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dim(v) + dim(orthogonal complement of v) = n

Showing that if V is a subspace of Rn, then dim(V) + dim(V's orthogonal complement) = n. Created by Sal Khan.

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  • blobby green style avatar for user bfu12
    if V is a basis, then doesn't that imply that its columns are linearly independent and that the dimension of V perp is 0 since the dim of A = dim of A transpose (so dim nulA = 0 = dim nulA transpose)
    (6 votes)
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    • blobby green style avatar for user BrianLHouser
      I think where you're confused is that V is not a basis. V is a subspace (see beginning of video). Then, Sal defined k vectors which form a basis for V.

      Also, V doesn't have columns because it's not a matrix, it's a subspace. To be fair, you could represent it as a matrix, but if k > 0 like in the example, it would have an infinite number of columns (all the linear combinations of the basis vectors v1 to vk)
      (10 votes)
  • hopper cool style avatar for user Cole Wyeth
    Is there such a thing as a 3-d matrix? Like a box of numbers in three dimensions?
    I feel like that should exist for some reason... a n x m x k matrix?
    (4 votes)
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  • mr pants teal style avatar for user Moon Bears
    Isn't this the rank and nullity theorem?
    (2 votes)
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    • aqualine ultimate style avatar for user Aaron Williams
      The result is essentially the rank-nullity theorem, which tells us that given a m by n matrix A, rank(A)+nullity(A)=n. Sal started off with a n by k matrix A but ended up with the equation rank(A transpose)+nullity(A transpose)=n. Notice that A transpose is a k by n matrix, so if we set A transpose equal to B where both matrices have the same sizes and entries, then you get the standard form for the rank-nullity theorem, which is rank(B)+nullity(B)=n.
      (5 votes)
  • blobby green style avatar for user Jason
    why do we have to use the column space of V to represent V instead of using the row space?
    (3 votes)
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    • leaf green style avatar for user Gobot
      He starts with V, and then wants to talk about a basis for it.. so declares the vectors that make up the basis. I'm not sure there is any more special reason than it is more common to use column vectors.

      If he declared them as row vectors he would get to the same conclusion via the rowspace of V being the orthogonal complement of the Nullspace N(V).
      (2 votes)
  • piceratops ultimate style avatar for user Luvneesh kumar
    rename this as rank-nullity theorem to make it easier to find
    (3 votes)
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  • male robot hal style avatar for user ledaneps
    When you are indicating the number of rows and columns in a matrix, you usually choose from k,m, and n. In a square matrix, obviously, you use the same letter for both rows and columns.
    Is there some rationale as to when you use an n x k matrix as opposed to an n x m matrix.
    In other words, why did you choose k? Does it imply a different relationship to n than does m?
    I hope I explained what confuses me clearly. If you understand what I am trying to say, can you help me word it better before you answer it?
    Thanks.
    (2 votes)
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    • leaf green style avatar for user Radosław Rusiniak
      Short answer: Usually m>=n>=k.
      Long answer: I think he chose k, because he wanted to concentrate on the vectors (columns) more and show that it doesn't have to be equal to n, but it really doesn't matter that much. Most notation in math is as it is, because someone wrote it that way and from that point everyone else started to follow him to not confuse people with different notation. Like in your case, usually less used k shows up and you start to be little confused or at least curious :)
      (2 votes)
  • leaf green style avatar for user ERNEST FOSU
    whats is n? Is it the number of rows of a matrix formed from column vectors of V or the number of column vectors of V ?
    (2 votes)
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  • blobby green style avatar for user Simen
    If A is visualized in n-dim space, the dimensions that A cannot reach is null(A-transpose).
    Is that what null(A-tran) really is?
    The rref(A) shows where it spans, why does A have to be transposed to know where it cannot span?
    (2 votes)
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    • eggleston blue style avatar for user DonKiin
      From what I understood, null(A-trans) just represents the part of the overall vector that includes the free variables (a set of redundant values), hence finishing the full "k" length of a vector (along with the pivot variables that are present).
      (1 vote)
  • marcimus pink style avatar for user Erika Osorio
    i am having a really hard time with this.
    (2 votes)
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  • blobby green style avatar for user Mohamad Shebli
    Why don't we say that the null space of A equals the orthogonal complement of the column space?
    i mean orthogonal complement is the set of all x'es that satisfy this equation x1v1+x2v1+...+xnvn ?where v1,v2....vn are the columns of A
    i am really confused
    (1 vote)
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    • primosaur ultimate style avatar for user Derek M.
      Let A be an m x n matrix. Null space vectors live in R^n. Vectors in the column space live in R^m.
      Vectors in the orthogonal complement of the column space still live in R^m. Unless m=n, there is no way to compare R^n vectors to R^m.

