Current time:0:00Total duration:9:27
0 energy points
Studying for a test? Prepare with these 5 lessons on Alternate coordinate systems (bases).
See 5 lessons

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

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.