36 - Bases and dimensions of subspaces

the concepts of a basis and the concept of a dimension both can be relevant for subspaces as well okay we have a subspace of a vector space we can ask what's it's dimension okay and it's dimension is gonna be the number of elements in a basis for that subspace and a basis is still gonna be a set that spans only the subspace and is linearly independent okay so let's start by doing some examples that's always a good way to relate to concepts do examples so example number one we're gonna take our space - V - V to be 2 by 2 matrices over R and what's the dimension of V right the dimension of V is 4 we know that everybody sure okay and here's a subspace of V I'm gonna call it u U is gonna be all the elements in V I'm calling them capital a because elements here are matrices right so all the A's in V satisfying that a equals a B minus B a plus B where a and B are scalars okay so U is the set of all the two-by-two matrices satisfying that the matrix is of a specific form okay so just a side remark this is a standard way of defining a subset of a set okay by imposing some conditions on the elements okay so I'm only taking those matrices that satisfy certain constraints certain conditions that if you have B here but you're gonna have - be here okay and then here you're gonna have whatever sits here plus whatever sits here okay and in order to define a subspace in this manner it doesn't really have to be a subspace it could be any subset okay if I want to define for example all the students in this class that's a set and a subset is gonna be all the redheads all the students whose hair is red or orange or anywhere in between okay so that's a way of defining a subset by listing a certain number of constraints on the elements okay and I'm emphasizing this because in example two we're gonna define a subspace in a different way which is you look two subspaces and cannot be done for subsets okay so is it clear that defining by listing some restrictions is is is is good for any set not necessarily a subspace we didn't use some the fact that we're living in a vector space to do this okay good okay so here's a spanning set for you so the set containing the matrices 1 0 0 1 and 0 1 negative 1 1 this set of two matrices is a spanning set for you for the subspace Y because any element in you since any element in you a B minus B a plus B that's the form of a general element in you it can be written as a linear combination of these two guys that's why I chose them is a times this plus B times this do you see that equals a times 1 zero zero 1 plus B times 0 1 negative 1 1 do you agree everybody good so this is a spanning set for you is it a basis for you well we have to check if it's linearly independent and are these two guys linearly independent yes they are because there is only two of them and one is not a scalar multiple of the other right so that's one way to say it if you really want to you can straighten them out in rows of a matrix and it's already in Ashland form and there's no zero rows okay so this is a spanning set for you let's give it a name let's call it B let's call it B sub u b for basis you for the what it is a basis of so beef B sub u bu is also linearly independent and therefore B U is a basis for you implying implying that what is the dimension of you two because we found the basis with two elements good ok here's another example here's another example let's take V to still be the two by two matrices over our same ambient vector space and let's take W to be the span of three guys one two three four 1 0 0 5 & 1 4 6 3 so W is the span of these three matrices namely all the linear combinations of these three matrices okay so this is another way of defining a subspace in a vector space which would not work for a subset I cannot define a subset of the students here as all the students that are linear combinations of you you and you because it's not defined what is a linear combination of students okay clear so there are no operations on a set and therefore you can do anything in the form of generators you can only do that when you have operations like in a vector space or in a group or in a field or anywhere where you have operations okay so what is gonna be the dimension of W right we need to check its it's tempting to say well 3 right there are three elements here but they're not necessarily linearly independent okay and it's not obvious just by looking at them because for three guys to be linearly independent it's not enough that one is not a multiple of the other maybe this guy is six times this minus five times this okay we don't see that automatically so only after we verify if they're linearly independent or maybe not will we know what is the dimension of W clear that's important so W these three guys are a spanning set for W but they're not yet a basis because we don't know linear independence okay so we have to check that so we need to check linear independence clear how do we check linear independence this is something we've done several times for example we can straighten these out as rows of a bigger matrix right of three by four matrix and start doing row operations until we reach in echelon form if there's a zero row a row of zeros then they're not linearly independent if not then they are let's do that so I'm straightening them out as rows of this matrix so it's one two three four one zero zero five and one four six three now I'm doing these row operations you can check our two is our two minus R one R 3 is transforms to R 3 minus R 1 and I get the matrix 1 2 3 4 0 negative 2 negative 3 1 & 0 2 3 negative 1 do you agree good and then one more step R 3 let's do it here are three I'm gonna take it to be R 3 plus R 2 and I get the 0 the the zero row right so since there's a zero row here only these two rows are linearly independent remember the nonzero rows in an in a matrix in Ashkelon form are linearly independent right you can see that directly because one is not a multiple of the other right so these two are linearly independent meaning that are three in the original one was extra right