Positive Definite Matrices and Minima

[@mekala]

We can also calculate C for the semi-specific case by taking the correct combinations of the first two columns or rows. Matrix will be positive semidefinite when det is zero. We know matrix is singular when determinant is zero. [2 -1 -1] [-1 2 -1] [-1 -1 2+C] If we take - 1 * column1 + (-1 * column2) : -2 - 1 = -1 1 - 2 = -1 1 + 1 = 2 So if matrix is singular 2 + c = 2 ----> c = 0

[@alperaslan5865]

If she was my lecturer, I would never miss any of her classes.

[@tonguc745]

MIT de ne okudun reis

[@ericawanja1749]

You remind me that I am here after missing a lecture 😅😌

[@emmanuelisichei-zr7mp]

Seriously dude

[@emmanuelisichei-zr7mp]

Seriously dude

[@alejandrosanchez3246]

Seriously dude

[@justpaulo]

It seems to me that the pivot test was much faster and easy. In addition it seems that from the U matrix you can read directly the final "complete the square" equation.

[@yigitsezer6696]

I think it depends on the matrix. For bigger matrices you are right of course.

[@WelsyCZ]

It was because the Instructor already knew how to complete the square. Finding out how to complete the square would take longer than calculating the determinants.

[@kimmielee3100]

You can see from the pivot test that the pivots in the echelon form of the matrix are also the coefficients of these squares, and when we use the distributive laws for each row to extract the pivots from each row, we left with the coefficients of x, y, and z in the squares.

[@kimmielee3100]

And that explains how the instructor can read the equation immediately from the results she got in the pivot test.

[@yasseralabaishi6693]

She really good in explaining 👍👍

[@user-or7ji5hv8y]

Very clear and concise. Thanks

[@Robocat754]

Overall very good explanation. But I don't see how the formulas in the upper right corner could help us with the completing the squares. It's not the same. Anyway it's very simple that you can actually do it in your head when you already know the pattern. Another point is that in the end you talked about the null space of positive definite matrix but you didn't come up with a conclusion. It's not mentioned in the last lecture either.

[@SIYA-PAL]

Very easy way अब अंपने बताया means u understand me very ejey

[@matthewjames7513]

is A = [ -1 0; 0 -4] negative definite? Method 1: Determinant method -1 < 0 (first sub matrix) -1*-4-0*0 = 4 > 0 (second sub matrix) Therefore this matrix is indefinite Method 2: eigenvalues lambda1 = -1 < 0 lambda2 = -4 < 0 Therefore this matrix is negative definite What am I doing wrong here!?

[@capturedart0]

pivots test is better than determinant

[@jonathannielsen1247]

When using the determinant for the negative definite case, you have to check if (-1)^k*Det(submatrix) > 0, where k is the dimension of the submatrix. Therefore in your case, you need to do as follows: D1 = -1*(-1)^1 = 1 > 0 and D2 = 4*(-1)^2 = 4 >0. Because D1>0 og D2>0, it is negative definite. Hope it makes sense, that you need to multiply with (-1)^k

[@SIYA-PAL]

Love from India mam thanks u 😇😇🙏😄🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳 u r very jenious because I injoy understand everything thanks

[@nemesis9080]

चुतिये अपनी मां की चुट पे टैटू करवाले ये इंग्लिश... नहीं आता तो सीधा THANK YOU दे

[@АлександрСницаренко-р4д]

when c=0, the matrix is a positive semidefinite?

[@sentyjessicaakok2066]

0 has no signs, hence semipositive definite.

[@kirillnovik8661]

I think that without any further information the scenario where c=0 would mean that it can be positive semidefinite, negative semidefinite or indefinite

[@reactioner2005]

I like how teaching assistants are smiling after 5 second being out of view

[@mathzone4893]

Very nice explain 👍👍

[@quirkyquester]

Thank you!

[@hsccbo32385]

You sholdn't expect determinant test to work for positive semi-definite matrices. Consider the matrix, [0 0] [0 -1] which certainly passes the ">=0" condition for all the principal submatrices yet its spectrum is {0, -1}.

[@mozhdehyazdanifard6565]

You are damn right! Thanks for your comment. However, she's still a pretty smart girl to me. She is a teaching assistant at MIT.

[@mozhdehyazdanifard6565]

I think the only reliable method is to follow "completing the square"! You barely can see what you are doing. I brought an example which indicates how possibly her methods lead to potential mistakes. I brought in another comment, not as reply under your comment.

[@hsccbo32385]

​@@mozhdehyazdanifard6565 Well, for positive definite matrices the test works fine. But note that it is fine because determinant of leading principal submatrices are incremental products of pivots. We know that matrix A is positive definite iff all of A's pivots are positive. If any zero pivot appears earlier than negative pivots, you won't detect that negative pivot. The test wouldn't even make sense if there's a zero in the middle, because you need to do the row-exchange in Gaussian elliminiation, but then what should the pivot be? (unless you skip that zero-pivot column directly, I'm not sure what would the natural constraint for this to work be.) You could, instead, require that all principal submatrices (not necessarily leading) having non-negative determinant.

