What does it mean to take the derviative of a matrix? --- Like, Subscribe, and Hit that Bell to get all the latest videos from ritvikmath ~ --- Check out my Medium: / ritvikmathematics My Patreon: www.patreon.co...
"Please, take a minute to pause and convince yourself that everything on this board is accurate." So difficult to do when I was in school ("several" moon ago) madly scribbling down everything before it got wiped off the board, but now with the internet, with videos, and most importantly with a person who wants you to learn, this is so much easier to absorb. I'm looking forward to teaching my children and using your wise words. Thank you!
@@berylliosis5250 dude in my line of work I got to know a lot of people who have been homeschooled and they all show anti social tendencies and varying degrees of depression, but most of all they all hate their parents for forcing them into being homeschooled. most of them have an extremely hard time making friends or socialising with others. how are you supposed to learn to work in a group with kids in your age, if you don't have the social construct of a school. Also why not trust people who have studied a subject for years to teach your kids, over your own superficial knowledge of science and literatur. Also it's almost always the parents who want this whole homeschooling thing never the children. Cause they have serious attachment problems with their kids and can't let go of them because they are so obsessive. please get over yourselves hoomschooling parents!
@@REALdavidmiscarriage I know a bunch of people who've been homeschooled too. They've been socially capable, intelligent, mentally healthy (in one case, far more so than when they were in public school), and completely educated - potentially more so than their peers. They started homeschooling by mutual consent with their parents. Anecdotes don't prove anything here. While I personally wouldn't want to be homeschooled or to homeschool myself, there are some people who thrive in that kind of system.
@@berylliosis5250 No shit. you just proved my point, exceptions prove the rule. Also you aren't bringing any evidence for it being as good as regular school or better. That's not how that works. You can't just say unicorns exist and ask me to disprove it. You are the one making a bold claim here in comparing homeschooling with regular schools you have to bring factual evidence but you are using anecdotes yourself. So why don't we just slow down a bit and treat this for what it is an argument based on anecdotes not some scientific research paper. Maybe 1 in 1000 students might thrive off of homeschooling. Yeah also maybe 1 in a few million people win the lottery ,so? Does that mean it is worth playing the lottery?
'Everyone' includes all persons, presumably... that would include the observer, so this sentence is inadmissible or meaninglesss.. ps i am only a minor student of logic so I praise the observer's meaning...peace
Okay, so he looks at the function x -> Ax. This is a linear transformation, and the jacobian of any linear transformation is the linear transformation itself. This makes sense because you can think of the Jacobian as the best linear approximation for any function between R^n and R^m, whether it be linear or not. Now, in some sense, yes you can say that the derivative of the matrix is the Jacobian, because a matrix, after all, represents a linear function. As already stated, the derivative of a linear function is basically the Jacobian. I think the moral of this video is that it is best to actually think in terms of function from R^n -> R^m, (vector-valued functions) Does this clarify things?
@Aletak 13 yeah, the Jacobian represents a local linear transformation, which describes how much you are stretching or squishing space. The determinant of the transformation gives you what the area is scaled by, which is why it comes up when you change variables :)
One question: why the derivative of the second example is a column vector? (9:35) I thought it was a row vector, similar to the form in 3:38 (the first row: [df/dx1 df/dx2]. A great video! (It is the same problem as Ravi Shankar’s two months ago)
I think you are correct, at 10:23 he does say that "if you had 3 different functions and 4 different variables you would have a 3 by 4 matrix, i.e. 3 rows and 4 columns". And the result would be 2*xt*A
yeah hes mixing numerator and denominator layout :/, in numerator layout a vector function by a scalar is a column vector, a scalar function by a vector is a row vector. In denominator layout a vector function by a scalar is a row vector and a scalar function by a vector is a column vector. (*By = derivatve with respect to)
When you calculated the derivative of A over the vector x you add the partial derivatives of the function f1 and f2 as row vectors in the matrix. Then, when you calculated the gradient of f1 = x^{T}Ax then over x the results was a column vector. Shouldn't be in this case the first result A^{T}?
