16
User123456789User123456789
457314
$endgroup$
$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
1
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
1
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24
|
show 2 more comments
$begingroup$
Lets consider a matrix $Ain mathbb R^{dtimes d}$, where $d=2$ or $3$,
$ A$ has entries $a_{ij}(x)$, where $a_{ij}in C^1(mathbb{R})$. Also, $A$ is positive definite. Given the eigenvalues $lambda_i$ of $A$ are the values that satisfy
$$Av_i = lambda_i v_i,$$
where $v_i$ is the corresponding eigenvector. I think the derivative
$$frac{mathrm dlambda_i}{mathrm dx}$$
exists, since $lambda_i$ are the roots of a polynomial with coefficients that are again multivariate polynomials in $a_{ij}$, therefore being continuous in $lambda_i$.
What I was wondering was if the eigenvectors $v_i$ are as well. Usually the eigenvectors are all vectors which satisfy
$$ (A-lambda_iI)v_i=0,$$
but the matrix $(A-lambda_iI)$ is singular, so I don't know whether the eigenvectors are still continuous.
calculus matrices eigenvalues-eigenvectors
asked Jan 3 at 11:16
User123456789User123456789
457314
$endgroup$
Lets consider a matrix $Ain mathbb R^{dtimes d}$, where $d=2$ or $3$,
$ A$ has entries $a_{ij}(x)$, where $a_{ij}in C^1(mathbb{R})$. Also, $A$ is positive definite. Given the eigenvalues $lambda_i$ of $A$ are the values that satisfy
$$Av_i = lambda_i v_i,$$
where $v_i$ is the corresponding eigenvector. I think the derivative
$$frac{mathrm dlambda_i}{mathrm dx}$$
exists, since $lambda_i$ are the roots of a polynomial with coefficients that are again multivariate polynomials in $a_{ij}$, therefore being continuous in $lambda_i$.
What I was wondering was if the eigenvectors $v_i$ are as well. Usually the eigenvectors are all vectors which satisfy
$$ (A-lambda_iI)v_i=0,$$
but the matrix $(A-lambda_iI)$ is singular, so I don't know whether the eigenvectors are still continuous.
calculus matrices eigenvalues-eigenvectors
calculus matrices eigenvalues-eigenvectors
asked Jan 3 at 11:16
User123456789User123456789
457314
asked Jan 3 at 11:16
User123456789User123456789
457314
edited Jan 3 at 11:40
User123456789
asked Jan 3 at 11:16
User123456789User123456789
457314
asked Jan 3 at 11:16
User123456789User123456789
457314
asked Jan 3 at 11:16
User123456789User123456789
457314
457314
$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
1
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
1
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24
|
show 2 more comments
$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
1
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
1
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24
$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
1
1
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
1
1
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $s$ denote $frac{sqrt{2}}{2}$, and let
$$
M = pmatrix{s & -s \ s & s}
$$
be rotation by $pi/4$.
Let
$$
X(u) = pmatrix{u & 0 \ 0 & 1}.
$$
Then for $u ne 1$, the matrix
$$
M X(u) M^t
$$
has eigenvectors in the directions $pmatrix{pm s\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $pmatrix{1\0}$ and $pmatrix{0\1}$ (the "axial directions").
Now consider
$$
H(t) = begin{cases}
M X(1 + t^4) M^t & t < 0 \
X(1 + t^4) & t ge 0
end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$.
(I admit I haven't written out every detail here, and maybe I need to put in something like $exp(-frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so).
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
$endgroup$
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $s$ denote $frac{sqrt{2}}{2}$, and let
$$
M = pmatrix{s & -s \ s & s}
$$
be rotation by $pi/4$.
Let
$$
X(u) = pmatrix{u & 0 \ 0 & 1}.
$$
Then for $u ne 1$, the matrix
$$
M X(u) M^t
$$
has eigenvectors in the directions $pmatrix{pm s\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $pmatrix{1\0}$ and $pmatrix{0\1}$ (the "axial directions").
Now consider
$$
H(t) = begin{cases}
M X(1 + t^4) M^t & t < 0 \
X(1 + t^4) & t ge 0
end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$.
(I admit I haven't written out every detail here, and maybe I need to put in something like $exp(-frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so).
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
$endgroup$
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
add a comment |
$begingroup$
Let $s$ denote $frac{sqrt{2}}{2}$, and let
$$
M = pmatrix{s & -s \ s & s}
$$
be rotation by $pi/4$.
