❖ The Span of a Set of Vectors ❖

all right in this video I want to start looking at the notion of what's called a span of a set of vectors so start off here with a little just some good old definitions and then from that we'll look at some more examples here so the idea is we've got a set of vectors V 1 V 2 up to you know V sub K a lot of times vectors the notation will be a little bold or have a little line over them I've kind of not done that here but again the V's will stand for vectors so what we do is we look at the set of all linear combinations of this set of vectors and that's what's known as the span so kind of the way I think about it is you know are we've got some set of vectors and we're basically just multiplying those by scalars and then adding them together and so these are kind of like the the building blocks that you're starting with and it's you're kind of thinking how can I take these vectors and basically create new vectors so again just to re-emphasize a linear combination again is just a C sub one times V sub 1 plus C sub 2 times V sub 2 up to C sub K times V sub K where the C sub i's are just real number scalars so again the C's are just scalars the V's again are vectors so just kind of at random I've picked a couple here nothing special about these at all but just to hopefully give you a little bit of intuition so suppose we just start with two vectors here suppose we have the vector 1 3 and the vector 2 5 so these would be vectors in r2 if we think about the span of those vectors we can write the span of those vectors as well C sub 1 times the vector 1 3 plus C sub 2 times the vector 2 5 so again I've kind of picked constants at random here 1 and 4 and I'm going to use these constants and just create some you know some new vectors so again these are kind of my initial two vectors I'm starting with and I can create new vectors so you know a new vector that we could create out of this you know out of this these two original well again I'm going to let C sub 1 equal 1 and we said we're going to let C sub 2 equal 4 and well recall if you multiply by a real number scalar you just multiply the entries respectively so if we multiply by 4 4 times 2 is 8 well 4 times 5 is 20 and then we can add the components together so if we add 1 plus 8 that will give us 9 3 plus 20 would give us 23 so we've got a new vector this vector 9 with components 9 23 this vector would be in the span of those original two vectors so again I started with two vectors hey I've multiplied by some numbers at random and created yet a new vector so you can do this obviously you know pick your two favorite numbers and go through this process and you can go through that and you know create all the vectors that would be in the span of again these original two vectors that we started with okay so a lot of times a very common thing that you'll want to do is you'll want to take a vector you know you'll have a vector in this case I'm taking the vector 20 with components 20 and 4 and I'm going to show it belongs to the span of those again those two vectors that I'm playing with in this example so so can I figure out a linear combination of these two vectors that will produce that new vector that's the the question that we're trying to address here ok so again this is a lot of times an important question can you start with certain vectors and create you know a new a new vector some vector in particular well you know again kind of the question what we have to do is we have to figure out some constant C sub 1 so that when I multiply that by 1 3 and then some other constant C 2 so that when I multiply that by 2 5 at the end of the day I want to get this vector 20 with components 20 and 4 well one way to do this is again you could multiply so we would have one c1 and then we would have 3 c1 + well 2 C 2 and then 5 c2 we're going to want that to equal the vector with components 20 and 4 and again what I'm going to do is a right this is a little system of equations so we've got C 1 plus 2 C 2 and then 3 C 1 plus 5 C 2 if we add the respective components and again we want that to equal 20 + 4 so again you can think about this just being a little you know linear system of equations that we have to solve so again just writing down the corresponding equations and what I'm going to do now is I'm going to use a little bit of row reduction to see what our constants C 1 and C 2 would have to equal in order to solve this so again if we take the coefficients 1 2 3 5 we'll put the 20 and the 4 on the other side so again obviously we could have done you know just jump to this stuff at the very beginning again we're just taking you know the components of each vector so 1 3 2 5 and again we're trying to produce this vector with components 20 and 4 so I'm going to do a little row reduction here so I think if we take negative 3 times Row 1 and add that to our Row 2 I'm going to use that to produce our new Row 2 so let's see the first row I'm going to leave a loop alone 1 to 20 so let's see negative 3 times 1 is negative 3 plus 3 is 0 let's see negative 3 times positive 2 is going to give us negative 6 negative 6 plus 5 is going to give us negative 1 and then let's see I guess we would have negative 60 plus 4 well let's see negative 60 plus 4 that's going to give us I guess how about negative 56 I can do my arithmetic here correctly that looks okay to me well what I'm going to do now because again we would like to have a 1 here I'm just going to take negative 1 times Row 2 and that's going to produce my new Row 2 and I'm not going to write this step out we can just change it so that'll be a positive 1 and a positive 56 and now what I'm going to do from here is I'm going to take negative 2 times our Row 2 and then I'm going to add that to Row 1 to produce our our new first row ok so we've still got 0 1 and 56 there at the bottom let's see so if I do negative 2 times 0 plus 1 that's still 1 negative 2 times 1 is negative 2 plus 2 is 0 let's see negative 2 times 56 that's going to be what I guess a negative 112 if we add that the positive 20 that's going to give us negative 92 so again remember this represented the coefficients C 1 and C 2 so it looks like if we let C 1 equal negative 92 and if we let C 2 equal positive 56 those are going to be the values of our constants that produce the this vector with components 20 comma 4 that we wanted so let's even check here and make sure so we've got let's see so C 1 we said is negative 92 times 1 and 3 plus we said our C 2 value should be positive 56 this is a problem when you pick numbers at random I guess they're not always the smallest numbers but that's okay so let's see well we would have again if we multiplied we'd have negative 92 let's see negative 92 times 3 that's going to be negative 276 if