The example at 12:30 is not really an inner product. It can happen that =0 but f is not the zero function. For Example you can take the function that is always 0 on [a,b) and that is 1 in b. This function is integrable on [a,b] and it's such that =0 but it's not the zero function. If you want to actually be an inner product you need to take the quotient of the space V with respect to the equivalence relation "almost everywhere". However the ideas of generalizing the concept of orthogonality can go very far and lead to wonderful concepts like Hilbert spaces. This abstract concept are the the bricks of the Fourier series with which many concrete problems can be solved. when I studied these topics I was impressed by all of this.

@BrollyyLSSJ

Жыл бұрын

It's also sufficient to restrict to the subspace of continous functions on [a, b], but you do lose completeness this way.

@mrnanisissa

Жыл бұрын

@@BrollyyLSSJ Yeah, that's true

@Alex_Deam

Жыл бұрын

@@BrollyyLSSJ What's meant by completeness?

@mrnanisissa

Жыл бұрын

@@Alex_Deam The fact that every cauchy sequence is convergent

@BrollyyLSSJ

Жыл бұрын

@@Alex_Deam What @Mr.Nani said. As an example, consider sequence of continuous functions on [-1, 1]: f_k(x) = {0 if -1

This was wild! I've always been given the idea of orthogonality from the top and then down - it's really enlightening and useful to see it from the other direction. In particular, it was amazing to see that the triangle directly led to the dot product. I can't believe I've never seen that before. And of course, naturally in the case of diff eqs and the like, polynomials and trig functions for Fourier series. Love it. I ran into Hermite polynomials while looking into interpolation with derivatives - incredible!

@pyropulseIXXI

Жыл бұрын

The triangle doesn't lead directly to the dot product. The dot product is a definition, and what you just saw can motivate someone to define the dot product as such. But that doesn't mean the triangle led directly to the dot product

Spoiler alert. For those who want to check their homework. Here's my working to find the family of matrices B that are orthogonal to A = ( 1 3 || 0 2) (I'm using || to indicate a new line inside the matrix): B is a 2x2 matrix, so we can set B = ( a b || c d ). Then we calculate transpose of A multiplied by B: (1 0 || 3 2) ( a b || c d ) = ( a b || 3a+2c 3b+2d ) and then its trace is a + 3b+2d which we set equal to 0. That gives a single condition, for example a = -3b-2d. So B = ( -3b-2d b || c d ) where b, c, d are free values.

@kaay8983

Жыл бұрын

With a little more work you can also find the orthogonal decomposition of this 3-dimentional space, for example: (0,2;2/sqrt(3),-3) (2,0;-2*sqrt(3),-1) and (3,-1;sqrt(3),0)

@maxmustermann3938

9 ай бұрын

c d eez nuts

This is nice. I like how you tied it into Legendre poly's. I did some work projecting droplet perimeters onto the Legendre poly's and fit the coefficients to a general linear model to predict new shapes.

great video! i love you algebra videos, they're really interesting!

Wow that was an incredibly dense video. Thank you. I'll definitely be rewatching it.

This was a very good video. I would have liked a few ideas of applications. I quite like the call outs. Thank you, professor.

That was a very nice presentation! Could you also introduce the complex inner product and elaborate on the use of the bilinear form as a broadened definition of the inner product/measure of orthogonality for instance when used in the context of an equipped/rigged Hilbert space?

Thanks

Maybe it's a good idea to connect the subject of this video to Lebesgue integration as it is a good stepping stone towards Fourier analysis etc.

Orthogonality plays an important role in engineering. For example, the correlation between two time series can be defined as an inner product, and if this is 0, then the time series are defined to be orthogonal. Two noise signals are an example of orthogonal signals. Another example is Orthogonal Frequency-Division Multiplexing (OFDM), which is used in your smart phone.

@sharpnova2

Жыл бұрын

rofl, so cute when engineers try to wax mathematical. it's like a baby's first words.

@1080lights

Жыл бұрын

@@sharpnova2 thanks, Mr. BS Mathematics. Can I get fries with my order?

@agustinmiranda3989

Жыл бұрын

​@@sharpnova2 This is a very toxic comment, SurfinScientist is just sharing some knowledge from their field.

@mathadventuress

Жыл бұрын

@@sharpnova2 they’re applying math The fuck is your problem? Get back to fast food

@charleyhoward4594

Жыл бұрын

then the time series are defined to be orthogonal -- what does that mean that "time series are orthogonal" ?

