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Using the Hessian Matrix to classify points
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From what I've gathered from my calculus supplements and the web, I want to know if I have the general computation procedure understood correctly.
Example: Given f such that f(x,y) = ___. Find and classify the critical points.
To begin, I find the first order partial derivatives of f. Which I will set to 0, and then use some factoring techniques. From this I will have critical point(s).
Then, I would find the second order partial derivatives of f, and place them accordingly into a 2x2 Hessian Matrix A (i.e. f_xx as a_11, f_xy as a_12, f_yx as a_21, fyy as a_22).
From here, I would sub in the critical point (granted there is only one C.P.) into the matrix A. Then, depending on how friendly the numbers of each entry turn out to be, I have one of two options to follow through to classify the points.
1.) If the matrix turns out to be a symmetric matrix, then I can just take the diagonal entries as corresponding eigenvalues, which can be used directly to classify critical points by the following...
Where all eigenvalues are positive is a positive definite which is local minimum.
Where all eigenvalues are negative is a negative definite which is local maximum.
Where it is not positive or negative definite, it is a saddle point.
or
2.) If the matrix is not a symmetric matrix, then I will need to verify that the each sub-matrices follows a certain condition to find it's classification such that
If det(a_1) > 0 and det(A) > 0, then it is a positive definite, which is a local minimum
If det(a_1) < 0 and det(A) < 0, then it is a negative definite, which is a local maximum
If det(a_1) > 0 and det(A) < 0 or the alternative, it is a saddle point.
Thus, the critical point is classified.
Does anybody know if this is the correct format for a general case of using Hessian to classify points? Thanks!
calculus partial-derivative maxima-minima hessian-matrix
asked 2 days ago
Donald Mayer
53
add a comment |
up vote
0
down vote
favorite
From what I've gathered from my calculus supplements and the web, I want to know if I have the general computation procedure understood correctly.
Example: Given f such that f(x,y) = ___. Find and classify the critical points.
To begin, I find the first order partial derivatives of f. Which I will set to 0, and then use some factoring techniques. From this I will have critical point(s).
Then, I would find the second order partial derivatives of f, and place them accordingly into a 2x2 Hessian Matrix A (i.e. f_xx as a_11, f_xy as a_12, f_yx as a_21, fyy as a_22).
From here, I would sub in the critical point (granted there is only one C.P.) into the matrix A. Then, depending on how friendly the numbers of each entry turn out to be, I have one of two options to follow through to classify the points.
1.) If the matrix turns out to be a symmetric matrix, then I can just take the diagonal entries as corresponding eigenvalues, which can be used directly to classify critical points by the following...
Where all eigenvalues are positive is a positive definite which is local minimum.
Where all eigenvalues are negative is a negative definite which is local maximum.
Where it is not positive or negative definite, it is a saddle point.
or
2.) If the matrix is not a symmetric matrix, then I will need to verify that the each sub-matrices follows a certain condition to find it's classification such that
If det(a_1) > 0 and det(A) > 0, then it is a positive definite, which is a local minimum
If det(a_1) < 0 and det(A) < 0, then it is a negative definite, which is a local maximum
If det(a_1) > 0 and det(A) < 0 or the alternative, it is a saddle point.
Thus, the critical point is classified.
Does anybody know if this is the correct format for a general case of using Hessian to classify points? Thanks!
calculus partial-derivative maxima-minima hessian-matrix
asked 2 days ago
Donald Mayer
53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From what I've gathered from my calculus supplements and the web, I want to know if I have the general computation procedure understood correctly.
Example: Given f such that f(x,y) = ___. Find and classify the critical points.
To begin, I find the first order partial derivatives of f. Which I will set to 0, and then use some factoring techniques. From this I will have critical point(s).
Then, I would find the second order partial derivatives of f, and place them accordingly into a 2x2 Hessian Matrix A (i.e. f_xx as a_11, f_xy as a_12, f_yx as a_21, fyy as a_22).
From here, I would sub in the critical point (granted there is only one C.P.) into the matrix A. Then, depending on how friendly the numbers of each entry turn out to be, I have one of two options to follow through to classify the points.
1.) If the matrix turns out to be a symmetric matrix, then I can just take the diagonal entries as corresponding eigenvalues, which can be used directly to classify critical points by the following...
Where all eigenvalues are positive is a positive definite which is local minimum.
Where all eigenvalues are negative is a negative definite which is local maximum.
Where it is not positive or negative definite, it is a saddle point.
or
2.) If the matrix is not a symmetric matrix, then I will need to verify that the each sub-matrices follows a certain condition to find it's classification such that
If det(a_1) > 0 and det(A) > 0, then it is a positive definite, which is a local minimum
If det(a_1) < 0 and det(A) < 0, then it is a negative definite, which is a local maximum
If det(a_1) > 0 and det(A) < 0 or the alternative, it is a saddle point.
Thus, the critical point is classified.
Does anybody know if this is the correct format for a general case of using Hessian to classify points? Thanks!
calculus partial-derivative maxima-minima hessian-matrix
asked 2 days ago
Donald Mayer
53
From what I've gathered from my calculus supplements and the web, I want to know if I have the general computation procedure understood correctly.
