This lecture is definitely a positive effect on my grasp of the matrix, and this lecture plays a pivotal role in the whole series. Thank you prof. Strang. Nobody has explained those concepts so clearly and coherently -- a whole new world is ahead of me. This is a must-see. A genuine human heritage.
The first 2 lines are the joke. He said positive definite as "definitely ... positive." For those who thought the last comment with the rimshot was explaining the original joke. For completeness sake, no shade.
This lecture blew my mind. That linear algebra can be used to solve multivariable functions and multivariable calculus problems... amazing. At one point he was dropping one mind blowing fact after another. Thank you for these awesome lectures.
After viewing the 3blue1brown video on Duality, I am just seeing xT*A*x as the results of applying a linear transformation to a vector x and then projecting that new vector back onto x; if the vectors still point "in the same direction" i.e. the projection is positive, then A is positive definite
Yes. In fact, there is another way to look at positive definite matrix; that is, the transformation of vector X is another vector that lies in the same quadrant as vector X. The angle between the vector and its transformation should always be acute.
I watched this once, taking all notes. Then watched it again (with break) without taking notes. Then I read my notes (after a break). The lecture's quite good! Btw for those that thought the second derivatives thing came out of left field, Du is the directional derivative operator, so you want Du( Du( f(x1,x2,...,xn) ) ), which gives the expressions for x^T*A*x for a connected f and A. This tells us that x^T*A*x is like a second order operation for the derivative of a function when A is the Hessian (matrix of second derivatives) of said function f(). This wasn't clear to me initially so I was kinda lost. Best!
If I had teachers like Gilbert Strang, I likely would've had a Ph.D in Math by now. No kidding! I love this guy! He made me fall in LOVE with Mathematics.
The funny thing is, I'm also doing a general relativity series and got to covectors. I understand what they are computationally, but couldn't "visualize" them, beyond the typical "stacks". Then I stumbled on the interpretation of covectors as linear maps, which served as a connection to linear algebra. I tried so many things to built a geometric interpretation all night but nothing formed in my head properly, so I was like, "meh, I'm 80% done with Professor Strangs lectures, might as well do another one". He led with x^H*A*x > 0 and for some reason, everything just started to click (x^H*A is an implicit map, so I have ideas about how to analyze things in my other GR class). Funny how his lecture was what I needed to get my head turning again. Thank you, and awesome lecture! ☺️🙌🏽 Stay safe during #COVID19
From this lecture, I really understand Positive Definite Matrices and Minima thanks to Dr. Gilbert Strang. The examples really help me to fully comprehend this important subject.
@15:16, I apologies.......................That's how a matchless, renowned and extraordinary person respect everyone, everywhere. Love you Prof. Gilbert Strange:)
31:40 In the case of A = [2, 6; 6, 7] for example, when setting z = 0, what we actually get is a cross (an 'X' shape) ; when z euqals other values, then we get a hyperbola.
It is my dream to meet prof. Gilbert strang . His voice , his words ,his action touch my soul . Please prof. Read my comment so that I can satisfy only by this And I pray You may live 1000 years .
Symmetric matrices represent ortho-scaling linear transformation; or geometrically, a n-by-n symmetric matrix [A] scales vectors along n # of orthogonal axis by factors of their corresponding eigenvalues. the axis are line spans of the orthonormal vectors of [Q], which are from diagonalization decomposition of [A] = [Q][D][Q_Transpose]. The quadratic form represented by [A] is a functional qA:R^n→R defined by qA(x)=[x_Tr][A]x. if we set qA(x) = 1, then {x:qA(x) = 1} = [A][{{x' : ∥x'→∥_. = r1 }] or the football outline is the image of linear transformation A for a circle of x' such that the length all equals to some real number r1.
the distinctions between the matrices and linear transformations and the interpretations of linear transformation are important foundations in understanding linear algebra and abstract linear algebra and normed vector spaces in the Analysis and the Point-Set Topology and differential Geometry.
sorta got the general idea of this whole lecture, im still not so good at the details. Reading the book and summary might be a good idea to solidate the knowledge if needed. Thank you Professor Strang and MIT! great lecture!