because we managed to annihilate it right so only these two were sufficient to span everything this one was a linear combination of the other two okay and in in previous in a previous example that we did some time ago we even saw how to trace the exact scalars by which this is a linear combination of these two remember we did that okay so the conclusion is that those three matrices that we started with are linearly dependent the third one is not necessary and BW which is the set containing only the first two one two three four and one zero zero five is a basis for W implying that the dimension of W is also too good it's a basis because these two are linear linearly independent okay you can take these two that would be just as good right so these two are linearly independent and they span everything good everybody good okay so what can we say in general about dimensions and Basie's for subspaces well we said one thing that if you take basis for a subspace for example this we can complete it into a basis of the entire space and if I'm not mistaken this is precisely the example we did in the previous lesson Latika take a quick glance and what at which example we did I think this is precisely what we did it was one two three four and one zero zero six oh well okay so we know how to do that but there's a theorem here that is useful and let's write it there they're a couple of them so here's the first steer-in very good very good so if u is contained in V if u and V is a subspace of vector subspace then the dimension of U is less than or equal to the dimension of V and furthermore and the dimension of U equals the dimension of V if and only if u is all of you okay so let's prove this okay it should sound extremely obvious it should make sense okay it should you should think wait there's no way that the dimension of you is greater than the dimension of e because that would mean that you lives inside V and somehow I took a basis for you inside V which has more than n elements right and it's a spanning set that's a contradiction we know that the greatest number of elements in a spanning set in V is at most n right do you agree that that makes sense okay that's the proof that's precisely the proof okay so let's write it down if the dimension of you were greater than the dimension of V okay and we're gonna reach a contradiction then a basis for you for you for you is a linearly independent set with more then N equals dim V elements right because it has dim U which is greater than three elements and that cannot be it's living inside feet we had a theorem a linearly independent set could have at most n elements okay so linearly with more than n equals then V elements in contradiction to the previous theorem the previous theorem in the previous lesson so it was part four in contradiction to part four of the previous theorem okay good okay so this proves up to here the dimension of you has to be less than or equal to because if it's greater than will you reach a contradiction but now why are they equal well this is an if and only if statement so we need to prove two things if u equals V obviously they have the same dimension because they're the same thing so this direction really is a trivial statement there's nothing to prove what we need to prove is this that if the dimension of U equals the dimension of V then they're the same space okay and this is this is it's not hard but it's not a trivial statement all we know is that these two things have the same number of elements in a basis that's all the dimension is why does it force them to be precisely the same thing why can't we find four dimensional sub space inside V which is not all of you okay clear is it clear why there is still something to say in this direction okay so so by part two of the previous theorem so part two of the previous theorem said that every linearly independent set with n elements is a basis okay so by part two of the previous theorem if if the dimension of u equals the dimension of V then a basis of you suppose that I mention of you equals the dimension of e take a basis for you it's gonna have n elements right the dimension of V we denoted by n it's gonna have n elements then a basis of you has n elements right and part two of the previous theorem said it's a basis it's linearly independent with n elements it's a basis for everything it's a basis for the entire space ok has n elements and is thus a basis for V good so we started with a basis for you it has n elements therefore it's a linearly independent set inside V you lives inside V with n elements therefore it's a basis for V so anybody in V is already in you right because it's generated by a basis for you okay so V equals u hence U and V coincide anybody in V is already in you Sophie is contained in U and a priori you was contained in V so the sets are equal good follow the logic ok so this is the first theorem that we have this completes the proof there are two statements here the first statement is this one the dimension of U is less than or equal to the dimension of you that's one statement and the proof ended at this point okay then there's a further statement which is not the same thing if they're the same then or if and only if they're the same then the spaces are the same meaning meaning what the practical meaning of this is that if you look back at our example for example here V was 2 by 2 matrices over R you cannot have a subspace of dimension 4 in V that's not everything you cannot have a strict subspace an honest subspace that is 4 dimensional if you found a subspace that's 4 dimensional it's everything ok clear ok ok so here's the next steer and the next theorem I'm not going to prove the proof is lengthy and has a lot of it's very technical and I'm gonna skip the proof over the next theorem however I do want to write it it's it's useful theorem and it relates to sums of subspaces remember we had the notion of a sum of subspaces ok ok and it's the following theorem before I write it I want to tell you the idea so let's start and then I'll draw a little sketch here let u and W in V be subspaces now think of sets