26 - Linear combinations and spans

okay so we want to discuss now two more notions which relate to vectors to vector spaces and two as we're going to see two subspaces so the notions are what is a linear combination and what is a spin so let's start with the first one so here's the definition definition V when I write V it's probably going to be a vector V which is alpha 1 times V 1 plus alpha 2 times V 2 plus dot dot dot alpha what did I use here K times V K where alpha 1 alpha 2 all the way up to alpha K are just scalars they're elements of the field is called a linear combination of v1 v2 all the way to V T so we start with a bunch of vectors K of them V 1 up to VK and we write anything that we can write as a sum of scalar multiples of these vectors okay five times the first one plus 13 times the second one minus 6 times the 8th one that's a linear combination the first one itself plus the second one itself v1 plus v2 that's a linear combination it's 1 times v1 plus 1 times v2 okay clear let's do an example quick example example let's take v1 to be the vector 1 2 3 so this is a vector in r3 okay let's take v2 to be the vector 2 negative 1 negative 2 let's take v3 this is a 3 this lorry think v3 to be the vector 3 1 1 and let's take v4 to be the vector 4 0 2 you agree that these are 4 vectors in r3 ok so um I'm gonna write something and you're gonna feel some nostalgic air ok so 13 8 13 recognize this dude so I'm claiming that this is a linear combination of these four vectors why I have to be able to write it as something times v1 plus something times v2 plus something times v3 plus something times before okay since here goes I'm going to take 0 times v1 minus 3 times v2 plus 5 times v3 plus 1 times V 4 and I claim that this is indeed this vector and you can check it's easy to check right the first coordinate is 0 minus 6 plus 15 is 9 plus 4 30 the second coordinate is 0 plus three plus five is eight plus 0 8 and the third coordinate is 0 6 + 5 11 + 2 13 do you agree ok so this is a linear combination of these guys is it clear what a linear combination is ok now this is not a coincidence that I took this specific vector and maybe you're not recognizing these yet but you have seen them before - okay so we'll get to that in a few minutes okay but before that I want to define the notion of a spin so this was called a linear combination linear combination and now I want to define a notion of a spin so suppose we take so let let's call them um use now let's call them you let u 1 u 2 all the way up to UK UK be K vectors in um V V is some vector space okay define W is going to be the set of all the guys you all the guys you which are linear combinations of these K given vectors such that maybe let's write it in in words first and then write it explicitly so W is going to be the collection collection of all linear combinations of u 1 u 2 up to u K this is W I can rewrite it as the set of all use such that u equals alpha 1 u 1 plus alpha 2 u 2 plus dot alpha K UK and all these alpha i's are elements in the field do you agree that this is the same thing just written in math rather than in English right and I can also write it I'm still going to write the exact same thing but I'm going to use some more compact notation namely the notation the Sigma notation so I can write it as all the guys u where u equals the sum of alpha i u I where I goes from 1 to K and all these alphas are in the field do you agree that this is again the same thing just written in Sigma notation good so W is just all the possible linear combinations of K given specific vectors and hinted by the fact that I called it W I want to prove that this W is not just a collection or a bunch of elements or a set of elements in fact it's a subspace okay so this is a theorem so this is W this is the W were going to refer to right now so theorem W is a subspace a V a collection of all possible linear combinations of given vectors is always a subspace now if you recall for a moment this is what we did in a previous example we had some a couple of tube of 3x3 matrices one was the identity matrix 1 1 1 under diagonal and the other one had a one one one on the diagonal going the other way remember that and we took alpha all possible alpha multiples of the first one plus all possible beta multiples of the second one and we said hey that's a subspace and we argued that it is but in fact that's exactly what we're doing now abstractly we're taking we're putting our hand into the big sack all the elements in G pulling out K specific ones you want up to UK and doing the minimum that we need in order to make them a subspace we're taking all their scalar multiples and all their possible sums okay so this is going to be a subspace and we're going to remark that it's the smallest possible subspace containing these exact K elements that we started with okay so let's first prove this and then we're going to write some remarks in another definition so proof so what do we need to prove we need to prove that the three necessary conditions in order for a subset to be a subspace hold okay so the first condition one has to be non-empty okay so one W is not empty since the UI's are there usually we check that zero is there here we don't know a priori that zero is there it's not easy it's not hard to to see that zero is there we just take all the alphas to be zero but even more straight for where it is well we started with a bunch of UI so it's not empty okay okay the second property is we want to show that if we take a multiple of somebody in W it's also in W okay so W is closed under scalar multiplication since let's take an element in W a general element in W is of the form alpha 1 u 1 plus alpha 2 u 2 plus dot alpha K u K this is a general element in W right we want to multiply it by a scalar here's a scalar let's call it alpha and we want to show that the product is still an element in W right but it is because this can be rewritten as alpha alpha 1 u 1 plus alpha alpha 2 u 1 sorry u 2 plus dot dot alpha alpha K u K