This lecture is a good one to have Matlab or Octave pulled up to look at some of these surfaces. x=-10:1:10; y=-10:1:10; [XX, YY]=meshgrid(x,y); Z=2*XX.*XX + 12*XX.*YY + 7*YY.*YY; surf(XX,YY,Z)
Intuitively, a positive definite matrix is one that has more stretching weight than a rotation weight. If it operates on a vector in the first quadrant, it guarantees that the output vector is also in the first quadrant. This is almost like a closure property under the transformation of a matrix. I need to examine the case when the x vector in Ax is not in the first quadrant. The entire underlying theme is creating something that is very close to an Eigen matrix with more weight around the diagonal. The important of Eigen matrices is well known in applications where the power matrix has the same effect of a power scalar function. With big matrices, there is nothing more valuable than decomposing a matrix in terms of simpler matrices among which the Eigen matrix is one of those decompositions. Diagonalizing a matrix and SVD is very related to this theme. Any analysis that attempts to transform the effect of a matrix to one that resembles a scalar transformation is the goal.
Question) If we complete the squares of the eq. (at 44:53), I don't think the squares get multiplied by the three pivots(2, 3/2, 4/3)... Can anyone tell me why?
There's one video on Khan academy multivariable calculus talking about this, didn't quite get it. Now after his explanation, I finally know what is that.XD
The Pivots ARE 2,3,4. He just took it one step further by dividing each row by the previous pivot. So the last row was [0,0,4] and he divided by pivot 3 to get [0,0,4/3]. He then took the 2nd row which was [0,3,-2] and divided by pivot 2 to get [0,3/2,-1]. If you take the product of the pivots 2,3,4 you get 24 which is the determinant of this new matrix. But by dividing by the previous pivot the product of the pivots [2, 3/2, 4/3]=4, which was the determinant of the original matrix.
Can someone please explain why the condition for pivot test is (ac-b^2)/a > 0? If I just do the elimination to figure out the pivot, shouldn't the second pivot just be （c-b^2)/a?
in order to make b (in 21 position) 0, you have to multiply first row with -b/a and add it to the second. When you do that with b (in 12 position) and add it to c, you get -b²/a + c, which is the same as (ac-b²)/a
@@pavlenikacevic4976 I know that it has 4 years, but I'd like to thank you, mr. Pavle. I'm curious about how did you get this insight? Greetings from Brazil!
@@douglasespindola5185 you're welcome. I don't remember where that is in the video, and I've also been out of maths for some time, so I cannot provide you any useful information at this point 😅
@@pavlenikacevic4976 thanks anyway. I'm studying for a job selection that will apply an exam about data science content. It'll be helpful. Why are you out of maths? Seems to me that you were a nice student.
@@douglasespindola5185 now I'm doing research in quantum chemistry. Doesn't really require math on a day to day basis. I needed math during my studies, to help me understand physics better Good luck with studying for the exam!
This max and min method makes intuitive sense but does anyone have a proof showing that if the gradient is zero and the matrix is positive definite then the function is a local minimum?
First of all, in order for a function to have a minimum, the gradient has to be zero. If this were not the case, there would be a direction next to the point in question where the function decreases - the direction opposite to the gradient - and this would contradict the fact that it is a minimum. If one takes an arbitrary function and carries out a Taylor expansion around the minimum, all terms of order higher than two can be neglected if we are close enough to the minimum (and we just need to know the function values right next to the minimum point to know that it is a minimum). Since the gradient is zero, all first-order derivatives are zero at the minimum point, and then we just have the constant term - the function value at the minimum - and the second order derivatives. The Taylor approximation around the minimum can thus be written as f(x_0 + x) ≈ f(x_0) + 1/2 (x^T A x), where A is the matrix of second derivatives (the Hessian) and x_0 is the minimum. Thus, if x^T A x > 0 for all x - i.e., the matrix A is positive definite - this is in particular valid for small x-values (such that x_0 + x is close to x_0), where the Taylor approximation is a good approximation of f(x_0 + x). Thus, we can conclude that in a neighbourhood of the point x_0 (i.e. if we are close enough so that the Taylor approximation is good), all other points give function values such that f(x_0 + x) > f(x_0), and thus x_0 is a local minimum, since the function has higher values for all points in a neighbourhood around this point.