Wow. I haven't taken calculus in years and this video made taking derivative of a matrix seem easy to do and understand. Well done as teaching well is an art form unto itself.
You are an absolute life-saver! I am a transfer student studying chemical engineering at UC Davis and your videos match up perfectly with what we are taught :) You have helped tremendously and have given me the knowledge to solve my overly complicated problem sets. Keep making videos and I'm certain you've helped many others as well. Brilliant instructor.
Hmm... Good question. I didn't notice that before I read your comment. It could because of the x' present at the beginning of x'Ax. I'm not sure though.
Maybe he is writing the d(Ax)/dx in matrix notation while d(x^(T)Ax)/dx in vector notation? He does use square brackets for the former and parentheses for the latter, but I'm not too sure myself.
Suppose 2×2 matrix=A has a characteristic polynomial = C.P(A) = λ² - bλ + c then dƒ/dλ = 2λ - b Cayley Hamilton: A² - b•A + c•I means dƒ/dA = 2A - b•A which looks an awful lot like 2λ - bλ Oh, that doesn't mean anything I'm just using power rule with A & λ instead of x.... right? Well what is rhe definition of a derivative? lim [ (ƒ(x+Δx)-ƒ(x))/Δx] = dƒ/dx Δx→0 What about this? lim [ (ƒ(λ+Δθ)-ƒ(λ))/Δλ] = dƒ/dλ? Δλ→0 What about this? lim [ (ƒ(A+ΔA)-ƒ(A))/ΔA] =dƒ/dA ? ΔA→0 Ok, fine Im doing the same thing again with limits now. but suppose you define a 2×2 matrix=A with actual numbers and then you say ƒ[A] = A² = AA and you speculate dƒ/dA = d/dA[A²] =2A Right??? I mean you actually write entries in the matrix in this limit below s.t. I = Identity matrix only instead of this: lim [ (ƒ(A+ΔA)-ƒ(A))/ΔA] ΔA→0 you cant ÷ a matrix, so you do this lim [ ((A+ΔAI)² -A²)(ΔA)⁻¹ ] = ΔA→0 lim [ A² + 2ΔAI + (ΔAI)² -A² (ΔA)⁻¹ ] ΔA→0 ΔAI = ΔA•Identity matrix = [ΔΑ 0] [0 ΔΑ] = ΔΑΙ (ΔΑΙ)⁻¹ = (1/det(ΔΑΙ))•adj(ΔΑΙ)= [1/ΔΑ 0 ] [ 0 1/ΔΑ] = (ΔΑΙ)⁻¹ We can get 2A.... right? Or is it, as you say, just like taking the derivative of a constant? (I leave this as an exercise for the reader to verify.) Just playing... I'M DOING THIS!! NO CONSTANT, BABY!! e.g. Claim: It is possible to take the derivative of at least one 2×2 matrix = A s.t. ƒ[A] = A² & d/dA [A²] = 2A according to "the limit definition of a derivative" and the definition of a function, ƒ. Proof of Claim: Let [ 1 1] [ 0 2] = A [ 1 3 ] [ 0 4 ] =A² [ 2 2 ] [ 0 4 ] = 2A [1/ΔΑ 0 ] [ 0 1/ΔΑ] = (ΔΑΙ)⁻¹ lim [ A² + 2ΔAI + (ΔAI)² -A² (ΔA)⁻¹ ] ΔA→0 lim [ A² + 2ΔAI + (ΔAI)² -A² (ΔA)⁻¹ ] ΔA→0 oh, look at that boy!! wait until that s**t cancels out (I kneew they wouldn't line up, but you see it!!) [1/ΔΑ 0]• ([1 3]+[2 2]+[ΔΑ 0]+[(ΔΑ)²0]-[1 3]) [0 1/ΔΑ] ([0 4] [0 4] [0 ΔΑ] [0(ΔΑ)²] [0 4]) as lim ΔA→0 Look at A² & -A² gone! canceled [1/ΔΑ 0]• ([2 2]+[ΔΑ 0]+[(ΔΑ)²0]) [0 1/ΔΑ] ([0 4] [0 ΔΑ] [0(ΔΑ)²]) as lim ΔA→0 now add those 3 matrices [1/ΔΑ 0][2+ΔΑ+(ΔΑ)² 2ΔΑ] [0 1/ΔΑ][ 0 4ΔΑ+(ΔΑ)²] as lim ΔA→0 Multiply [1/ΔΑ 0][2ΔΑ+(ΔΑ)² 2ΔΑ] [0 1/ΔΑ][ 0 4ΔΑ+(ΔΑ)²] as lim ΔA→0 = [(2ΔΑ+(ΔΑ)²)/ΔΑ 2ΔΑ/ΔΑ] [ 0/ΔΑ (4ΔΑ+(ΔΑ)²)/ΔΑ] as lim ΔA→0 = [(2+ΔΑ 2] [ 0 4+(ΔΑ)] as lim ΔA→0 = [(2+0 2] [ 0 4+0] = [ 2 2 ] [ 0 4 ] = 2A = d/dA[A²] therefore, it is possible to take the derivative of at least one 2×2 matrix = A s.t. ƒ[A] = A² & d/dA [A²] = 2A according to "the limit definition of a derivative" and the definition of a function, ƒ. ■ edit : I knew these wouldn't all line up, lol