Example: Given f such that f(x,y) = ___. Find and classify the critical points.
To begin, I find the first order partial derivatives of f. Which I will set to 0, and then use some factoring techniques. From this I will have critical point(s).
Then, I would find the second order partial derivatives of f, and place them accordingly into a 2x2 Hessian Matrix A (i.e. f_xx as a_11, f_xy as a_12, f_yx as a_21, fyy as a_22).
From here, I would sub in the critical point (granted there is only one C.P.) into the matrix A. Then, depending on how friendly the numbers of each entry turn out to be, I have one of two options to follow through to classify the points.
1.) If the matrix turns out to be a symmetric matrix, then I can just take the diagonal entries as corresponding eigenvalues, which can be used directly to classify critical points by the following...
Where all eigenvalues are positive is a positive definite which is local minimum.
Where all eigenvalues are negative is a negative definite which is local maximum.
Where it is not positive or negative definite, it is a saddle point.
or
2.) If the matrix is not a symmetric matrix, then I will need to verify that the each sub-matrices follows a certain condition to find it's classification such that
If det(a_1) > 0 and det(A) > 0, then it is a positive definite, which is a local minimum
If det(a_1) < 0 and det(A) < 0, then it is a negative definite, which is a local maximum
If det(a_1) > 0 and det(A) < 0 or the alternative, it is a saddle point.
Thus, the critical point is classified.
Does anybody know if this is the correct format for a general case of using Hessian to classify points? Thanks!
calculus partial-derivative maxima-minima hessian-matrix
calculus partial-derivative maxima-minima hessian-matrix
asked 2 days ago
Donald Mayer
53
asked 2 days ago
Donald Mayer
53
edited 2 days ago
asked 2 days ago
Donald Mayer
53
asked 2 days ago
Donald Mayer
53
asked 2 days ago
Donald Mayer
53
53
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The test is not quite right. First, the diagonal entries of a symmetric matrix are rarely equal to its eigenvalues. For example
$$
begin{pmatrix}
1 & 2 \
2 & 1
end{pmatrix}
$$
is a symmetric matrix whose eigenvalues are $3$ and $-1$. I think you may be confusing the terms "symmetric" and "diagonal", as the Hessian will always be symmetric (with light assumptions).
Second, the test is not correct. You are correct that if the Hessian is positive definite then you have a local minimum, and if the Hessian is negative definite then you have a local maximum. However, a $ 2 times 2 $ matrix is negative definite if $a_1$ is negative and the determinant is positive. E.g.
$$
begin{pmatrix}
-1 & 0 \
0& -1
end{pmatrix}
$$
is negative definite and has determinant equal to one. Also, if the determinant of a $2 times 2$ matrix is negative then you have a nondefinite matrix regardless of the sign of $a_1$.
A third comment is, you can have a local max or min at a point where the Hessian is not positive definite. The Hessian must be positive semidefinite at a local min, but you cannot gaurantee that the eigenvalues are nonzero. E.g. if $f(x,y)=x^4+y^4$ then $(0,0)$ is the global minimum but the Hessian at $(0,0)$ is equal to zero.
And finally, while determinant is good for testing in the two variable case, if you go to three or more variables testing the determinant and the sign of $a_1$ is not enough. It doesn't seem you need to worry about this for the course you are taking, but its nice to keep in mind for the future.
answered 2 days ago
Eric
915
Thanks dude_____
– Donald Mayer
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The test is not quite right. First, the diagonal entries of a symmetric matrix are rarely equal to its eigenvalues. For example
$$
begin{pmatrix}
1 & 2 \
2 & 1
end{pmatrix}
$$
is a symmetric matrix whose eigenvalues are $3$ and $-1$. I think you may be confusing the terms "symmetric" and "diagonal", as the Hessian will always be symmetric (with light assumptions).
Second, the test is not correct. You are correct that if the Hessian is positive definite then you have a local minimum, and if the Hessian is negative definite then you have a local maximum. However, a $ 2 times 2 $ matrix is negative definite if $a_1$ is negative and the determinant is positive. E.g.
$$
begin{pmatrix}
-1 & 0 \
0& -1
end{pmatrix}
$$
is negative definite and has determinant equal to one. Also, if the determinant of a $2 times 2$ matrix is negative then you have a nondefinite matrix regardless of the sign of $a_1$.
A third comment is, you can have a local max or min at a point where the Hessian is not positive definite. The Hessian must be positive semidefinite at a local min, but you cannot gaurantee that the eigenvalues are nonzero. E.g. if $f(x,y)=x^4+y^4$ then $(0,0)$ is the global minimum but the Hessian at $(0,0)$ is equal to zero.