Probably not the best instructor. There are a lot of instructors on RUclips who are much much better teachers than him. Imagine the students at MIT have to through this .
The absolute lack of rigor and the hand-waviness of the lesson makes this video a popular science segment rather than a math lesson. A nice popular science segment, but I certainly hope MIT students of Physics and engineers are directed to take something with a higher level, significantly so. Physics undergrads don't need total rigor, but such a total lack of it doesn't develop mathematical thinking.
He struggles to get the idea across, the intuition, which he does so masterfully. Very few teachers have this talent.... most math teachers are too pedantic and they lose their students along that way.
You can only write so many proofs in 50 minutes. The proofs are very well laid out in the book. It's a good read (and a required read for students of the class)
This is intended to be an applied Linear Algebra class at MIT, for a more rigorous course on Linear Algebra (with proofs) Axler's or Friedberg are the best out there for the undergraduate level.
This lecture is definitely a positive effect on my grasp of the matrix, and this lecture plays a pivotal role in the whole series. Thank you prof. Strang. Nobody has explained those concepts so clearly and coherently -- a whole new world is ahead of me. This is a must-see. A genuine human heritage.
Don't you mean... a positive definite effect? *cymbal crash*
The first 2 lines are the joke. He said positive definite as "definitely ... positive."
For those who thought the last comment with the rimshot was explaining the original joke. For completeness sake, no shade.
my calculus 3 professor taught us, think of a saddle point as a point on a "pringles chip" a lot of people know exactly what it looks like.
or a saddle ?
@@matiasmoanaguerrero8095 lmao
(S)He’s the smartest person ever.
@@matiasmoanaguerrero8095 most American kids have seen way more pringles than they have saddles. Sorry John Wayne
This lecture blew my mind. That linear algebra can be used to solve multivariable functions and multivariable calculus problems... amazing. At one point he was dropping one mind blowing fact after another.
Thank you for these awesome lectures.
After viewing the 3blue1brown video on Duality, I am just seeing xT*A*x as the results of applying a linear transformation to a vector x and then projecting that new vector back onto x; if the vectors still point "in the same direction" i.e. the projection is positive, then A is positive definite
Yes. In fact, there is another way to look at positive definite matrix; that is, the transformation of vector X is another vector that lies in the same quadrant as vector X. The angle between the vector and its transformation should always be acute.
Thank you!
This is maybe the best lecture in the entire course.
Lets see! gonna do it! There is also a recent close up video of gilbert doing a 20 minute bit on positive definite matrices!
I agree with you! What an incredible way to show how different branches of mathematics actually refers to the same thing.
beautiful how comes together.
I watched this once, taking all notes. Then watched it again (with break) without taking notes. Then I read my notes (after a break). The lecture's quite good! Btw for those that thought the second derivatives thing came out of left field, Du is the directional derivative operator, so you want Du( Du( f(x1,x2,...,xn) ) ), which gives the expressions for x^T*A*x for a connected f and A. This tells us that x^T*A*x is like a second order operation for the derivative of a function when A is the Hessian (matrix of second derivatives) of said function f(). This wasn't clear to me initially so I was kinda lost. Best!
If I had teachers like Gilbert Strang, I likely would've had a Ph.D in Math by now. No kidding! I love this guy! He made me fall in LOVE with Mathematics.
