Mixing
About complex eigenvectors and proof of Sylvester's criterion
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Hello my question is referring to the assertion of the user 1551 in that question:
Characterization of positive definite matrix with principal minors
Assertion itself :"in particular, det(A)>0. It follows that if A is not positive definite, it must possess at least two negative eigenvalues " - why 2, and not 1? And why that eigenvalues can't be just imagine, so they won't be as positive as negative(and is that possible? Or they always be strictly positive or negative?)
linear-algebra matrices operator-theory matrix-equations matrix-decomposition
asked Jan 31 at 7:50
user2952487user2952487
224
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add a comment |
$begingroup$
Hello my question is referring to the assertion of the user 1551 in that question:
Characterization of positive definite matrix with principal minors
Assertion itself :"in particular, det(A)>0. It follows that if A is not positive definite, it must possess at least two negative eigenvalues " - why 2, and not 1? And why that eigenvalues can't be just imagine, so they won't be as positive as negative(and is that possible? Or they always be strictly positive or negative?)
linear-algebra matrices operator-theory matrix-equations matrix-decomposition
asked Jan 31 at 7:50
user2952487user2952487
224
$endgroup$
add a comment |
$begingroup$
Hello my question is referring to the assertion of the user 1551 in that question:
Characterization of positive definite matrix with principal minors
Assertion itself :"in particular, det(A)>0. It follows that if A is not positive definite, it must possess at least two negative eigenvalues " - why 2, and not 1? And why that eigenvalues can't be just imagine, so they won't be as positive as negative(and is that possible? Or they always be strictly positive or negative?)
linear-algebra matrices operator-theory matrix-equations matrix-decomposition
asked Jan 31 at 7:50
user2952487user2952487
224
$endgroup$
Hello my question is referring to the assertion of the user 1551 in that question:
Characterization of positive definite matrix with principal minors
Assertion itself :"in particular, det(A)>0. It follows that if A is not positive definite, it must possess at least two negative eigenvalues " - why 2, and not 1? And why that eigenvalues can't be just imagine, so they won't be as positive as negative(and is that possible? Or they always be strictly positive or negative?)
linear-algebra matrices operator-theory matrix-equations matrix-decomposition
linear-algebra matrices operator-theory matrix-equations matrix-decomposition
asked Jan 31 at 7:50
user2952487user2952487
224
asked Jan 31 at 7:50
user2952487user2952487
224
asked Jan 31 at 7:50
user2952487user2952487
224
asked Jan 31 at 7:50
user2952487user2952487
224
asked Jan 31 at 7:50
user2952487user2952487
224
224
add a comment |
add a comment |
1 Answer
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The eigen values of a Hermitian matrix are are real. If there is only one negative eigen value then the determinant, which is product of the eigen values would be $leq 0$.
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
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$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
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
$begingroup$
The eigen values of a Hermitian matrix are are real. If there is only one negative eigen value then the determinant, which is product of the eigen values would be $leq 0$.
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
add a comment |
$begingroup$
The eigen values of a Hermitian matrix are are real. If there is only one negative eigen value then the determinant, which is product of the eigen values would be $leq 0$.
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
add a comment |
$begingroup$
The eigen values of a Hermitian matrix are are real. If there is only one negative eigen value then the determinant, which is product of the eigen values would be $leq 0$.
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
The eigen values of a Hermitian matrix are are real. If there is only one negative eigen value then the determinant, which is product of the eigen values would be $leq 0$.
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 31 at 7:54
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
add a comment |
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
$begingroup$
But why did he use 2 instead of 1 eigenvalues?
$endgroup$
– user2952487
Feb 1 at 1:19
add a comment |
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