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Current time:0:00Total duration:19:00

Video transcript

let's say I have the subspace V 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 so this is going to be 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 are linearly independent so v1 v2 all the way to VN this set of vectors are linearly independent 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 is represents all of the possible linear combinations of all of these all of these vectors so you know I could have all of the combinations for all of the different seas so c1 times v1 plus c2 times v2 all the way to CN times VN for all of the possible for all the possible seas in the real numbers you take all 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 equaling the zero vector maybe I should put a little vector sign up there is when all of these terms are equal to zero c1 is equal to c2 is equal to all of these CN is all all of them are equal to 0 or kind of a more a 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 this set of vector the span of the 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 we can call maybe we call this we could call the set of vectors s where let me 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 a basis for V and this is the definition I wanted to make if something is a basis for a a set that means that those vectors if you take the span of those vectors you can construct you can get you can get to any of the vectors in that subspace and that those vectors are linearly independent so the linear 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 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 no no let me call it the V special vector so it's going to be essentially the set s plus 1 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 asked you what the span of if I asked you what the span of T is the span of T is still going to be this subspace is still going to be this subspace V but I have this extra vector in here that made it nonlinear nonlinearly independent this set is not linearly independent so this is linearly so T is linearly dependent so in this case T is not a basis basis for V and I'd showed you this example because the way my head thinks about basis is the basis is really the 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 the minimum I'll put it in quotes because I haven't defined that the minimum set of vectors of vectors that spans 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 I'm not going to prove it just yet but you can see that look this one this set of vectors right here it dis span the subspace but it's clearing up 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 and the span of what's left over is still going to get me to my subspace is still going to get to it's still going to span my subspace V so this guy right here was redundant and a basis in a basis you have no redundancy each one of these guys is needed to be able to form to be able to construct any of the vectors in the Subspace V let me do some examples let me do some examples so let's just take let's just take some vectors here let's say I have let's say I define my set of vectors and I'll deal in r2 so let's say I have the vector 2 + 2 3 and I have the other vector let's say I have the other vector 7 0 so first of all let's just think about the span the span of the 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 our two so if it's all of our two that means the linear combination of this could be could be all we could always construct anything in r2 with a linear combination of this so if we have c1 times 2/3 plus c2 times seven zero if it is true that this fans all of our two then we should be able to construct we should always be able to find a c1 and a c2 to construct any point in r2 and let's see if we can show that so we get to c1 plus seven c2 is equal to x1 and then we get 3 c1 3 c1 plus 0 C 2 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 X 2 over 3 and then if we substitute that back into this first equation we get 2/3 2/3 I'm just substituting c1 in there so 2/3 X 2 right 2 times X 2 over 3 is 2/3 X 2 plus 7 c2 is equal to X 1 and then what can we do we can subtract the 2/3 X 2 from both sides let me do it right here so we get 7 c2 is equal to x1 minus 2/3 X 2 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 give me any X 1 and any x2 where either of you know x1 or x2 are a member of the real numbers we're talking about about our revolver everything we're going to be dealing with right now is real numbers you give me any two real numbers I just put them into I take my X 2/3 and I'll give you your c1 and you I take the x1 divided by 7 and subtract 221st times your x2 and I'll get you your c2 this will never break there's no division by any of these so you'll have to worry about these equal to zero these two formulas will always work so you give me any X 1 and any X 2 I can always find you a c1 or c2 which is essentially finding a linear combination that will equal your vector so the span of s is r2 the span of s is equal to r2 now the second question is is are these two vectors linearly independent are they linearly independent and linear independence means that the only solution to the equation C let me switch colors the only solution to the equation C 1 times the first vector plus C 2 times the second vector equaling the zero vector that the only solution to this is when both of these equal zero so let's see if that's true we've already solved for it so if X 1 in this case X 1 is equal to 0 and X 2 is equal to 0 right this is just a special case where I'm making them equal to 0 vector if they're 1 if I want to get the zero vector C 1 is equal to 0 over 3 so C 1 must be equal to 0 and C 2 is equal to 0 over 7 minus 220 first times 0 so C 2 must also be equal to 0 so the only solution to this was setting both of these guys equal to 0 so these two vectors so s is also s is also a linearly independent set linearly independent so it's fans are two it's linearly independent so we can say definitively that s the set s the set of vectors s is a basis is a basis for r2 now is this the only basis for r2 well I could draw a trivial trivially simple vector set of vectors I could do this one let me call it let me call it T if I define T to be the set one zero and zero one does this span r2 can I you know let's say I want to I want to generate the I want to get to X 1 and X 2 how can i construct that out of these two vectors well if I always just do X 1 times 1 0 plus X 2 times 0 1 that'll always give me X 1 X 2 so this definitely does span r2 so it definitely does span r2 is it linearly independent well I could show it to you know if if you wanted to make this equal to the 0 vector so if you want 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 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 so T also a basis for r2 and I wanted to show you this to show that if I if I look at a vector subspace and r2 is a valid subspace of itself you can 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 and this is this is the standard standard basis and this is what you're used to dealing with in just you know regular calculus or physics class and if you remember from physics class this is the unit vector this is a unit vector I and then this is the unit vector J and it's the standard basis for two-dimensional Cartesian coordinates and what's useful what's useful about a basis is that you can always and it's not just drove the standard basis is that you can represent any vector 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 let's say the set v1 v2 all the way to VN let's say that this is say a basis for for I don't know just you know some some subspace u so this is a subspace subspace u 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 the possible components all of the different members of U now what I want to show you is each member of you can only be uniquely defined by a unique set of kind of a unique combination of these guys let me be clear about that let's say that a let's say my vector a is a member of our subspace u that means that that a can be represented by some linear combination of these guys right these guys ban 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 right now I want to show you that this is a unique combination let's and to show that I'm going to prove I'm going to prove let's I'm going to proof by contradiction let's say that there's another combination let's say I can also represent a by some other combination d1 times v1 plus d2 times v2 plus all the way to DN times VN right now what happens if I subtract a from a I'm going to get the zero vector let me subtract these two things if I subtract a from a a - a is clearly the zero vector it's clearly the zero vector and if I subtract this side from that side what do we get we get hold it in a different color we get C 1 minus D 1 times V 1 plus C 2 minus D 2 times v2 all the way to my I'm at the point on my board where it starts to malfunction all the way to C you don't you can't see it CN minus V n it's showing up so on CN minus V CN minus it's messing me up let me rewrite it on the left right here where it's less likely to mess up the zero vector and the zero vector I'll write it like that is equal to C 1 minus D 1 times V 1 plus all the way all the way up to CN minus DN times V n right I just subtracted the vector from by itself now I told you that these are a basis there's two things that meet when you say a basis it says that these guys the span of these guys makes this subspace or the span of these guys is this subspace and also tells you that these guys are linearly independent so if they're linearly independent the only solution to this equation right this is just a constant times V 1 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 equals 0 so all of those constants have to be equal to 0 over here before it messed up this has to equal 0 this Ackles equals 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 C 1 if this is equal to 0 then C 1 is equal to D 1 C 2 is equal to D 2 all the way to C n is equal to D n 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 force them to be the same 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 I told you that this was a basis for r2 and my next question is and I just want to kind of backtrack a bad if I just added another vector here if I just added the vector 1 0 is now 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 this guy is in r2 and I already told you that that these two guys Alone's fan 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 linearly dependent and because it's linearly 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