The funny thing is, I'm also doing a general relativity series and got to covectors. I understand what they are computationally, but couldn't "visualize" them, beyond the typical "stacks". Then I stumbled on the interpretation of covectors as linear maps, which served as a connection to linear algebra. I tried so many things to built a geometric interpretation all night but nothing formed in my head properly, so I was like, "meh, I'm 80% done with Professor Strangs lectures, might as well do another one". He led with x^H*A*x > 0 and for some reason, everything just started to click (x^H*A is an implicit map, so I have ideas about how to analyze things in my other GR class). Funny how his lecture was what I needed to get my head turning again. Thank you, and awesome lecture! ☺️🙌🏽 Stay safe during #COVID19
This lecture is just amazing, what a beautiful thing, all coming together...
This lecture is especially beautiful..
Positive Definite connecting matrices, algebra, geometry, and calculus: priceless!
Oh Yeah, It's All Coming Together.
From this lecture, I really understand Positive Definite Matrices and Minima thanks to Dr. Gilbert Strang. The examples really help me to fully comprehend this important subject.
At 41:54, Strang gets the eigenvalues correct from memory. I'm impressed!
This lecture is a true masterpiece. I was in awe with the similar feeling of watching suspense thrillers.
Never Seen someone like this.... Amazing!!!!
What I feel just like you....
the explanation starting at 25:10 helped me understand what a positive definite matrice is
@15:16, I apologies.......................That's how a matchless, renowned and extraordinary person respect everyone, everywhere.
Love you Prof. Gilbert Strange:)
31:08 how everything beautifully connects together
31:40 In the case of A = [2, 6; 6, 7] for example, when setting z = 0, what we actually get is a cross (an 'X' shape) ; when z euqals other values, then we get a hyperbola.
"had better be more than 18" was the correct answer
Hahahah nice
From 1,012,547 views in the first lecture to 203,355 views in 27th...You belong to 20% who made it this far, Congrats!
Very good lecture Professor Strang thank you from Amazon Forest.
It is my dream to meet prof. Gilbert strang . His voice , his words ,his action touch my soul . Please prof. Read my comment so that I can satisfy only by this And I pray You may live 1000 years .
fancy tie!!!!
Symmetric matrices represent ortho-scaling linear transformation; or geometrically, a n-by-n symmetric matrix [A] scales vectors along n # of orthogonal axis by factors of their corresponding eigenvalues. the axis are line spans of the orthonormal vectors of [Q], which are from diagonalization decomposition of [A] = [Q][D][Q_Transpose]. The quadratic form represented by [A] is a functional qA:R^n→R defined by qA(x)=[x_Tr][A]x. if we set qA(x) = 1, then {x:qA(x) = 1} = [A][{{x' : ∥x'→∥_. = r1 }] or the football outline is the image of linear transformation A for a circle of x' such that the length all equals to some real number r1.
the distinctions between the matrices and linear transformations and the interpretations of linear transformation are important foundations in understanding linear algebra and abstract linear algebra and normed vector spaces in the Analysis and the Point-Set Topology and differential Geometry.
Wow.....how beautiful it is! How beautiful!!!
sorta got the general idea of this whole lecture, im still not so good at the details. Reading the book and summary might be a good idea to solidate the knowledge if needed. Thank you Professor Strang and MIT! great lecture!
Dr. Gilbert Strang is really a master!
49:42 eigen value tell length of axis eigenvextors tell
My favourite sports team is the Seattle Submatrices
Thanks prof. This is one of the best lectures of the course!
This lecture is a good one to have Matlab or Octave pulled up to look at some of these surfaces.
x=-10:1:10;
y=-10:1:10;
[XX, YY]=meshgrid(x,y);
Z=2*XX.*XX + 12*XX.*YY + 7*YY.*YY;
surf(XX,YY,Z)
thanks prof. now I can imagine what eigenvectors and eigenvalues are like.