@Roman Koval literally the very existence of an electron in a place in space is a probability, and electrons are building blocks for literally every material object
Sure we can take the derivative of a matrix! It just depends on what the function is. In this example shown in the video the function output is a vector. But it could have also been a matrix output. In that case we would have a rank 4 matrix as the derivative assuming inputs are two 2 dimensional tensors each. The main idea is to understand what a Jacobian matrix is and then you will see how all these are various special cases of that general idea. To rephrase, yes, we don' take a derivative of just any matrix as it makes no sense in the same way it doesn't make sense to take derivative of a vector. Derivative is defined for a function. But no matter what the output of a function is, be it scalar, vector, tensor or matrix, there is always a way to define its derivative.
A matrix inherently has discrete, integral indices, so it can't be differentiated, but you can differentiate a function whose coefficients are expressed by a matrix
I'd even go further and just say that you can identify a matrix with a vector in R^{nm} and use the idea of the Jacobian matrix like you talk about. I don't think it is necessary to go into the idea of rank unless you specifically care about tensor calculus. Even still, In that case, it is still basically vectors, except you might be taking tensor products with elements in the dual space. I absolutely agree that the main idea is to understand what a Jacobian matrix is
@@astrobullivant5908 The fact that the indices are discrete doesn't matter- a vector also has discrete indices! You don't differentiate with respect to the index number, but with respect to whatever variable each component depends on. If the matrix is constant, like [ 1 2 ; 3 4], then the derivative would just be the zero matrix.
Shouldn't the Derivate of x^T A x be a row vector? Since the derivative can be defined as the linear part of an approximation of the initial function. Here the function takes values from K^(2×1) to K, therefore a corresponding linear operator on x € K^(2×1) should be in K^(1×2) not K^(2×1)
At@ 9:41, could you tell me why you took a column vector and not a row vector? Is it a rule that we should take it as a column vector ? How to know what would be the shape of the matrix or vector?
@@NoahElRhandour Ding dong you're mr. wrong go back to zero. At 3:34 he writes (df_1/dx_1) etc. but uses normal d:s when he's writing out a derivate of a multi-variable function with respect to one of the parameters. This is known at a partial derivative and is written with a squiggly d, not a normal d. You could interpret his d:s as squiggly but in that case, he wrote out partial derivatives of single-variable functions with a squiggly d which is also incorrect notation. Really rude to call people who have a lesser degree of education morons (this isn't simple mathematics) and even worse to call people morons when they're right and you're wrong. Maybe there is a special notation that uses normal d:s when talking about partial derivatives of multi-variable matrix functions but I doubt it... And if it is the case, no one with that minor misunderstanding is a moron. Don't be a prick.
