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# Basis of a subspace

Video transcript
Let's say I have the subspace v. And this is a subspace and we learned all about subspaces in the last video. And it's equal to the span of some set of vectors. And I showed in that video that the span of any set of vectors is a valid subspace. It's going to be the span of v1, v2, all the way, so it's going to be n vectors. So each of these are vectors. Now let me also say that all of these vectors are linearly independent. So v1, v2, all the way to vn, this set of vectors are linearly independent. Now before I kind of give you the punchline, let's review what exactly span meant. Span meant that this set, this subspace, represents all of the possible linear combinations of all of these vectors. So you know, I could have all of the combinations for all of the different c's. So c1 times v1 plus c2 times v2, all the way to cn times vn for all of the possible c's and the real numbers. If you take all of the possibilities of these and you put all of those vectors into a set, that is the span and that's what we're defining the subspace v as. Now, the definition of linear independence meant that the only solution to c1, v1, plus c2, v2 plus all the way to cn, vn, that the only solution to this equally the 0 vector-- maybe I should put a little vector sign up there-- is when all of these terms are equal to 0. c1 is equal to c2, is equal to all of these. All of them are equal to 0. Or kind of a more common sense way to think of it is that you can't represent any one of these vectors as a combination of the other vectors. Now, if both of these conditions are true that the span of this set of vectors is equal to this subspace or creates this subspace or it spans this subspace, and that all of these vectors are linearly independent, then we can say that the set of vectors-- maybe we call this, we could call this set of vectors s. Where we say s is equal to v1, v2, all the way to vn. It's equal to that set of vectors. We can then say and this is the punchline. We can then say that S, the set S is a basis for v. And this is the definition I wanted to make. If something is a basis for a set, that means that those vectors, if you take the span of those vectors, you can construct-- you can get to any of the vectors in that subspace and that those vectors are linearly independent. So there's a couple of ways to think about it. One is there's a lot of things that might span for something. For example, if this spans for v, then so would-- let me add another vector. Let me define another set. Let me define set T to be all of set S: v1, v2, all the way to vn. But it also contains this other vector. I'm going to call it the v special vector. So it's going to be essentially, the set S plus one more vector. Where this vector I'm just saying is equal to v1 plus v2. So clearly, this is not a linearly independent set. But if I had asked you what the span of T is, the span of T is still going to be this subspace, v. But I have this extra vector in here that made it non-linearly independent. This set is not linearly independent. So T is linearly dependent. So in this case, T is not a basis for v. And I had showed you this example because the way my head thinks about basis is, the basis is really the minimum set of vectors that I need, the minimum set-- and I'll write this down. This isn't a formal definition, but I view a basis-- let me switch colors-- as really the-- let me get a good color here. As a basis is the minimum-- I'll put it in quotes because I haven't defined that. The minimum set of vectors that spans the space that it's a basis of, spans the subspace. So in this case, this is the minimum set of vectors. And I'm not going to prove it just yet, but you can see that, look. This set of vectors right here, it does span the subspace, but it's clearly not the minimum set of vectors. Because the span of this thing, I could still remove this last vector here. I could still remove that guy and still-- and then the span of what's left over is still going to span my subspace v. So this guy right here was redundant. In a basis, you have a no redundancy. Each one of these guys is needed to be able to construct any of the vectors in the subspace v. Let me do some examples. So let's just take some vectors here. Let's say I had to find my set of vectors, and I'll deal in r2. So let's say I have the vector 2, 3. Let's say I have the other vector 7, 0. So first of all, let's just think about the span, the span of this set of vectors. This is a set of vectors. So what's the span of S? What's all of the linear combinations of this? Well, let's see if it's all of r2. So if it's all of r2 that means the linear combination of this could be-- we could always construct anything in r2 with the linear combination of this. So if we have c1 times 2, 3 plus c2 times 7, 0. If it is true that this spans all of r2, then we should be able to construct-- we should always be able find a c1 and a c2 to construct any point in r2. And let's see if we can show that. So we get 2c1 plus 7c2 is equal to x1. And then we get 3c1 plus 0c2. Plus 0 is equal to x2. And if we take this second equation and divide both sides by 3 we get c1 is equal to x2 over 3. And then if we substitute that back into this first equation, we get 2/3-- I'm just substituting c1 in there. So 2/3 x2. 2 times x2 over 3 is 2/3 x2. plus 7c2 is equal to x1. And then, what can we do? We can subtract the 2/3 x2 from both sides. Let me do it right here. So we get 7c2 is equal to x1 minus 2/3 x2. Divide both sides of this by 7 and you get c2. Let me do it in yellow. You get c2 is equal to x1 over 7 minus 2 over 21 x2. So if you've given me any x1 and any x2 where either x1 or x2 are a member of the real numbers, we're talking about-- well, everything we're going to be dealing with right now is real numbers. You give me any two real numbers. I take my x2 divided by 3 and I'll give you your c1. And I'll take the x1 divided by 7 and subtract 2/21 times your x2. And I'll get you your c2. This will never break. There's no division by any of these. You don't have to worry about these equaling a 0. These two formulas will always work. So you give me any x1 and any x2, I can always find you a c1 or a c2. Which is essentially finding a linear combination that will equal your vector. So the span of S is r2. Now the second question is, is are these two vectors linearly independent? And linear independence means that the only solution to the equation c-- let me switch colors. The only solution to the equation c1 times the first vector plus c2 times the second vector equaling the 0 vector, that the only solution to this