Absolutely well done and definitely keep it up!!! 👍👍👍👍👍
Intuitively, a positive definite matrix is one that has more stretching weight than a rotation weight. If it operates on a vector in the first quadrant, it guarantees that the output vector is also in the first quadrant. This is almost like a closure property under the transformation of a matrix. I need to examine the case when the x vector in Ax is not in the first quadrant. The entire underlying theme is creating something that is very close to an Eigen matrix with more weight around the diagonal. The important of Eigen matrices is well known in applications where the power matrix has the same effect of a power scalar function. With big matrices, there is nothing more valuable than decomposing a matrix in terms of simpler matrices among which the Eigen matrix is one of those decompositions. Diagonalizing a matrix and SVD is very related to this theme. Any analysis that attempts to transform the effect of a matrix to one that resembles a scalar transformation is the goal.
The mathematics fills my heart with beauty.
Love these lectures
A great lecture.
Question) If we complete the squares of the eq. (at 44:53), I don't think the squares get multiplied by the three pivots(2, 3/2, 4/3)... Can anyone tell me why?
Like watching someone like Aristotle teach
Saddle Points: sometimes you gotta get up to get down.
damn subtitle, my exam is in a couple of days and I need to watch this seriously, but the subtitle just keeps making me giggle ^^haha
this man just related a rugby ball to the eigenvectors of a matrix. damn!
This was amazing!!!!!!!!!!!!!!!!
I think my mind just exploded !
The last function f(x1,x2,x3)>=0 can be proved by creating squares.
Quite possibly the hardest lecture, so far.
There's one video on Khan academy multivariable calculus talking about this, didn't quite get it. Now after his explanation, I finally know what is that.XD
prof. nice tie 👍😁
u just sumarized my 12 hours of studying in 50 mins
Is a hyperbola in high dimensional spaces called a hyperhyperbola?
Damn he's good at teaching!
Dove chocolate is not smooth as silk. This lecture is. indeed
tremendous explanation thank you for sharing ...
How will we define negative definate. Will it have everything negative like
1) All pivots are negative
2) All e-values negative ?
Yes for both
It's a .... superbowl
Thank you.
could someone tell me 6:48 why is this true? how can you calculate lambda with trace? Thank youooouuu!
The trace of a matrix is the sum of its (complex) eigenvalues, and it is invariant with respect to a change of basis.
?? I understood everything except my pivots are 2, 3, 4 :(
I love that last bit. Kind of like SVD.
maybe this (with Sagemath) could help you ?
sagecell.sagemath.org/?q=dfqzji
The Pivots ARE 2,3,4. He just took it one step further by dividing each row by the previous pivot. So the last row was [0,0,4] and he divided by pivot 3 to get [0,0,4/3]. He then took the 2nd row which was [0,3,-2] and divided by pivot 2 to get [0,3/2,-1]. If you take the product of the pivots 2,3,4 you get 24 which is the determinant of this new matrix. But by dividing by the previous pivot the product of the pivots [2, 3/2, 4/3]=4, which was the determinant of the original matrix.
Is that video available or not? I'm following the course on you tube but I can't visualize this particular lecture (n° 27).
amazing!
47:25 drawing an eye 101
Thanks!
Can someone please explain why the condition for pivot test is (ac-b^2)/a > 0? If I just do the elimination to figure out the pivot, shouldn't the second pivot just be （c-b^2)/a?
in order to make b (in 21 position) 0, you have to multiply first row with -b/a and add it to the second. When you do that with b (in 12 position) and add it to c, you get -b²/a + c, which is the same as (ac-b²)/a
@@pavlenikacevic4976 I know that it has 4 years, but I'd like to thank you, mr. Pavle. I'm curious about how did you get this insight? Greetings from Brazil!
@@douglasespindola5185 you're welcome. I don't remember where that is in the video, and I've also been out of maths for some time, so I cannot provide you any useful information at this point 😅
@@pavlenikacevic4976 thanks anyway. I'm studying for a job selection that will apply an exam about data science content. It'll be helpful. Why are you out of maths? Seems to me that you were a nice student.
@@douglasespindola5185 now I'm doing research in quantum chemistry. Doesn't really require math on a day to day basis. I needed math during my studies, to help me understand physics better
Good luck with studying for the exam!