for a minute it's hard to draw subspaces because it's hard to draw the operations how do we draw the fact that a + b lives inside the subspace it's okay it's it's easy to draw a and B live inside the subspace but how do we draw that a + b where where do we draw a + b ok but it's easy to draw sense so suppose we have a big set here's V as a set ok let's not call it V let's call it E as a set and suppose we have two subsets here this is the subset B and here we have a subset C ok and now we're just thinking of the number of elements ok suppose B has B has X elements and C has let's see Y elements okay so how many elements are there in B Union C no not or less necessarily less because you're counting these guys twice do you agree I'm asking how many elements are in the Union in in this entire figure 8 here ok so there are all the elements the X elements of B no no these are just set there are some guys here's one here's another here's a third one there are some elements that belong to both that's what this drawing indicates that there is an intersection ok so the number of elements so the number of elements let's use this notation the number of elements in B Union C equals number of elements in B plus the number of elements in C but if we just do that we're counting these guys twice do you see that minus the number of elements in B intersection C do you agree is is this clear ok and here we're just looking at sets with elements that's it so if this is the number of elements in B Union C is X plus y minus the number of elements in the intersection clear everybody so this is precisely the statement that's going to be written here but it's not going to be in terms of the elements of the vector spaces it's gonna be in terms of the generators in terms of the dimensions if you take a big space V and you take a subspace you in a subspace W the dimension of U and the dimension of W this big thing that the union well the Union is not necessarily a subspace remember we we said that ok but the sum is a subspace and the number of element that the dimension of the sum is the dimension of U plus the dimension of W minus the dimension of the intersection which I'm counting twice ok so that's where the idea comes from and you'll see formulas like this appear in many places in mathematics ok and they're they always have the same flavor but here it's going to be in terms of the dimension ok so is the idea clear again I'm not going to prove it because the proof is technical the proof requires a lot of time and a lot of concentration and I'm I really am not convinced that it's worth that time in as far as your needs go ok you're not mad patience after all just a number of elements okay so let u and V u and W be subspaces then the dimension of u plus W this is a subspace okay it's all the guys of the form little u plus little W right equals the dimension of U plus the dimension of W minus the guys that I counted twice the dimension of you intersection W and this again is a subspace it makes sense to talk about its dimension okay clear okay so this is part one of the theorem and part two remember we had the notion of a direct sum a direct sum of spaces we denoted it with a circle around the plus was the situation where there was no intersection okay so by definition a direct sum was a sum where each vector can be written in only a unique way remember that and then we had a theorem that a sum is a direct sum if and only if there's nobody in the intersection except the zero vector remember that okay so u plus W is a direct sum denoted this is just a reminder u plus W if and only if and this is going to be pretty obvious if you believe part one part one is what requires a lot of proof if it only if the dimension of u plus W equals the dimension of you plus the dimension of W because if it's a direct sum here's the proof of two I'm saying the proof of two out loud a sum is direct if and only if the dimension of the intersection the intersection is just the zero vector and therefore the dimension of the C of the intersection is zero right there's no nonzero vector in a basis for for the zero subspace okay so this becomes nothing and you get this okay so sometimes you'll see it as a statement that the dimension of u plus W is the dimension of you maybe we'll write that as two so this goes along together okay if u plus W is not a direct sound and this does not hold okay only we do need the intersection okay good clear okay so I'm not I'm not proving these things but I do want to do examples of course so let's do an example so in fact I used you rather than W okay let's erase this and do an example so example let's take Z again to be two by two matrices over R and let's take two subspaces so u U is gonna be the subspace that we had before all the matrices of the form a B minus B a plus B so this is an example we considered earlier okay and W is gonna be all the matrices satisfying that they call inside with their transpose do you remember what we called a matrix like this symmetric all the symmetric matrices okay so let's study let's let's understand what is u what's the dimension of you what is W what is the dimension of W what is u plus W what's it's dimension what is U intersection W what's it's dimension okay let's study all the different possibilities here so let's start with you we know we have seen just not a long time ago that the dimension of U is 2 and that the set 1 0 0 1 and 0 1 negative 1 1 is not just spans as a basis for you and we even called it bu let's call it you again good remember this okay what about W what about W so W is all the matrices of the form a equals a transpose this I can I can rephrase in a similar manner to this okay so a general element in W so this is U W so a general element a general matrix in W is what a and C can be anything and being symmetric imposes that B here has to be the same as B here do you agree that's the property of being symmetric when you do the transpose you get the same thing do you agree okay therefore I can write this as a times 1 0 0 0 plus B times 0 1 1 0 plus C times 0 0 0 1 do you agree so a basis for W