And finally, while determinant is good for testing in the two variable case, if you go to three or more variables testing the determinant and the sign of $a_1$ is not enough. It doesn't seem you need to worry about this for the course you are taking, but its nice to keep in mind for the future.
answered 2 days ago
Eric
915
Thanks dude_____
– Donald Mayer
2 days ago
add a comment |
up vote
1
down vote
accepted
The test is not quite right. First, the diagonal entries of a symmetric matrix are rarely equal to its eigenvalues. For example
$$
begin{pmatrix}
1 & 2 \
2 & 1
end{pmatrix}
$$
is a symmetric matrix whose eigenvalues are $3$ and $-1$. I think you may be confusing the terms "symmetric" and "diagonal", as the Hessian will always be symmetric (with light assumptions).
Second, the test is not correct. You are correct that if the Hessian is positive definite then you have a local minimum, and if the Hessian is negative definite then you have a local maximum. However, a $ 2 times 2 $ matrix is negative definite if $a_1$ is negative and the determinant is positive. E.g.
$$
begin{pmatrix}
-1 & 0 \
0& -1
end{pmatrix}
$$
is negative definite and has determinant equal to one. Also, if the determinant of a $2 times 2$ matrix is negative then you have a nondefinite matrix regardless of the sign of $a_1$.
A third comment is, you can have a local max or min at a point where the Hessian is not positive definite. The Hessian must be positive semidefinite at a local min, but you cannot gaurantee that the eigenvalues are nonzero. E.g. if $f(x,y)=x^4+y^4$ then $(0,0)$ is the global minimum but the Hessian at $(0,0)$ is equal to zero.
And finally, while determinant is good for testing in the two variable case, if you go to three or more variables testing the determinant and the sign of $a_1$ is not enough. It doesn't seem you need to worry about this for the course you are taking, but its nice to keep in mind for the future.
answered 2 days ago
Eric
915
Thanks dude_____
– Donald Mayer
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The test is not quite right. First, the diagonal entries of a symmetric matrix are rarely equal to its eigenvalues. For example
$$
begin{pmatrix}
1 & 2 \
2 & 1
end{pmatrix}
$$
is a symmetric matrix whose eigenvalues are $3$ and $-1$. I think you may be confusing the terms "symmetric" and "diagonal", as the Hessian will always be symmetric (with light assumptions).
Second, the test is not correct. You are correct that if the Hessian is positive definite then you have a local minimum, and if the Hessian is negative definite then you have a local maximum. However, a $ 2 times 2 $ matrix is negative definite if $a_1$ is negative and the determinant is positive. E.g.
$$
begin{pmatrix}
-1 & 0 \
0& -1
end{pmatrix}
$$
is negative definite and has determinant equal to one. Also, if the determinant of a $2 times 2$ matrix is negative then you have a nondefinite matrix regardless of the sign of $a_1$.
A third comment is, you can have a local max or min at a point where the Hessian is not positive definite. The Hessian must be positive semidefinite at a local min, but you cannot gaurantee that the eigenvalues are nonzero. E.g. if $f(x,y)=x^4+y^4$ then $(0,0)$ is the global minimum but the Hessian at $(0,0)$ is equal to zero.
And finally, while determinant is good for testing in the two variable case, if you go to three or more variables testing the determinant and the sign of $a_1$ is not enough. It doesn't seem you need to worry about this for the course you are taking, but its nice to keep in mind for the future.
answered 2 days ago
Eric
915
The test is not quite right. First, the diagonal entries of a symmetric matrix are rarely equal to its eigenvalues. For example
$$
begin{pmatrix}
1 & 2 \
2 & 1
end{pmatrix}
$$
is a symmetric matrix whose eigenvalues are $3$ and $-1$. I think you may be confusing the terms "symmetric" and "diagonal", as the Hessian will always be symmetric (with light assumptions).
Second, the test is not correct. You are correct that if the Hessian is positive definite then you have a local minimum, and if the Hessian is negative definite then you have a local maximum. However, a $ 2 times 2 $ matrix is negative definite if $a_1$ is negative and the determinant is positive. E.g.
$$
begin{pmatrix}
-1 & 0 \
0& -1
end{pmatrix}
$$
is negative definite and has determinant equal to one. Also, if the determinant of a $2 times 2$ matrix is negative then you have a nondefinite matrix regardless of the sign of $a_1$.
A third comment is, you can have a local max or min at a point where the Hessian is not positive definite. The Hessian must be positive semidefinite at a local min, but you cannot gaurantee that the eigenvalues are nonzero. E.g. if $f(x,y)=x^4+y^4$ then $(0,0)$ is the global minimum but the Hessian at $(0,0)$ is equal to zero.
And finally, while determinant is good for testing in the two variable case, if you go to three or more variables testing the determinant and the sign of $a_1$ is not enough. It doesn't seem you need to worry about this for the course you are taking, but its nice to keep in mind for the future.
answered 2 days ago
Eric
915
answered 2 days ago
Eric
915
answered 2 days ago
Eric
915
answered 2 days ago
Eric
915
915
Thanks dude_____
– Donald Mayer
2 days ago
add a comment |
Thanks dude_____
– Donald Mayer
2 days ago
Thanks dude_____
– Donald Mayer
2 days ago
Thanks dude_____
– Donald Mayer
2 days ago
add a comment |
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