26:02 b, c, d in R, matrix with such elements: (-3b -2d, b, c, d)

[Comment for the engagement algorithm] love it! I never realized this simple transition from Pythagoras to inner product of vectors. For the homework, I thought you would be asking about the Pauli spin matrix, or will that be the exam ? ;)

Good use of the board tap at ~9:30!

Finally A Nice Explanation 😍😍😍

Fun, thanks

Oh, that's why we care about the inner product. It always felt like it came out of nowhere.

I read Gram-Schmidt as Granny Smith.

Could you do more on deriving the Chebyshev/Hermite polynomial using gram-schmidt?

That was an action packed video. Linear Algebra in a minute

Could you in theory find a weight such that the orthogonal polynomials will be some unrelated polynomials, say Bernoulli or Stirling? I do understand that it's probably hard if not impossible to calculate it, but is the existence of one sure?

15:25 if you didn't want to start with the assumption that the orthogonal function had to be linear you'd have to do a series representation of a polynomial function where n is in the integers. Then you'd have to do a bunch of special cases.

awesome

Why did the answer need to be linear? Couldn’t it have an arbitrary dimension?

26:09

In a matrix of the form (a b || c d), c can be any constant, while the other 3 must obey the property of a + 3b + 2d = 0. This equation says little more than the definition of the inner product itself, but it is nicely represented as an equation for a plane in 3D space. One thing I'm a little sad wasn't mentioned is how by asserting the Pythagorean theorem, you automatically disqualify hyperbolic and spherical geometry. In the latter, it's completely legal to make an equilateral triangle using 3 right angles. In hyperbolic spacetime, if you try to find the "inner product" of a light-like velocity or momentum vector with itself, you get a value of 0 despite the momentum itself not being the 0 vector. In addition while it's likely impossible in physical spacetime, it's also mathematically possible to describe what would likely be tachyons by using a momentum vector with a _negative_ "inner product" with itself.

@angeldude101

Жыл бұрын

@@angelmendez-rivera351 So everything we do with inner products and orthogonality can't actually apply to physical spacetime? That seems like a serious deficiency. It doesn't seem right to try to generalize things and then still exclude very useful applications that are missing only 1 arbitrary property.

I feel like I'm back in quantum, except that I wish I'd seen this first!

What do you do for physical fitness? Are you only climbing, or also in the gym?

24:48 The question here is why spend time even calculating all the entries in the matrix. We should probably just calculate the diagonal entries and add them up.

@abrahammekonnen

Жыл бұрын

Lol as soon as I say it you mention it. Great minds think alike ;)

Can a vector space have multiple distinct products defined for it which meet the definition of an inner product? And can two functions which both meet the definition of an inner product disagree on whether a pair of vectors are orthogonal?

This was interesting. How could we generalize orthagonality to non Euclidean geometries?

Isn’t the distance formula derived from the Pythagorean theorem?

@cH3rtzb3rg

Жыл бұрын

Once you define what a scalar product is, you can easily show that it defines a norm, via |v| = sqrt , and then you can trivially proof Pythagoras' Theorem: |a|²+|b|² = |a-b|² + 2, i.e., |a|²+|b|² = |a-b|² iif. =0.

@minwithoutintroduction

Жыл бұрын

تذكير رائع كالعادة.

@CM63_France

Жыл бұрын

I agree with you @James Mosher, the Pythagorian theorem not is really a "good place to start" 😄

Nice

I kinda want to see how this works out in characteristic 2 (I mean, I'm too lazy to work it out for myself... though maybe that would be a good exercise)

@pyropulseIXXI

Жыл бұрын

In characteristic 2, 2 = 0, so 2(ac+bd)=0 means that does not have to equal 0 for it to be orthogonal. This means that any values of a, b, and c would lead to a 90 degree angle in characteristic 2.

how does this relate to tensors?

I want to make sure I'm not misunderstanding: is the idea that "orthogonal" for lines can be different from "perpendicular" (where, in the example, I would have expected b=-1/6)

@Eye-vp5de

Жыл бұрын

It depends on the definition of inner product. In the video it was defined with an integral, which doesn't seem to have a nice geometrical meaning for me, but you can define inner product as an inner product of vectors of length which lie on the lines to get orthogonality = perpendicularity

Why is the 1st degree Laguerre polynomial written as -(-x-1) instead of just x+1?

14:56 Think there's a mistake here, because if you integrate (4-6x)*(1+6x^2), you will also get 0 i.e. 1+6x^2 (and multiples thereof) is an example of non-linear polynomial that's orthogonal to 4-6x.