we take 56 times 2 that's going to give us 112 56 times 5 let's see so that's what 280 so the question is does that give us 20 comma 4 well let's see negative 92 plus 112 that to me definitely looks like it gives us 20 negative 276 did I do my math wrong here I sure hope not okay everything is perfect losing my mind so negative to tip 76 plus 280 is definitely positive for so hey in fact we have found our correct constants c1 c2 so we've now shown we've now shown that this vector with components 20 and 4 belongs in the span of those original two vectors okay so you know kind of a more general question and this is a lot of times you know where you're headed can we take any vector in r2 and write that as a linear combination of these vectors with components 1 3 & 2 5 so the short answer which I'm going to show right now and we can we'll generalize this later on down the road but it says any two vectors in r2 that are not scalar multiples of each other will span all of our two so not quite proving that but that's something that in fact is true so based on that it says we should be able to write any single vector in r2 using these two vectors notice they're not a multiple of each other you know if I multiply the first vector by say positive 2 how we get to 6 well that doesn't produce this vector okay so these are certainly not multiples of each other so based on that result we should be able to write any vector and r2 as a linear combination of these two vectors so okay so to justify this there's definitely different ways to do that okay so there's way there's a way to use inverses and square matrices but I'm going to show a slightly different way just a slightly different way so you know so I what I want to do is I'm going to take c1 times 1 and 3 and c2 times our vector 2 comma 5 and again what we want to do is we want to be able to produce any vector let's suppose it has components X&Y okay so if I can somehow figure out you know my scalars c1 and c2 that will produce this generic vector well then it says that any vector in r2 can be written as a linear combination ok so again different ways to do this I'm going to write a little system of equations so we would have C 1 times 1 plus 2 C 2 equals x and then we would have I guess 3 C 1 plus 5 C 2 equals y so what I want to do is I'm just going to find some condition you know for our are scalars that at the end of the day will produce these components x and y all I'm going to do in this case again you could use row reduction I'm just going to use a little bit of substitution here as all I'm going to do so notice if we take the first equation and solve for C 1 it says that C 1 will equal well X minus 2 times C 2 and I'm going to substitute that into my next equation so it says 3 times C 1 which is X minus 2 C 2 plus 5 C 2 that's going to equal Y and what I'm going to do is I'm trying to find conditions for my constants my scalars so I'm going to solve for C 2 so it says we have 3x minus 6 C sub 2 plus 5 C sub 2 that equals y well let's see I'm going to subtract the Y over so we'll have 3x minus y notice if we can by negative 6 C 2 plus 5 C 2 that would give us a negative C 2 and I'm going to add that to the right side so there's my constant C 2 and now I'm just going to plug that back in to find the restriction for my constant C 1 so it says C 1 is going to be X minus 2 times C 2 which is 3 X minus y well this is going to be X minus it looks like 6 X plus 2y and I guess that gives us a negative 5x plus 2y so what it says is it says if you want to produce you know any vector with components a vector with components and why it says you now have a little formula to determine what your your scalars need to be it says plug them in here that'll tell you see one plug them into the other that'll give you C 2 so maybe for example we want to produce I don't know let's produce the vector I don't know how about four seven again just something at random well again so it says C one would have to equal okay so we said that's negative 5x plus 2y which in this case we'll get negative 5 times 4 so again I'm just multiplying plus 2 times y which is the value 7 so let's see that's negative 20 plus 14 that's going to give us negative 6 to get our value for C 2 we said we're D'Arcy to go that's 3x minus y so we've got 3 times X which again is positive 4 minus y which in this case is 7 okay so that's 12 minus 7 which is going to give us positive 5 so let's in fact see if this works ok so C sub 1 we said that should be negative 6 and then I'm going to multiply that by the vector 1 comma 3 plus C 2 okay so our scalar there we decide it should be 5 and we'll multiply that by the vector with components 2 and 5 and again the vector that we were trying to produce was the vector with components 4 and 7 and let's see if it works it looks like we get well negative 6 plus 10 negative 6 plus 10 is definitely going to give us 4 and then we get negative 18 plus 25 well negative 18 plus 25 does give us positive 7 so lo and behold we have figured out you know what our real number scalars would need to be so that when we multiply by our original two vectors we do produce that new vector so you know we've kind of gone through a generic argument here to basically show if you give me any vector X and Y again we've now got nice little formulas that will produce those those scalars so that when we perform our linear combination we do in fact get that desired vector okay so other ways to address this question for sure and we'll talk about that and some other examples but this is the basic idea with the span of vectors obviously I'm using vectors in r2 just to kind of keep the arithmetic a little bit manageable you know you can do you can obviously extend this to to any vector space R sub N and go through the same argument I'll be it a bit more tedious so you know another thing that we'll want to generalize so you know I said any two vectors in r2 that are not scalar multiples of each other will span all of our two we have to be a little bit more careful about this as we go into higher dimensions and we'll we'll talk about this in some other examples so again this is the notion of a span of vectors just you're kind of starting with some number of vectors again I started with two vectors originally you could certainly start with more than just two vectors but the idea is if you start with a certain number of vectors by going through these linear combinations you know what new vectors can you produce from those that's that's the basic idea

Channel: patrickJMT

Views: 1,142,800

Rating: 4.9118762 out of 5

Keywords: span, vector, set, linear, algebra, combination, basis, independence, row, reduction, system, of, equations

Id: ivP-6oicIWU

Length: 17min 5sec (1025 seconds)

Published: Fri Aug 26 2011