Let
$$
X(u) = pmatrix{u & 0 \ 0 & 1}.
$$
Then for $u ne 1$, the matrix
$$
M X(u) M^t
$$
has eigenvectors in the directions $pmatrix{pm s\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $pmatrix{1\0}$ and $pmatrix{0\1}$ (the "axial directions").
Now consider
$$
H(t) = begin{cases}
M X(1 + t^4) M^t & t < 0 \
X(1 + t^4) & t ge 0
end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$.
(I admit I haven't written out every detail here, and maybe I need to put in something like $exp(-frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so).
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
$endgroup$
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
add a comment |
$begingroup$
Let $s$ denote $frac{sqrt{2}}{2}$, and let
$$
M = pmatrix{s & -s \ s & s}
$$
be rotation by $pi/4$.
Let
$$
X(u) = pmatrix{u & 0 \ 0 & 1}.
$$
Then for $u ne 1$, the matrix
$$
M X(u) M^t
$$
has eigenvectors in the directions $pmatrix{pm s\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $pmatrix{1\0}$ and $pmatrix{0\1}$ (the "axial directions").
Now consider
$$
H(t) = begin{cases}
M X(1 + t^4) M^t & t < 0 \
X(1 + t^4) & t ge 0
end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$.
(I admit I haven't written out every detail here, and maybe I need to put in something like $exp(-frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so).
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
$endgroup$
Let $s$ denote $frac{sqrt{2}}{2}$, and let
$$
M = pmatrix{s & -s \ s & s}
$$
be rotation by $pi/4$.
Let
$$
X(u) = pmatrix{u & 0 \ 0 & 1}.
$$
Then for $u ne 1$, the matrix
$$
M X(u) M^t
$$
has eigenvectors in the directions $pmatrix{pm s\s}$ (the "diagonal directions") while the matrix $X(u)$ has eigenvectors in the directions $pmatrix{1\0}$ and $pmatrix{0\1}$ (the "axial directions").
Now consider
$$
H(t) = begin{cases}
M X(1 + t^4) M^t & t < 0 \
X(1 + t^4) & t ge 0
end{cases}
$$
Then $H$ has the form required by your problem (although my variable is $t$ rather than $x$), but for $t > 0$, the eigenvectors are axial, and for $t < 0$, the eigenvectors are diagonal, and no choice of eigenvectors for $t = 0$, where the matrix is the identity and every vector is an eigenvector, will make either eigenvector a continuous (let alone differentiable) function of $t$.
(I admit I haven't written out every detail here, and maybe I need to put in something like $exp(-frac{1}{t^2})$ instead of $t^4$ to make everything smooth....but I don't think so).
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
answered Jan 3 at 13:18
John HughesJohn Hughes
63.3k24090
63.3k24090
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
add a comment |
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
NB: You could make the "eigenvectors" be continuous functions of $t$ if you were (a) willing to let them not have unit magnitude, and (b) let them be zero vectors for the identity. But the "eigenvectors are nonzero" thing is an important part of the definition.
$endgroup$
– John Hughes
Jan 3 at 13:30
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
$begingroup$
I have some more information about $A$ then I could say in my question, I actually know a lot. This answer gives me more then enough information to put the pieces together. Thank you!
$endgroup$
– User123456789
Jan 3 at 13:33
add a comment |
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$begingroup$
$lambda_i$ is a function of $a_{ij}$, which is a function of $x$.
$endgroup$
– User123456789
Jan 3 at 11:19
1
$begingroup$
It's confusing that you denote both the eigenvectors and the position variable by $x$. I would use the notation $v_i$ for the eigenvectors.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:21
$begingroup$
Good point, will fix it now
$endgroup$
– User123456789
Jan 3 at 11:21
1
$begingroup$
Also, you should define an ordering of the eigenvalues, otherwise there is no reason that $lambda_i$ would refer to the "same" eigenvalue for different $x$.
$endgroup$
– Roberto Rastapopoulos
Jan 3 at 11:22
$begingroup$
I don't see why this is important I want to know about all eigenvectors, given that I know $lambda$ and $lambda'$. $v_i$ is only dependent on the corresponding eigenvalue, if I am not mistaken.
$endgroup$
– User123456789
Jan 3 at 11:24