@vinvic1578

Жыл бұрын

why would that be a mistake ? the task is to find a polynomial thats orthogonal to 4-6x. In the video he finds a first degree one, and you find a second degree one. But because R[X] is an infinite dimensional space, every polynomial will have an infinite number of linearly independent orthogonal polynomials. You can therefore find polynomials of any degree that are orthogonal to it :) Does that make sense or did I misunderstand your point ?

@Alex_Deam

Жыл бұрын

@@vinvic1578 It's a mistake because Michael says that "we're going to assume it's linear because if you work through it with a polynomial of arbitrary degree, you'll see it has to be linear" which is what I did do and found it wasn't true. It doesn't have to be linear.

@vinvic1578

Жыл бұрын

@@Alex_Deam Oh yes sorry ! That makes sense. Yeah I'm a bit confused as to why he said that now ! Thanks for clearing it up for me :)

@polyhistorphilomath

9 ай бұрын

@@vinvic1578 probably because the orthonormal bases enumerated are straightforwardly independent if the initial basis vectors are of strictly increasing degree. IIRC -1, ax+b, cx^2+hx+k showed up at least 3 or 4 times in the examples.

Interesting

Since matrices are not commutative, it is not obvious that = in this case.

@APaleDot

Жыл бұрын

In order for it to be an inner product must equal . If the given operation does not always satisfy that condition, it is not an inner product. Since he states that tr(A^T B) _is_ an inner product, then it must satisfy that condition, although it is a good exercise to check for yourself.

@byronwatkins2565

Жыл бұрын

@@APaleDot Frobenius said it before he did. Neither of these facts make it obvious that the statement is true. A better demonstration would have addressed this unexpected result explicitly... however briefly.

@APaleDot

Жыл бұрын

@@byronwatkins2565 Like I said, it's good to check for yourself, but you can't expect a lecturer to prove every little statement they make especially when he's just running down a list of examples. There's nothing wrong with simply stating that such an operation is an inner product, which necessarily includes the symmetric condition. There is such a thing as an irrelevant detail.

@bjornfeuerbacher5514

Жыл бұрын

That = in that case is rather obvious - one only has to know that tr(A^T) = tr(A) and that (A^TB)^T = B^TA. However, what is not obvious (at least to me) is that is always non-negative in this case and that = 0 only for v = 0.

interesting

The PT is only for right angled triangles.

Something about Fourier...

The flow chart mentions a process for generating families of functions, but what are those family definitions and how are they used? Recreational math fan here.

@bjornfeuerbacher5514

Жыл бұрын

The families are defined exactly as given in the flowchart: start with the scalar product given there, do a Gram-Schmidt orthonormalization of the polynomials (starting with x^0, then x^1, then x^2 and so on). The polynomials you will get by this process are exactly these families of functions. (There are other ways to define them, too, but that's one way to define them.)

@BridgeBum

Жыл бұрын

@@bjornfeuerbacher5514 Cool, ok. But what separates these starting values or these families from any other starting values? That is, why are they interesting?

@bjornfeuerbacher5514

Жыл бұрын

@@BridgeBum I don't know about the Chebyshev polynomials, but all the others one are important in physics: the Legendre polynomials for the spherical harmonics, which are used to describe many spherical symmetric systems (among other things, the angular dependence of electromagnetic waves produced by antenna and the angular dependence of atomic orbitals); the Hermite polynomials for the harmonic oscillator in quantum mechanics; and the Laguerre polynomials for the radial dependence of atomic orbitals.

@BridgeBum

Жыл бұрын

@@bjornfeuerbacher5514 Thank you, that's a helpful perspective.

Ah, but why is orthogonality so _important_ in Maths? Is it because it hugely simplifies the problem of representing vector in different bases?