At 10:25 you said for 3 different functions and 4 different variables you'd have a 3x4 matrix. But the one you solved above only had 1 function and 2 variables x1 and x2. Why then did you create a 2x1 matrix instead of a 1x2?
Actually, the calculations for \frac{\mathrm{d} Ax}{\mathrm{d} x} you use the numerator-layout notation and the result is A, but when you compute \frac{\mathrm{d} x^T Ax}{\mathrm{d} x}, you use the denominator-layout notation which the result is 2Ax, and if you use the numerator-layout notation, the result should be 2 x^T X. Reference: en.wikipedia.org/wiki/Matrix_calculus
I found the same thing. xTAx = 2xTA instead of 2Ax. The difference is the result is a row vector instead of a column vector. I also used the same wikipedia resource for definitions.
Came here looking for LOWESS algorithm, and it turns out that the the derivative of xTAx plays a role in it. You helped me understand what matrix derivation is, plus solved my very particular need. Thanks.
Could you clarify my confusion with notation and shapes? This is critically important to understanding matrix implementations of backpropagation. At 10:25 you mentioned 3 functions and 4 vars will be 3x4 matrix. This is consistent with the Ax example where you wrote different f down the rows and different x across columns. However at the xTAx example it's 1 function with 2 variables so i was expecting a 1x2 matrix, but you arranged it as 2x1, why so? If I had set up the result to be 1x2 following your rule at 10:25, then the resulting derivative will not be Ax but xTA. I know when doing by hand we can transpose/setup anyhow as long as matrices multiply correctly but when it comes to implementing in programs, the direction of multiplication of the whole chain, with the matrices shapes (transpose or not) must be correct relative to each other, and this I can't find a good teaching source. From another video (2:50-3:03) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iWxY7VdcSH8.html&ab_channel=BenLambert , it emphasizes that the shape of resulting derivative must be same as the vector you're differentiating wrt. I do see how your xTAx example then ends up with a 2x1 which is same dimension as how x started (2x1 too), but then this "same shape rule" fails to apply to your Ax example, where the output shape became 2x2 which is not same as x shape of 2x1. Please help! I don't know who or what is right or wrong and which are conventions or rules.
Hi RItwik! One doubt ---> At 9:04 of the video, shouldn't the resultant matrix be 1*2 and not 2*1 as we are multiplying 1*2 and 2*2, so result should be 1*2 and not 2*1?. Please correct me if I am wrong. A fan of your videos!
Very impressive! I like how the total derivative "emerges" from the xtAx form. It really shows how effectively linear algebra notation can be used to assemble new structures. One comment I would make is that as far as I know when taking the partial derivative it is common to use ∂ instead of d.
With the xT A x case, we can let A be symmetric without loss of generality. If not, replace aij and aji with their mean, you get the exact same function. In fact you would not get that derivative to be 2Ax if A wasn't symmetric
He didn't say "let A be symmetric without loss of generality". Instead, he said that he will only focus on the case when A is symmetric because that is the case which we care about and can apply to "principal component analysis"
@@seanki98 I think you didn't understand what I'm saying. I mean, it is a FACT the A can be assumed symmetric without loss of generality. Thus, only considering the symmetric case invokes NO LOSS OF INFORMATION. for each square matrix A, define f_A(x) = x^T A x. it hold that "for all" such A, symmetric or not, "there exists" symmetric B, so that f_A, and f_B are identical functions. In fact B=(1/2)(A^T+A). DO YOU UNDERSTAND NOW YOU DUMB DIMWIT?
@@Grassmpl Pretty sure there's only one DUMB DIMWIT here, and it certainly isn't Sean. Oh, wait, I'm here too, that makes two. Seriously, man, insulting somebody for not automatically understanding your poorly-worded and difficult-to-read comment isn't cool.
