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Introduction to orthonormal bases

Looking at sets and bases that are orthonormal -- or where all the vectors have length 1 and are orthogonal to each other. Created by Sal Khan.

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  • blobby green style avatar for user durval.menezes
    What are the prerequisites for this lesson? How do I determine what other videos I need to watch in order to understand this one?

    I have < 1 week (for a Quantum Computing course), it mentions specifically this and one other Linear Algebra topic (eigenvalues/vectors). I've been serially watching every video in the "Linear Algebra" section from the beginning, but there will not be enough time.

    So, how to determine what videos I can skip in order to reach this one and be able to understand it?
    (3 votes)
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  • leaf green style avatar for user sst
    All these concepts are directly applied to electrons in atoms. In that sense, if I consider Vi be the wave function of i^th electron, is it correct to consider as follow:

    Normalized vector Vi : the total probability of finding the electron is 1 (Vi.Vi=1)
    No two electrons can be in the same place (becasue Vi.Vj=0)
    Vi and Vj are Linearly independent : One electron does not cross each other electrons position.

    Please give corrections and suggestions for further reading of basic level for this.
    (6 votes)
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  • leaf orange style avatar for user Nicholas Eckert
    If you have a set of 30 vectors in r2 how can they all be orthogonal to each other? It seems like you could have at most 2?
    (1 vote)
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    • leaf green style avatar for user Johink
      That's correct; you could never have more than two vectors in R2 and have them all be orthogonal to one another. To see a visual example of this, try drawing three straight lines (vectors in R2) such that each line intersects the origin and is perpendicular to the other two lines.
      (4 votes)
  • blobby green style avatar for user Kyle Olson
    If you have a orhonormal basis set u, then is their inner product <u|u> defined to be 1?
    (2 votes)
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    • mr pants teal style avatar for user Moon Bears
      I think you're confusing sets and their elements. An orthonormal basis is a set of vectors, whereas "u" is a vector. Say B = {v_1, ..., v_n} is an orthonormal basis for the vector space V, with some inner product defined say < , >. Now <v_i, v_j> = d_ij where d_ij = 0 if i is not equal to j, 1 if i = j. This is called the kronecker delta. This says that if you take an element of my set B, such as v_1 and consider <v_1 , v_1> then this value must be 1. If the subscript isn't 1 then you will always get zero! The short answer is yes, but you had a slight conceptual mishap in your question.
      (1 vote)
  • blobby green style avatar for user NateJCho
    Must a scalar multiple of an orthogonal matrix orthogonal as well? Is this answered in another video?
    (2 votes)
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    • leaf green style avatar for user Lucas Van Meter
      Do you mean that if "M" is an orthogonal matrix is "kM" orthogonal? If so lets check the definition. I would recommend trying some examples.

      "kM" is orthogonal if all of its columns are unit vectors. But if "M" was orthogonal and we multiply a "k" into "M" somewhere it will multiply one of the columns by a scaler that is not 1 so that column will no longer be a unit vector.
      (1 vote)
  • male robot hal style avatar for user Andrew Yang
    Is it called "Orthonormal bases" or "Orthonormal basis"?
    It was "bases" in the title, but he said and wrote (as at ) "basis"
    (1 vote)
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  • piceratops seedling style avatar for user monikilla_am
    For expressing the dot product of the vectors, shouldn't we put the first vector transposed?
    (1 vote)
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    • male robot hal style avatar for user Yamanqui García Rosales
      If you treat the vectors as 1-column matrices, then yes, in order to do the dot product you have to put express your first vector as a 1-row matrix. But if you are using normal vector notation (as most of the video does) then you are not committed to the matrix representation of vectors, and as such each vector can be seen as either a 1-column matrix, a 1-row matrix, a tuple of numbers or even as an arrow in space.