[@mozhdehyazdanifard6565]

​@Taylor Huang You are right! Now, I see why your example is special. can you say why mine is not in compatible with what she said? I think my matrix can be considered as counterexample to take into question all what she said, not just the "determinant test" method. My example matrix which is brought in the first comment on this video. [2 -2 -2 ] [0 2 -2 ] [0 0 2+c] 1st Method) using "completing square" we derive the same expression as what she achieved for her own example; so we have: C > 0 2* x^2 + 2* y^2 + (2+c)* z^2 - 2 xy -2xz - 2yz 2nd, 3rd methods) while following "determinant test" and "pivot test" methods we end up with a result like: c> -2

[@hsccbo32385]

@@mozhdehyazdanifard6565 everything is under assumption that the matrix is Hermitian (in real matrices, it's symmetric)

[@radicalpotato666]

Are they all necessary and sufficient tests, anyone?

[@islamhaouas7015]

thank you very much

[@SIYA-PAL]

Aanp bahoot axcha study krati hai mujhe anpka sab kuch samajh aata hai it's hindhi but wriiten in English 😄😄😄😄😄🇮🇳🇮🇳

[@S_hubham_singh]

She is so cute😊

[@SIYA-PAL]

I Lives in India but i ever study by u bicouse u r a English girl so I can to talk English by hearing your voice 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳 i lives in up loknow India 🇮🇳

[@bettypassion5164]

Good

[@oualidbenamar6468]

thanks

[@mozhdehyazdanifard6565]

Thanks for your quite useful video. Your explanations are pretty clear for dummies like me! You are pretty smart, and also a good teacher. You are also beautiful. My brother was distracted by your beauty.. He could not take his eyes off you, to look at the whiteboard. He had to watch your video some couple of times. Do not worry. I'll explain it for him.

[@quirkyquester]

loll hahhahaha thats so funny

[@shabnamahmed9136]

Are you talking about your brother or yourself coz it's black board 😂

[@mozhdehyazdanifard6565]

@@shabnamahmed9136 It was long time ago that I posted this comment. No I was distracted by my brother's eccentric behavior.

[@mozhdehyazdanifard6565]

By the way, You are right; that is actually a blackboard, not whiteboard! I'm not sure why I did this mistake, even if I was distracted by him. Maybe I have to practice some more English!

[@noemilaszlo8822]

the cameramen is making me sea sick

[@yashas9974]

I didn't even notice it.

[@strossicro1]

Oklen si mala.

[@akagamishanks-q3o]

Lookin good :)

[@mozhdehyazdanifard6565]

I think what you said is all incorrect!!! I'll bring another matrix that takes into question all what you have mentioned. That is why I'm dummy :( ------------------------------------------------- 1) Completing the Square: ------------------------------------------------- Using the method "completed squared" does not seem to be compatible with other test methods! However, this is the most reliable method, so other methods cannot be used! For example, we know that the following matrix should have equal results with what you presented: [2 -2 -2 ] [0 2 -2 ] [0 0 2+c] Using "completed squared" we have: 2* x^2 + 2* y^2 + (2+c)* z^2 - 2 xy -2xz - 2yz This is the same expression as what you have derived for your matrix. --> As you said: C > 0 ---------------------------------- 2) determinant test: ---------------------------------- What we see is that the determinant of the whole matrix is 2 * 2 * (2+c) = 8 + 4C > 0 ---> C> -2 !!!! ------------------------- 3) pivot test: ------------------------- It's an upper triangle matrix. All that is required is to divide the rows by 2, except the third one: 2 * [1 -1 -1 ] 2* [0 1 -1 ] (2+c) [0 0 1 ] ---> 2( x-y-z)^2 + 2 * (y-z)^2 + (2+c)* z^2 !!!! --> C > -2 , but that is not equal to what you derived for "completing the square"!

[@rhversity5965]

Matrices are only positive definite if they are symmetric. Your matrix is not symmetric so it fails the test.

[@martinpesek8377]

@@rhversity5965 I disagree. A non symmetric matrix may be positive definite, but the so-called determinant test is not applicable and may show wrong results. E.g. ((2,0)(2,2))

[@mgk4908]

Very few women can do maths the way she's doing.

[@hugofidelcamposespinoza4913]

en.wikipedia(dot)org/wiki/List_of_women_in_mathematics

[@ga7073]

Not "can do", but "would do". Despite all the freedom to conquer STEM disciplines, only a minority of women are attracted by this ..

[@mimialaa4980]

Thank you ❤️😊