Some advice: The kids that need this video most have likely learned about gradients. They may have heard of this concept as a 'hyper-gradient,' which is a less common way they can be taught in some schools. In either case, I've found that introducing it to them as a stack of gradients, one of f1 and one of f2, can help a lot. This puts things in terms that many kids would have already learned. Also, it may help to determine the ideal background of the viewers your targeting before making the video, just to crystallize the constraints you should be working with in making this video. If you do this, it's not apparent, and maybe identifying the ideal background explicitly can help. Finally, many concepts, especially differential operators like derivatives, may have other names. In this case, Jacobian is an obvious one. Listing these aliases may help students that need additional resources.
If someone could please clarify for me, why in the first example when we take d/dx(Ax) are the derivative functions in the rows of the matrix (i.e. row 1 was [df1/dx1, df1/dx2], and row 2 was [df2/dx1, df2/dx2] but when we take the derivative d/dx(x^TAx) the final matrix has the derivative functions of f1 in a column rather than a row (i.e. column 1 is [df1/dx1, df1/dx2]^T? Does it have something to do with x^T in the second derivative?
For the derivative of the transpose does that mean A has to be a square Matrix? or else the answer will have as many elements as there are rows in A, and then you couldnt go from there
How do we know whether to take df/dx1 and df/dx2 as a row or a column in the derivative? By your defination where you had f1 and f2, you wrote df1/dx1 and df1/dx2 along the columns. By transferring the same to f, shouldn't it be written along columns as well? In that case, the answer wont be 2Ax, but rather 2 * x.T * A.T
Sorry if my question is lame but at 10:24 you say that "If you have 3 different functions and 4 different variables, you have 3X4 matrix." Since we have 1 function and 2 different variables in the example, why don't we have 1X2 matrix instead of 2X1 matrix?
I'm new to your channel, I hace studies in engineering and economics so i'm not new to math per sé but I could use some order in reviewing this subjects, where should i start on your channel? i'm specializing in data science
I thought of it on this semester. If we consider the properties of the linearity of the derivative, I supposed that it must be distributed on the matrix. I'm still taking the subject of Linear Algebra but the video showed off some neat tricks for this type of problem
Could someone help to understand why the first matrix (2:31) was treated like two functions and the second matrix (9:07) was treated like a single function? :( I tried to pause and "convince myself" like ritvik said, but I still got no clue about it. Thanks!
Some intuition in me tell me that this can be applied to understand amplitude spilit up by a crystal by lasing and the third one interact to give interferometric functional as phase changes involved. S.Nandakumar
if i have a simple x^2 do i need to express it into x times x, where x is just a simple unknown, and then i need to express one of the x in terms of matrix Ax like in this video and the other x in term of vector ? so that i can define what differentiation is to the computer, or do i have other option?
Who are the 226 people who didn't like the video? Maybe the ones who didn't understand why the derivative of kx = k, and the derivative of kx^2 is 2kx. This is mind-blowingly intuitive. I've never heard a matrix being called a bunch of scalars in a box. All the videos made by ritvikmath are excellent videos. Although I have used Eigenvalues, Eigenvectors, and derivatives of linear combinations extensively, it never made this kind of intuitive sense.
If A is a scalar function it's derivative with respect to a vector x is a vector. It's like taking the gradient of a scalar field, you'll get a vector. The physical meaning is that you're computing how that scalar function "evolves" along the "directions" specified by the vector along which you're computing the derivative
@@UmbertoBettinardi thanks! I actually got a lot of answers from this video on the subject ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-IgAr5kzza78.html It's pretty good and talks about different kinds of differentiation.
According to mathematical equation event are fixed after calculation of all types of possibilities so how we found that equation which shows all types of stability ???
In first example u take one row for one function and one column for one variable, but in second example one row for one variable .Not consistent.please advise.
Hi, I think there is a slight mistake on the last part. The results of the derivative should've been a 1x2 matrix instead of a 2x1. Therefore the derivative is equal to 2x'A instead of 2Ax. cmiiw, thanks.