      In notation, there is no difference between:
          ⎡a_x⎤
      a = ⎥a_y⎥
      ⎣a_z⎦

      a = [a_x a_y a_z]
      (2 votes)
  • blobby green style avatar for user jafis2374
    What are the coordinates for the translation of a triangle given the matrix addition ? Or yet how can I solve the problem?
    (1 vote)
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  • blobby green style avatar for user Austin Chen
    If we state that Vi and Vj are orthogonal to each other, how can we say that they are linearly dependent? (I'm confused how we can say Vi = cVj, if the two vectors are orthogonal)
    (1 vote)
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    • blobby green style avatar for user Nausheen Qaiser
      We wanted to prove that Vi and Vj are linearly independent, and to prove it we used "proof by contradiction", in which we take initial assumption and find some contradiction to that assumption and then negate our initial assumption. That's why here we first assumed that let's say Vi and Vj are NOT linearly independent (that is they are linearly dependent) by writing Vi = cVj and then finding a contradiction which disproved our initial assumption that they are linearly dependent, and that's how we proved that Vi is not equal to cVj and thus Vi and Vj are linearly independent.
      (1 vote)
  • blobby green style avatar for user stasis.trap
    I dont understand what goes on here
    at
    you wrote vi.vj=0 (which is true for orthogonal or linearly independent vectors)

    then you supposed that vi and vj are linearly dependent and substituted the value of vi=cvj in the equation where the result is only true when the vectors are linearly independent
    It will never come out to be true
    Where am i going wrong?
    (1 vote)
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    • aqualine ultimate style avatar for user Arcanine
      That is the point. He is constructing a proof by contradiction, the point of which is to prove that something is true by showing it cannot be false. Sal arrived at 0 = 1 by assuming that vi and vj were L.D., demonstrating that assumption to be impossible.
      (1 vote)

