How do I find eigenvalues of an hessian matrix that depends on (x,y,z)?
$begingroup$
For example:
$f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$
Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$
Well, how do I calculate the eigen values?
Is it going to be in function of y and z? So they won't be fixed?
If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?
I'm a little bit confused:
my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.
Thank you all!!
linear-algebra analysis
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
$endgroup$
add a comment |
$begingroup$
For example:
$f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$
Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$
Well, how do I calculate the eigen values?
Is it going to be in function of y and z? So they won't be fixed?
If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?
I'm a little bit confused:
my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.
Thank you all!!
linear-algebra analysis
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
$endgroup$
add a comment |
$begingroup$
For example:
$f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$
Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$
Well, how do I calculate the eigen values?
Is it going to be in function of y and z? So they won't be fixed?
If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?
I'm a little bit confused:
my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.
Thank you all!!
linear-algebra analysis
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
$endgroup$
For example:
$f(x,y,z)=1+z^2+sin(z)+x^2+|y|^3$
Hf $(x,y,z) = diag(2, 6y, 2- sin(z))$
Well, how do I calculate the eigen values?
Is it going to be in function of y and z? So they won't be fixed?
If whenever I find $eig(x,y,z) > 0$ for every $(x,y,z)$, only then I can be sure it is convex?
I'm a little bit confused:
my teacher says that this function has a unique global minimum, but I'm quite sure it doesn't.
Thank you all!!
linear-algebra analysis
linear-algebra analysis
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
edited Nov 29 '18 at 15:13
greedoid
41.2k1150102
41.2k1150102
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
asked Nov 29 '18 at 14:53
Lucas TononLucas Tonon
445
445
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A quick comment first,
$$
frac{partial |y|^3}{partial y} not= 2y
$$
as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
A quick comment first,
$$
frac{partial |y|^3}{partial y} not= 2y
$$
as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
add a comment |
$begingroup$
A quick comment first,
$$
frac{partial |y|^3}{partial y} not= 2y
$$
as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
add a comment |
$begingroup$
A quick comment first,
$$
frac{partial |y|^3}{partial y} not= 2y
$$
as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
$endgroup$
A quick comment first,
$$
frac{partial |y|^3}{partial y} not= 2y
$$
as a matter of fact at $y = 0$ the whole expression just break. You can follow this other post to see more details. That being said, it is not a problem if the Hessian and its eigenvalues explicitly depend on the coordinates, it just tells you that the geometry changes with location.
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
answered Nov 29 '18 at 15:02
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
add a comment |
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
actualy it is 6y, sorry my bad, but anyways the problem stands still
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:05
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
$begingroup$
Well, I can see that the geometry changes, but can I define the eigenvalues as functions of (x,y,z)?
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:06
1
1
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
@LucasTonon Yes, absolutely. There's not a problem with that
$endgroup$
– caverac
Nov 29 '18 at 15:06
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
$begingroup$
it's 6ysign(y) actually cause y^3 is smooth in the origin, so it is derivable in zero and has derivative zero, so |y|^3 derived may be sign(y)*3y^2 and that derived is 6ysign(y), my bad
$endgroup$
– Lucas Tonon
Nov 29 '18 at 15:11
add a comment |
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