At 14:32, I think there is some error in the approach, because if we visualise the graph on 3d calculator, the function value is positive for non zero x_1 and x_2 , hence it can't be negative definite. For example, x_1 =0 and x_2 = 2 , f(x_1,x_2) = 4
Many books have given this wrong information as well. Actually, for positive semidefinite we need to check sign of "leading principle minors" while for positive semidefinite we need to check sign of all "principle minors". Hermitian matrix is the most generalized case of symmetric matrices. So, 1. a Hermitian matrix M is positive-semidefinite if and only if all "principal minors" of M are nonnegative. 2. a Hermitian matrix M is positive-definite if and only if all the "leading principal minors" are positive. For more details please : en.wikipedia.org/wiki/Sylvester%27s_criterion#:~:text=Sylvester's%20Criterion%3A%20The%20real%2Dsymmetric,minors%20of%20A%20are%20positive.
If I take a Symmetric Matrix, such that, a11=4, a22=1, a33= -1 ,a12= 2, a13= -1, a23= -0.5. Then the first leading principal minor is 4, while other two leading principal minors are 0. Then by the principle minor rule (Sylvester's Rule) this matrix should be positive semidefinite. But actually this is indefinite.
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Very nice
Thank you so much.
Please continue to upload more videos.
Worth the time spending here.
At 14:32, I think there is some error in the approach, because if we visualise the graph on 3d calculator, the function value is positive for non zero x_1 and x_2 , hence it can't be negative definite.
For example, x_1 =0 and x_2 = 2 , f(x_1,x_2) = 4
Many books have given this wrong information as well. Actually, for positive semidefinite we need to check sign of "leading principle minors" while for positive semidefinite we need to check sign of all "principle minors". Hermitian matrix is the most generalized case of symmetric matrices. So,
1. a Hermitian matrix M is positive-semidefinite if and only if all "principal minors" of M are nonnegative.
2. a Hermitian matrix M is positive-definite if and only if all the "leading principal minors" are positive.
For more details please : en.wikipedia.org/wiki/Sylvester%27s_criterion#:~:text=Sylvester's%20Criterion%3A%20The%20real%2Dsymmetric,minors%20of%20A%20are%20positive.
I think you wrote "semi" definite for both Leading principal minor and principal minor by mistake. BTW thanks !
Great job!
If I take a Symmetric Matrix, such that, a11=4, a22=1, a33= -1 ,a12= 2, a13= -1, a23= -0.5. Then the first leading principal minor is 4, while other two leading principal minors are 0. Then by the principle minor rule (Sylvester's Rule) this matrix should be positive semidefinite. But actually this is indefinite.
Thank you very much sir, your explanation is really helpful for my exam tomorrow, it's really nice you gave a lot of examples
My pleasure ... Kindly share with other too
Really good
Well explained
Thanks sir , I came across your video it is great help and super clear for me . Kindly provide some links to read further on same topic
Got it. Thank you sir
Thank you so much, the best video on the subject!
Thanks Andre ... Kindly share with others too
So much helpful ,😊
Thank you sir
thank you sir !
Thank you 😊it helped a lot.
Really great explanation sir 🙏🏻
Thanks.
Nicely explained 😊😊😊😊
My pleasure... Keep watching.
Tq
Welcome
Could you simulink the queen honey bee migration algorithms please?
love u
Thank you ❤
Tau 2x2 ka bhi to batao
D1, d2 aur d3 kya hoga?
Why to give half knowledge?