His definition of saddle Prof Gilbert 😂
Nice tie prof
"What's the long word for bowl?"
+Lobezno Meneses eliptic paraboloid z=(x/a)^2 + (y/b)^2
It should be a function of three variables, not two variables.
I would have just called it a hyper bowl.
@@usmanhassan1887 f(x_1,x_2,x_3)
Yes, it is, the long word for bowl is a function of 3 variables. Thank you.. @@marcinkovalevskij5820
35:00
recap
It seems that in mathematics everything at some point connects
Where is lecture no 28?
Arslan Ali usually after 27 and before 29
I believe the graph of 2x^{2}+12xy+20y^{2}=0 do not exist
Very nice! :)
am i the only one watching the lectures in 1.5 or 1.25 playback speed?
This max and min method makes intuitive sense but does anyone have a proof showing that if the gradient is zero and the matrix is positive definite then the function is a local minimum?
First of all, in order for a function to have a minimum, the gradient has to be zero. If this were not the case, there would be a direction next to the point in question where the function decreases - the direction opposite to the gradient - and this would contradict the fact that it is a minimum.
If one takes an arbitrary function and carries out a Taylor expansion around the minimum, all terms of order higher than two can be neglected if we are close enough to the minimum (and we just need to know the function values right next to the minimum point to know that it is a minimum). Since the gradient is zero, all first-order derivatives are zero at the minimum point, and then we just have the constant term - the function value at the minimum - and the second order derivatives. The Taylor approximation around the minimum can thus be written as f(x_0 + x) ≈ f(x_0) + 1/2 (x^T A x), where A is the matrix of second derivatives (the Hessian) and x_0 is the minimum.
Thus, if x^T A x > 0 for all x - i.e., the matrix A is positive definite - this is in particular valid for small x-values (such that x_0 + x is close to x_0), where the Taylor approximation is a good approximation of f(x_0 + x). Thus, we can conclude that in a neighbourhood of the point x_0 (i.e. if we are close enough so that the Taylor approximation is good), all other points give function values such that f(x_0 + x) > f(x_0), and thus x_0 is a local minimum, since the function has higher values for all points in a neighbourhood around this point.
It's real algebraist hours, my dudes.
that tie looks cool tho
A rugby ball has 2 out of three eigen values the same.
Hope we become neighbors in heaven Mr Strang.
awsome
I Really Like The Video Positive Definite Matrices and Minima From Your
Thank you Prof.Strang 🇨🇳
Ellipsoid :D
Its like the UFO from the Movie Independence Day :D
how about calling it an egg...
We're down from 1.2mill views on the first video :D
man that was tragic...
Strang is so funny
@2:50 Seattle submatrices lol
4:34
Trump learned from Him how to wear a tie.
21:10
35:15
What's a long word for a bowl? lol
.
Probably not the best instructor. There are a lot of instructors on RUclips who are much much better teachers than him. Imagine the students at MIT have to through this .
The absolute lack of rigor and the hand-waviness of the lesson makes this video a popular science segment rather than a math lesson. A nice popular science segment, but I certainly hope MIT students of Physics and engineers are directed to take something with a higher level, significantly so. Physics undergrads don't need total rigor, but such a total lack of it doesn't develop mathematical thinking.
there is a 300+ pages book that come with that course (Strang wrote it) used in the majority of top level schools. I suggest you buy it and/or gtfo.
He struggles to get the idea across, the intuition, which he does so masterfully. Very few teachers have this talent.... most math teachers are too pedantic and they lose their students along that way.
You can only write so many proofs in 50 minutes. The proofs are very well laid out in the book. It's a good read (and a required read for students of the class)
I have 4th edition as well, but I believe this was with 2nd edition. I can't recall but I believe I saw that on the OCW course page.
This is intended to be an applied Linear Algebra class at MIT, for a more rigorous course on Linear Algebra (with proofs) Axler's or Friedberg are the best out there for the undergraduate level.
autodidact
Please prove the facts.
35:50