A problem on positive definite matrix
I am studying a chapter on positive definites and there is this question in which I have to find whether the quadratic forms are positive definite or not. I have to confirm my answer for this quadratic form: $x^{2}+8xy+y^{2}$. I made it's matrix representation which is $$A = begin{bmatrix} 1&4\4&1 end{bmatrix}$$
The eigenvalues of this matrix are 5 and -3. Since one of the eigenvalue is negative, it must not be a positive definite?
My second doubt is regarding this form: $x^{2}+6xy$. The matrix representation is $$B=begin{bmatrix} 1&3\3&0 end{bmatrix}$$
The eigenvalues of this matrix are $$lambda = frac{1 pm sqrt{37}}{2}$$
So this quadratic form is also not positive definite, right?
eigenvalues-eigenvectors quadratic-forms positive-definite
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
|
show 2 more comments
I am studying a chapter on positive definites and there is this question in which I have to find whether the quadratic forms are positive definite or not. I have to confirm my answer for this quadratic form: $x^{2}+8xy+y^{2}$. I made it's matrix representation which is $$A = begin{bmatrix} 1&4\4&1 end{bmatrix}$$
The eigenvalues of this matrix are 5 and -3. Since one of the eigenvalue is negative, it must not be a positive definite?
My second doubt is regarding this form: $x^{2}+6xy$. The matrix representation is $$B=begin{bmatrix} 1&3\3&0 end{bmatrix}$$
The eigenvalues of this matrix are $$lambda = frac{1 pm sqrt{37}}{2}$$
So this quadratic form is also not positive definite, right?
eigenvalues-eigenvectors quadratic-forms positive-definite
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
1
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02
|
show 2 more comments
I am studying a chapter on positive definites and there is this question in which I have to find whether the quadratic forms are positive definite or not. I have to confirm my answer for this quadratic form: $x^{2}+8xy+y^{2}$. I made it's matrix representation which is $$A = begin{bmatrix} 1&4\4&1 end{bmatrix}$$
The eigenvalues of this matrix are 5 and -3. Since one of the eigenvalue is negative, it must not be a positive definite?
My second doubt is regarding this form: $x^{2}+6xy$. The matrix representation is $$B=begin{bmatrix} 1&3\3&0 end{bmatrix}$$
The eigenvalues of this matrix are $$lambda = frac{1 pm sqrt{37}}{2}$$
So this quadratic form is also not positive definite, right?
eigenvalues-eigenvectors quadratic-forms positive-definite
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
I am studying a chapter on positive definites and there is this question in which I have to find whether the quadratic forms are positive definite or not. I have to confirm my answer for this quadratic form: $x^{2}+8xy+y^{2}$. I made it's matrix representation which is $$A = begin{bmatrix} 1&4\4&1 end{bmatrix}$$
The eigenvalues of this matrix are 5 and -3. Since one of the eigenvalue is negative, it must not be a positive definite?
My second doubt is regarding this form: $x^{2}+6xy$. The matrix representation is $$B=begin{bmatrix} 1&3\3&0 end{bmatrix}$$
The eigenvalues of this matrix are $$lambda = frac{1 pm sqrt{37}}{2}$$
So this quadratic form is also not positive definite, right?
eigenvalues-eigenvectors quadratic-forms positive-definite
eigenvalues-eigenvectors quadratic-forms positive-definite
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
edited Nov 23 '18 at 14:59
Himanshu Sharma
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
asked Nov 23 '18 at 14:41
Himanshu SharmaHimanshu Sharma
113
113
Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
1
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02
|
show 2 more comments
Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
1
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02
Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
1
1
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02
|
show 2 more comments
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Sylvester's criterion. By the way the eigenvalues for second matrix can be found using quadratic formula.
– Yadati Kiran
Nov 23 '18 at 14:43
Please could you clarify a little bit. It would be of great help.
– Himanshu Sharma
Nov 23 '18 at 14:45
Sylvester's criterion says "Symmetric (Hermitian) matrix M is positive-definite if and only if the principal minors are positive: i.e. determinant of the upper left $ktimes k$ matrix is positive for $1leq kleq n$.
– Yadati Kiran
Nov 23 '18 at 14:48
Why are you unable to find the eigenvalues of $B$. Using the standard method of finding the roots of $det (lambda I - B) $ works just fine.
– Eric
Nov 23 '18 at 14:49
1
Yes thats correct. For future- the fastest way to check it in this case is using the fact that the determinant is negative. The determinant is equal to the product of eigenvalues (can be proved using Jordan form), so if you have a negative determinant then you must have a negative eigenvalue. Sylvester's criterion expands on this idea.
– Eric
Nov 23 '18 at 15:02