this is the it's not exact well it's the distributive property yeah it's precisely the distributive property right but now I can do another step on the same thing and write this because originally it was alpha times alpha 1 u 1 but that is the same thing as listen to what I'm saying this was originally alpha times alpha 1 u 1 but it's the same thing as alpha times it's the same thing as alpha alpha 1 times u 1 do you agree so do you agree to this step which included two properties of being in a vector space ok the distributive property and the fact that scalars are multiplying by two scalars I can either first multiply the scalars or multiply them one by one okay I don't remember if that was property 8 or 9 or whatever good and now look it's it's a scalar times u 1 plus a scalar times u 2 plus a scalar times u k right so this is clearly an element of W do you agree ok so I took a general element in W multiplied it by any scalar and again got an element in W so it's closed under scalar multiplication clear let's show that it cool that it's closed under addition if we show that we're done right so I need to take two guys of this form add them and claim that it's again an element in W let's do it using the the Sigma notation let's see if we're not scared so here's a guy so let's write it first W is closed under addition so let's take two guys some alpha I you I I equals 1 to K this is a general element in W a linear combination of the UI's and I want to add it to another general element in W again it's going to be a sum I equals 1 to K of a linear combinations of the UI is I can't call them alpha because they're different guy different scalars so let's call them beta I UI this is these are two general elements in w ok how do I add Sigma's like this and if you're not used to this you have to open it to write it like this and see how you add okay so if it if it if it if if it's a bit scary or you feel that you're a bit not confident when doing this open it up right write it like this rather than like this if you're good with a Sigma notation it's it's good to get used to the Sigma notation it's useful ok it makes life easier usually so this is the same as Sigma alpha I plus beta I you I do you agree and this is an element in w it's a bunch of scalars it's a linear combination of the U is do you agree good so that's it this proves it W is a linear subspace a vector subspace of V good everybody good so let's go back to this board here where we defined W and AD um so definition W is called the span of u R u 1 u 2 up to u K the span of these elements or sometimes the set spanned by these guys or due to the serum the subspace spanned by these guys okay and denoted W equals and this is just how we write it span of u 1 u 2 up to u K okay sometimes you may see instead of the word span just SP abbreviated SP but I'm probably going to write the entire word usually okay so W is the span of these element everything that can be obtained from these elements by means of linear combinations okay and it's called the span it's a linear subspace a vector subspace that we just proved okay it obviously contains all of them and it contains the minimal stuff we need in order to be a subspace to contain them right it contains all there are scalar multiples and all sums of elements in there right so let's maybe remark that so for we remarked that I want to add one more piece of notation here the you eyes the you eyes are called a spanning set okay they're the set that spans W okay so you eyes are called um yeah we need a new board right let's take a new board oh I don't want to erase this example um I can erase here so W equals span of u1 u2 up to UK is not just a subspace it's the smallest subspace which contains u1 up to u K okay in the sense that if you take any other subspace which contains these elements it's going to contain W okay in the sense that if let's call it a W bar is any subspace which contains which includes which includes u1 up to u K then W bar is included in W sorry vice versa then W is included in W bar okay W bar can include them and be bigger right if you throw in more stuff it could still be a subspace but all of W is going to be there not just these guys all their linear combinations are necessarily going to live there okay and another comment that I made so this is one comment and another one u1 up to u K are called a spanning set sometimes you'll see them called generators okay but I'm going to use this terminology okay um good I want to go back to D questions about this is it clear what is a linear combination and what is a span and why is it a subspace and so on everything okay okay I want to go back to this example for a minute so we started with four vectors here just four vectors in r3 and showed that this specific vector is a linear combination of the fort okay and when you saw this you probably thought hey there was the system of linear equations that we are studied in several examples but saying different things about it and it had this as its be right as it's a constant factor but I want to I want to claim something else recall recall the system the non-homogeneous system of linear equations that we have been looking at and I'm going to remind you what it was it was this matrix one yuck one two three four two negative one one zero remember this three negative two one two times X 1 X 2 X 3 X 4 equals 13 8 13 remember this was the system do you see that these four vectors are in fact the columns of a which was the coefficient matrix of this system do you see that okay so what just happened here what just happened here how did we get this 13 this 13 is 1 times X 1 plus 2 times X 2 plus 3 times X 3 plus 4 times X 4 do you see that if we take the vector 0 negative 3 5 1 and plug it here it's going to solve this system in fact it's going to be precisely this statement right this 13 is 0 so plug in 1 negative 3 5 1 here 1 negative 3 5 1 ok this 13 is precisely 0 times 1 I'm just writing it the opposite way right 0 times 1 minus 3 times to the minus 3 