      For example, there is no notion of adding a triple (1, 0, 2) to the pair
      (5, -6), or asking how we could compare the two vectors.
      (2 votes)

Video transcript

Let's say I've got some subspace of Rn called V. So V is a subspace of Rn. And let's say that I know its basis. Let's say the set. So I have a bunch of-- let me make that bracket a little nicer-- so let's say the set of the vectors v1, v2, all the way to vk, let's say that this is equal to-- this is the basis for V. And just as a reminder, that means that V's vectors both span V and they're linearly independent. You can kind of see there's a minimum set of vectors in Rn that span V. So, if I were to ask you what the dimension of V is, that's just the number of vectors you have in your basis for the subspace. So we have 1, 2 and we count to k vectors. So it is equal to k. Now let's think about, if we can somehow figure out what the dimension of the orthogonal complement of V can be. And to do that, let's construct a matrix. Let's construct a matrix whose column vectors are these basis vectors. So let's construct a matrix A, and let's say it looks like this. First column is v1. This first basis vector right there. v2 is the second one, and then you go all the way to vk. Just to make sure we remember the dimensions, we have k of these vectors, so we're going to have k columns. And then how many rows are we going to have? Well, as a member of Rn, so these are all going to have n entries in each of these vectors, there's going to be an n-- we're going to have n rows and k columns. It's an n by k matrix. Now, what's another way of expressing the subspace V? Well, the basis for V is-- or V is spanned by these basis vectors, which is the columns of these. So if I talk about the span-- so let me write out this-- V is equal to the span of these guys, v1, v2, all the way to vk. And that's just the same thing as the column space of A. Right? These are the column vectors, and the span of them, that's equal to the column space of A. Now, I said a little while ago, we want to somehow relate to the orthogonal complement of V. Well, what's the orthogonal complement of the column space of A? The orthogonal complement of the column space of A, I showed you-- I think it was two or three videos ago-- that the column space of A's orthogonal complement is equal to-- you could either view it as the null space of A transpose, or another way you call it is the left null space of A. This is equivalent to the orthogonal complement of the column space of A, which is also going to be equal to, which is also since this piece right here is the same thing as V, you take it's orthogonal complement, that's the same thing as V's orthogonal complement. So if we want to figure out there orthogonal complement of-- if we want to figure out the dimension-- if we want to figure out the dimensional of the orthogonal complement of V, we just need to figure out the dimension of the left null space of A, or the null space of A's transpose. Let me write that down. So the dimension-- get you tongue-tied sometimes-- the dimension of the orthogonal complement of V is going to be equal do the dimension of A transpose. Or another way to think of it is-- sorry, not just the dimension of A transpose, the dimension of the null space of A transpose. And if you have a good memory, I don't use the word a lot, this thing is the nullity-- this is the nullity of A transpose. The dimension of your null space is nullity, the dimension of your column space is your rank. Now let's see what we can do here. So let's just take A transpose, so you can just imagine A transpose for a second. I can just even draw it out. It's going to be a k by n matrix that looks like this. These columns are going to turn into rows. This is going to be v1 transpose, v2 transpose, all the way down to vk transpose vectors. These are all now row vectors. So we know one thing. We know one relationship between the rank and nullity of any matrix. We know that they're equal to the number of columns we have. We know that the rank of A transpose plus the nullity of A transpose is equal to the number of columns of A transpose. We have n columns. Each of these have n entries. It is equal to n. We saw this a while ago. And if you want just a bit of a reminder of where that comes from, when you take a-- if I wrote A transpose as a bunch of column vectors, which I can, or maybe let me take some other vector B, because I want to just remind you where this, why this made sense. If I take some vector B here, and it has got a bunch of column vectors, b1, b2, all the way to bn, and I put it into reduced row echelon form, you're going to have some pivot columns and some non-pivot columns. So let's say this is a pivot column. You know, I got a 1 and a bunch of 0's, let's say that this is one of them, and then let's say I got one other one that's out, and it would be a 0 there, it's a 1 down there, and everything else is a non-pivot column. I showed you in the last video that your basis for your column space is the number of pivot columns you have. So these guys are pivot columns. The corresponding column vectors form a basis for your column space. I showed you that in the last video. And so, if you want to know the dimension of your column space, you just have to count these things. You just count these things. This was equal to the number of, well, for this B's case, the rank of B is just equal to the number of pivot columns I have. Now the nullity is the dimension of your null space. We've done multiple problems where we found the null space of matrices. And every time, the dimension, it's a bit obvious, and I actually showed you this proof, it's related to the number of free columns you have, or non-pivot columns. So, if you have no pivot columns, then you are -- if all of your columns are pivot columns, and none of them have free variables or are associated with free variables, then you're null space is going to be trivial. It's just going to have the 0 vector. But the more free variables you have, the more dimensionality your null space has. So the free columns correspond to the null space, and they form actually a basis for your null space. And because of that, the basis for your null space vectors, plus the basis for your column space, is equal to the total number of columns you have. I showed that to you in the past, but it's always good to remind ourselves where things come from. But this was just a bit of a side. I did this with a separate vector B. Just to remind ourselves where this thing right here came from. Now, in the last video, I showed you that the rank of A transpose is the same thing is the rank of A. This is equal to, this part right here, is the same thing as the rank of A. I showed you that in the last video. When you transpose a matrix, it doesn't change its rank, or it doesn't change the dimension of its column space. So we can rewrite this statement, right here, as the rank of A plus the nullity of A transpose is equal to n, and the rank of A is the same thing as the dimension of the column space of A. And then the nullity of A transpose is the same thing as the dimension of the null space of A transpose-- that's just the definition of nullity-- they're going to be equal to n. Now what's the dimension-- what's the column space of A? The column space of A, that's what's spanned by these vectors right here, which were the basis for V. So this is the same thing as the dimension of V. The column space of A is the same thing as the dimension of my subspace V that I started this video with. And what is the null space of A transpose? The null space of A transpose, we saw already, that's the orthogonal complement of V. So I could write this as plus the dimension of the orthogonal complement of V is equal to n. And that's the result we wanted. If V is a subspace of Rn, that n is the same thing as that n, then the dimension of V plus the dimension of the orthogonal complement of V is going to be equal to n.