Video transcript

Let's say I've got me a set of vectors. So let me call my set B. And let's say I have the vectors v1, v2, all the way through vk. Now let's say this isn't just any set of vectors. There's some interesting things about these vectors. The first thing is that all of these guys have length of 1. So we could say the length of vector vi is equal to 1 for i is equal to-- well we could say between 1 and k or i is equal to 1, 2, all the way to k. All of these guys have length equal 1. Or another way to say it is that the square of their lengths are 1. The square of a vi whose length is equal to 1. Or vi dot vi is equal to 1 for i is any of these guys. Any i can be 1, 2, 3, all the way to k. So that's the first interesting thing about it. Let me write it in regular words. All the vectors in B have length 1. Or another way to say is that they've all been normalized. That's another way to say that is that they have all been normalized. Or they're all unit vectors. Normalized vectors are vectors that you've made their lengths 1. You're turned them into unit vectors. They have all been normalized. So that's the first interesting thing about my set, B. And then the next interesting thing about my set B is that all of the vectors are orthogonal to each other. So if you dot it with itself, if you dot a vector with itself, you get length 1. But if you take a vector and dot it with any other vector-- if you take vi and you were to dot it with vj. So if you took v2 and dotted it with v1, it's going to be equal to 0 for i does not equal j. All of these guys are orthogonal. Let me write that down. All of the vectors are orthogonal to each other. And of course they're not orthogonal to themselves because they all have length 1. So if you take the dot product with itself, you get 1. If you take a dot product with some other guy in your set you're going to get 0. Maybe I can write it this way. vi dot vj for all the members of the set is going to be equal to 0 for i does not equal j. And then if these guys are the same vector-- I'm dotting with myself-- I'm going to have length 1. So it would equal length 1 for i is equal to j. So I've got a special set. All of these guys have length 1 and they're all orthogonal with each other. They're normalized and they're all orthogonal. And we have a special word for this. This is called an orthonormal set. So B is an orthonormal set. Normal for normalized. Everything is orthogonal. They're all orthogonal relative to each other. And everything has been normalized. Everything has length 1. Now, the first interesting thing about an orthonormal set is that it's also going to be a linearly independent set. So if B is orthonormal, B is also going to be linearly independent. And how can I show that to you? Well let's assume that it isn't linearly independent. Let me take vi, let me take vj that are members of my set. And let's assume that i does not equal j. Now, we already know that it's an orthonormal set. So vi dot vj is going to be equal to 0. They are orthogonal. These are two vectors in my set. Now, let's assume that they are linearly dependent. I want to prove that they are linearly independent and the way I'm going to prove that is by assuming they are linearly dependent and then arriving at a contradiction. So let's assume that vi and vj are linearly dependent. Well then that means that I can represent one of these guys as a scalar multiple the other. And I can pick either way. So let's just say, for the sake of argument, that I can represent vi-- let's say that vi is equal to sum scalar c times vj. That's what linear dependency means. That one of them can be represented as a scalar multiple of the other. Well if this is true, then I can just substitute this back in for vi. And what do I get? I get c times vj-- which is just another way of writing vi because I assumed linear dependence. That dot vj has got to be equal to 0. This guy was vi. This is vj. They are orthogonal to each other. But this right here is just equal to c times vj dot vj which is just equal to c times the length of vj squared. And that has to equal 0. They are orthogonal so that has to equal 0. Which implies that the length of vj has to be equal to 0. If we assume that this is some non-zero multiple, and this has to be some non-zero multiple-- I should have written it there-- c does not equal 0. Why does this have to be a non-zero multiple? Because these were both non-zero vectors. This is a non-zero vector. So this guy can't be 0. This guy has length 1. So if this is a non-zero vector, there's no way that I can just put a 0 here. Because if I put a 0 then I would get a 0 vector. So c can't be 0. So if c isn't 0, then this guy right here has to be 0. And so we get that the length of vj is 0. Which we know is false. The length of vj is 1. This is an orthonormal set. The length of all of the members of B are 1. So we reach a contradiction. This is our contradiction. Vj is not the 0 vector. It has length 1. Contradiction. So if you have a bunch of vectors that are orthogonal and they're non-zero, they have to be linearly independent. Which is pretty interesting. So if I have this set, this orthonormal set right here, it's also a set of linearly independent vectors, so it can be a basis for a subspace. So let's say that B is the basis for some subspace, v. Or we could say that v is equal to the span of v1, v2, all the way to vk. Then we called B-- if it was just a set, we'd call it a orthonormal set, but it can be an orthonormal basis when it's spans some subspace. So we can write, we can say that B is an orthonormal basis for v. Now everything I've done is very abstract, but let me do some quick examples for you. Just so you understand what an orthonormal basis looks like with real numbers. So let's say I have two vectors. Let's say I have the vector, v1, that is-- say we're dealing in R3 so it's 1/3, 2/3, 2/3 and 2/3. And let's say I have another vector, v2, that is equal to 2/3, 1/3, and minus 2/3. And let's say that B is the set of v1 and v2. So the first question is, is what are the lengths of these guys? So let's take the length. The length of v1 squared is just v1 dot v1. Which is just 1/3 squared, which is just 1 over 0. Plus 2/3 squared, which is 4/9. Plus 2/3 squared, which is 4/9. Which is equal to 1. So if the length squared is 1, then that tells us that the length of our first vector is equal to 1. If the square of the length is 1, you take the square root, so the length is 1. What about vector 2? Well the length of vector 2 squared is equal to v2 dot v2. Which is equal to-- let's see, two 2/3 squared is 4/9-- plus 1/3 squared is 1/9. Plus 2/3 squared is 4/9. So that is 9/9, which is equal to 1. Which tells us that the length of v2, the length of vector v2 is equal to 1. So we know that these guys are definitely normalized. We can call this a normalized set. But is it an orthonormal set? Are these guys orthogonal to each other? And to test that out we just take their dot product. So v1 dot v2 is equal to 1/3 times 2/3, which is 2/9. Plus 2/3 times 1/3, which is 2/9. Plus 2/3 times the minus 2/3. That's minus 4/9. 2 plus 2 minus 4 is 0. So it equals 0. So these guys are indeed orthogonal. So B is an orthonormal set. And if I have some subspace, let's say that B is equal to the span of v1 and v2, then we can say that the basis for v, or we could say that B is an orthonormal basis. for V.