is living here plus 5 times 3 5 times 3 this is X 3 right plus 4 times 1 this is X 4 do you see that okay so let's let's write it again let's write it again um let's rewrite this fact so the fact that I could write 13 8 3 as a linear as a linear combination of those four vectors which were the columns of the coefficient matrix a was not a coincidence it was because that system had a solution and the solution was precisely those coefficients that I needed in order to write 1383 as a linear combination of those vectors okay so let's write that so um another way of writing a system of linear equations a x equals B is the following let's denote denote the columns of a by the columns of a are vectors right are our column vectors let's denote them by a1 a2 all the way up to however many columns there are how many columns are there well for in this example but in general in write a n there are n columns no no in an M by n matrix there are n columns right so let's denote them by a 1 a 2 up to a n okay then a different way of writing the exact same system is X 1 times a 1 this is a factor this is a number a scalar plus x2 times a2 plus dot dot plus X n times a n equals B do you see that this is precisely the same calculation and when does the system have a solution when is there an X that satisfies this precisely when there's an X whose components are x1 through xn that satisfies this do you agree clear do you see that these are two ways of expressing the exact same thing it's a bit confusing but it's just two ways of saying the same things in an example here abstractly okay so so the system so conclusion conclusion the system a x equals B has a solution there is an X which satisfies this if and only if this has a solution but what does it mean for this to have a solution we can now sing it in in more professional words the system has a solution if and only if B is a linear combination of the columns of a do you see that okay do the words register let's write it if and only if this is an abbreviation for if and only if I don't know if you've encountered it if and only if B is a linear combination of the columns of a is this statement clear if you can find a way to express B as a linear combination of the columns of a then those components those alphas of the linear combination are precisely if you take them all as a vector they're precisely going to be a solution vector to this system okay in the example is this clear in the example in the example here I'm back to the example zero negative three I'm reading the coefficients here 5 1 is a solution it solves the system if you put it here you get this system satisfied it enabled us to write the columns the constant vector the solutions as a linear combination of the columns of the coefficient matrix okay good let's say the same thing again in different words no no not not the other way the same way but just in different words here I said that B is a linear combination of the columns of a can you say it differently can you say it using for example the word spin I want to say that B is some linear combination of these eight eyes it's not the span the span is a subspace the span is the entire subspace that's the span it belongs to the spin right it's an L it all the guys that you can write like this is precisely the span of the columns and the statement is that B belongs to the spin okay so I can add if and only if this is just rephrasing the same the same statement it's not a theorem or something it's just saying it in different words B belongs to the span of these columns a 1 a N we're eight one two ANR the columns of a clear okay and now there's a definition that that we use and the reason is that it's going to be useful so we have a notation for all the vectors spanned by the columns of a given matrix okay so notation notation we write in Co L from the word columns call of a equals the span of let's write the columns of a and the reason I'm not writing a 1 a 2 2 a n is because I'm going to define something else here Rho of a is the span of the rows of a the rows of a are again vectors ok these live in two totally different worlds right let's look at our example again over here on this board here's a the columns of a are vectors of length three right there are three tuples whereas the rows of a are four tuples okay so these live in totally different worlds row of a is anything you can get as a linear combination of these three rows column of a is anything you can get as a linear combination as as a linear combinations of these four columns this is a subspace of the DA call a is space of r3 in this case Rho a is a subspace of R for do you agree there in totally different worlds we saw an example of how the notion of call a is useful namely this conclusion here so this we can now rewrite as column of a right tell me if you agree over on this board good saying that a system has a solution is the same as saying that B belongs to call a to the span of the columns of a okay what is Rho of a useful for this we haven't discussed yet but it's going to be okay so we didn't say anything about Rho of a yet what what does what do the what does the span of the rows how does it relate we're not there yet we'll say it soon very easy good okay so um I guess we'll stop this one here so we now know the notion of a linear combination and we know that the notion of a span okay what I want to do next is by means of a couple of examples to show you how we decide suppose somebody gives you a subspace W and the subspace W is the span of some elements is given by its generators by its spanning set and you pick out an element in V and you want to know is it in W can it be written as a linear combination of the given you eyes okay I want to show you how we do that how we decide that so that's coming up next

Channel: Technion

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Keywords: Technion, Algebra 1M, Dr. Aviv Censor, International school of engineering

Id: -mGAxFDIHyg

Length: 35min 20sec (2120 seconds)

Published: Thu Nov 26 2015