@APaleDot

Жыл бұрын

I've been thinking about this for a while and here's the best explanation I can come up with: Orthogonality is important because it represents the _complete_ independence of two things. We say that a set of vectors is linearly independent so long as each one adds something "new" to the set. As soon as you add a vector which can be represented as a linear combination of the other vectors, the set is no longer linearly independent. This new vector can be represented in terms of the other vectors, so it isn't adding any "new" dimension to the set and thus is not independent from the set in an important way. What is this "newness" I'm talking about? Well, any time you have two vectors, you can represent one of the vectors as a combination of components, one which is parallel and one which is orthogonal to the other vector. The parallel component is obviously just a scalar multiple of the other vector, so it's really only the orthogonal component which is "different" from the other vector. This is the "newness" that is added to the set, it is a new direction _orthogonal_ to the rest of the set. This complete independence is very useful when trying to analyze real world events. Consider rolling two dice. In order for the dice to be fair, the two rolls _must_ be independent from each other. The outcome of the rolls should be orthogonal in a very real sense. This is what makes calculating probabilities of such independent events so easy. Calculating probabilities of events which depend on each other is much more difficult because you first have to account for the effect of one event on the other (i.e, you have to remove the parallel component of the "vectors"). In math, this independence means that if you have two subspaces which are orthogonal to each other, you can move a point parallel to one of them without affecting it's component in the other one at all. This obviously makes calculations much simpler when you can consider only the actions happening in a particular subspace because they are completely independent from the rest of the space.

@hybmnzz2658

Жыл бұрын

Spectral theorem

@stephenhamer8192

Жыл бұрын

@@APaleDot Very good answer and nice 2-dice example. Only (the part of) an element orthogonal to the set adds new information - how true that is, not just in Maths but in life generally

@stephenhamer8192

Жыл бұрын

@@hybmnzz2658 Study of. Adding it to my bucket list

By characteristic 2 I'll assume mod 2.

@super_pigeon

Жыл бұрын

He meant this on the field of coefficients: en.wikipedia.org/wiki/Characteristic_(algebra)

@abrahammekonnen

Жыл бұрын

@@super_pigeon oh I see. Thank you

@schweinmachtbree1013

Жыл бұрын

@@abrahammekonnen the integers mod 2 are the simplest example of a field of characteristic 2 though

It would of been easier to understand if you drew a right angle triangle.

I got B = {a, b, // c , d}, a = -3t - 2s, b = t, c = w (free), d = s.

eRRRmeet not her-mite

Hi, I think the easiest way to define a right angle is to say it's half of a flat angle. 15:59 : missing dx 18:27 : what is this Schmidt orthogonalization process? (I am not sure to remember). 18:56 [-1,1] instead of [-1,-1]

@sillymel

Жыл бұрын

(18:27) I would recommend just searching the internet for “Gram-Schmidt orthogonalization process.” It’s a bit too convoluted to properly explain in a comment.

@sillymel

Жыл бұрын

Since it’s not letting me edit my comment: That’s not to say that the Gram-Schmidt orthogonalization process is particularly difficult, just long.

@bjornfeuerbacher5514

Жыл бұрын

The basic idea of the Gram-Schmidt orthonormalization procedure is that given a set of already orthonormal basis vectors, one takes a new vector and subtracts every parts from it which are parallel to all the given basis vectors. Hence the vector one arrives at will be orthogonal to all the other given basis vectors. Finally, normalize it. Then take another new vector and do the process again etc.

@CM63_France

Жыл бұрын

About the GSOP : ok, thank you for your answers, it comes back to me now. About the video: would it be possible to demonstrate the Pythagorean theorem starting from this definition of the right angle: half of a flat angle.

Aren’t you using Pitagora to prove Pitagora in the vector example?

@scp3178

Жыл бұрын

Pitagora = Pythagoras?

@wafikiri_

Жыл бұрын

It is not proving Pythagoras's theorem that is sought, but a definition of what is orthogonal, based on the above theorem in Euclidean space.

@Macisordi

Жыл бұрын

@@wafikiri_ how do you define distance in a plane without using Pythagoras, p.s. my Italian iPad autocorrect Pitagora

@wafikiri_

Жыл бұрын

@@Macisordi I wouldn't. The Cartesian co-ordinate system, with orthogonal co-ordinates, practically requires Pythagoras to define a metric. To renounce to Cartesian co-ordinates in a plane would mean to use a tougher metric, whose distance formulae would surely be cumbersome.

I thought he was going to use the language of quantum mechanics.... he is REALLY close to doing applied mathematics here.

@schweinmachtbree1013

Жыл бұрын

I disagree - inner product spaces, bilinear forms, function spaces, etc. are all things that are taught in courses on a pure math degree. Some of the examples were close to measure theory and Fourier series, but again these are pure math theories; it is when you apply the theories, e.g. to quantum mechanics, that it becomes applied mathematics

/ɛʁ.mit/, not /həɹmait/, and "Chebyshev" should be yoficated and end in stressed "shov". If the field has √-1 and the vector space has at least 2 dimensions, a line can be perpendicular to itself. This is called a null line.