is when both of these equal 0. So let's see if that's true. We've already solved for it, so if x1-- in this case, x1 is equal to 0 and x2 is equal to 0. This is just a special case where I'm making them equal to 0 vector. If I want to get the 0 vector, c1 is equal to 0/3. So c1 must be equal to 0. And c2 is equal to 0/7 minus 2/21 times 0. So c2 must also be equal to 0. So the only solution to this was settings both of these guys equal to 0. So S is also a linearly independent set. So it spans r2, it's linearly independent. So we can say definitively, that S-- that the set S, the set of vectors S is a basis for r2. Now, is this the only basis for r2? Well I could draw a trivially simple vector, set of vectors. I could do this one. Let me call it T. If I define T to be the set 1, 0 and 0, 1, does this span r2? Let's say I want to generate the-- I want to get to x1 and x2. How can I construct that out of these two vectors? Well, if I always just do x1 times 0, 1 plus x2 times 0, 1, that'll always give me x1, x2. So this definitely does span r2. Is it linearly independent? I could show it to you. If you wanted to make this equal to the 0 vector. If this is a 0 and this is 0, then this has to be a 0 and this has to be a 0. And that's kind of obvious. There's no way that you could get one of the other vectors by some multiple of the other one. There's no way you could get a 1 here by multiplying this by anything and vice versa. So it's also linearly independent. And the whole reason why I showed you this is because I wanted to show you that look, this set T spans r2. It's also linearly independent, so T is also a basis for r2. And I wanted to show you this to show that if I look at a vector subspace and r2 is a valid subspace of itself. You can verify that. But if I have a subspace, it doesn't have just one basis. It could have multiple bases. In fact, it normally has infinite bases. So in this case, S is a valid basis and T is also a valid basis for r2. And actually, just so you know what T is, the situation here, this is called a standard basis. This is the standard basis. And this is what you're used to dealing with in just regular calculus or physics class. And if you remember from physics class, this is the unit vector i and then this is the unit vector j. And it's the standard basis for two-dimensional Cartesian coordinates. What's useful about a basis is that you can always-- and it's not just true of the standard basis, is that you can represent any vector in your subspace. You can represent any vector in your subspace by some unique combination of the vectors in your basis. So let me show you that. So let's say that the set v1, v2, all the way to vn, let's say that this is a basis for-- I don't know-- just some subspace U. So this is a subspace. So that means that these guys are linearly independent. And also means that the span of these guys, or all of the linear combinations of these vectors, will get you all of the vectors, all of the possible components, all of the difference members of U. Now what I want to show you is each member of U can only be uniquely defined by a unique set of-- a unique combination of these guys. Let me be clear about that. Let's say my vector a is a member of our subspace U. That means that a can be represented by some linear combination of these guys. These guys span U. So that means that we can represent our vector a as being c1 times v1 plus c2 times v2. These are vectors. All the way to cn times vn. Now I want to show you that this is a unique combination. And to show that I'm going to prove by contradiction. Let's say that there's another combination. Let's say I could also represent a by some other combination, d1 times v2 plus d2 times v2 plus all the way to dn times vn. Now, what happens if I subtract a from a? I'm going to get the 0 vector. Let me subtract these two things. If I subtract a from a, a minus a is clearly the 0 vector. It's clearly the 0 vector and if I subtract this side from that side, what do we get? I'll do it in a different color. We get c1 minus d1 times v1 plus c2 minus d2 times v2, all the way to-- I'm at the point on my board where it starts to malfunction. All the way to c-- you can't see it. cn minus vn. It's showing up somehow. cn minus-- no, it's messing me up. Let me rewrite it on the left right here where it's less likely to mess up. The 0 vector, I'll write it like that. Is equal to c1 minus d1 times v1 plus all the way up to cn minus dn times vn. I just subtracted the vector by itself. Now, I told you that these are a basis. There's two things that-- when you say a basis, it says that the span of these guys makes the subspace. Or the span of these guys is the subspace. And it also tells you that these guys are linearly independent. So if they're linearly independent, the only solution to this equation-- this is just a constant times v1 plus another constant times v2, all the way to a constant times vn. The only solution to this equation is if each of these constants equal 0. So all of those constants have to be equal to 0. Over here before it messed up, this has to equal 0, this has to equal 0. That was a definition of linear independence. And we know that this is a linearly independent set. So if all of those constants are equal to 0, then we know that c1-- if this is equal to 0, then c1 is equal to d1, c2 is equal to d2, all the way to cn is equal to dn. So by the fact that it's linearly independent, all of these-- each of these constants have to be equal to each other. And that's our contradiction. I assume they're different, but the linear independence forced them to be the same. So if you have a basis for some subspace, any member of that subspace can be uniquely determined by a unique combination of those vectors. And just to hit the point home, I told you that this was a basis for r2. And my next question is, and I just want to kind of backtrack a bit. If I just added another vector here, if I just added the vector 1, 0, is S now a basis for r2? Well no, it clearly will continue to span r2, but this guy is redundant. This guy is in r2. And I already told you that these two guys alone span r2. That anything in r2 can be represented by a linear combination of these two guys. This guy is clearly in r2, so he can be represented by a linear combination of these two guys. So therefore, this is not a linearly independent set. This is linearly dependent. And because it's literally dependent, I have redundant information here. And then this would no longer be a basis. So in order for these to be a basis I kind of have to create the minimum set of vectors that can span, or the most efficient set of vectors that can span, in this case, r2.