is the set 1 0 0 0 0 1 1 0 and 0 0 0 1 okay well I said this is a basis let's call it BW but that requires some justification because just having written it like this is only the statement that it's a spanning set right every element here is a times this plus B times this plus C times this so this set spans W in order for it to be a basis for W I need to show that they're linearly independent right so this set is a spanning set under spans W spans W and why is it linearly independent well we know how to do it we straightened them out as rows in a bigger matrix and then it's already in a chillon form let's show that and is linearly independent since okay that's true but completely unnecessary right all we care is is is it an echelon form is there a zero row that's all we care about in this case so it's linearly independent since when we straighten it out as rows of a bigger matrix 1 0 0 0 0 1 1 0 and 0 0 0 1 since this is in a Shalon form with no 0 rows good so BW is indeed a basis for W so therefore BW is a basis for W and what is the dimension of the you write good because the basis has three elements good everybody ok so now let's talk about u plus W so what is u plus W it's all the elements in V all the 2x2 matrix which can be written as a sum of a matrix in u of that specific form which defined u plus a matrix and W of that specific form which define W do you agree ok I can start understanding what a matrix like that would look like by kind of intersecting those restrictions ok let's look back one board what would a matrix have to satisfy in order to simultaneously be of this form and of this form I can start understanding those conditions and that's a that's a legitimate approach but I can say something else ok I can say any element and u plus W is spanned by the union of these two bases because it's something from you plus something from W that's an element in u plus some W something from you is a linear combination of these two guys something from W is a linear combination of these three guys so something from u plus something from W is precisely a linear combination of these five guys do you agree so be you together with B W is a spanning set for u plus W does everybody agree let me say it again something from you is a linear combination of these two guys because that's a basis for you something from W is a linear combination of these three guys cuz this is a basis for W something from u plus something of from W is a linear combination of these five guys clear okay so let's write that let's give them names okay let's make life a bit easier by giving them names so let's guide this let's call this guy a 1 let's call this guy a 2 this guy is gonna be a 3 a 4 and a 5 ok and now let me write the set a 1 a 2 a 3 a 4 and a 5 a 1 and a 2 or the basis for you a 3 through a 5 or the basis for W spans u plus W does everybody agree to this can it be a basis write it automatically is too big it cannot be a basis because it has five elements and we're working in an ambient space which is four-dimensional so a linearly independent set in V has at most four elements so we automatically know that at least one of these is extra at least maybe to our extra okay we don't know that yet but at least one is extra do you agree so this spans u + w it is not linearly independent since the dimension of the year sorry sorry the dimension of the ambient space is 4 and this has five elements ok good ok so how do we extract from this four guys or maybe only three that are linearly independent what do we do right we straighten these guys in two rows of a big matrix and start doing grow operations until we reach the Echelon form there's going to be at least one row of zeros there okay so let's do that so okay so what you're saying is that if sometimes instead of just diving into the the algorithm or the method for doing something you can take a step back look at it and sometimes you can kind of make observations that would save you a lot of calculations that's what you're saying right and you're saying let's see if I agree you're saying let's look at these five guys and what are you saying a one is a three plus a five does everybody agree so we can right away throw a one out would that mean that necessarily these four are linearly independent no maybe still there is some dependents going on here do you agree okay so there's still but I agree this is always a good approach I want to write the the the full matrix with the five rows it's very easy they're all zeroes and ones here so it's not going to be a lot of work but I want to do it the long way with your permission okay so here are the the five rows in in the order that they appeared a 1 a 2 a 3 etc so it's 1 0 0 1 0 1 negative 1 1 1 0 0 0 0 1 1 0 and 0 0 0 1 this is a 1 a 2 a 3 4 a 5 straightened out as row vectors in a bigger matrix and now I want to do the row operation and reach the H longs for him so I'm gonna do R 3 goes to R 3 minus R 1 and R 4 goes to r4 minus R 2 in any lesson in in Algebra one once pearl s you should do row operations on a matrix right it's it's kind of a therapy even if it's not necessary if you're talking about something completely abstract just take a matrix and do some row operations to calm down right if you're nervous because of something totally unrelated that has to do with you know personal relations politics your car broke down take a matrix and do some row operations you'll you'll feel called right okay what do we get tell me if I'm not making any mistakes here 1 0 0 1 0 1 negative 1 1 0 0 0 negative 1 0 0 2 negative 1 & 0 0 0 1 and then I'm gonna do our 5 goes to our 5 + R 3 and our 3 I'm gonna replace with exchange locations with R 4 and I get 1 0 0 1 0 1 negative 1 1 0 0 2 negative 1 & 0 0 0 negative 1 and then 0 goes good stakes good ok so this is s lon 4 there's one row of zeros right so these guys are a basis for u + W we started with a spanning set the rows of a and the rows of B span have the same span that's a theorem that we had this one's not necessary so the span of these four rows is exactly the same as those and this is a basis I don't need to figure out which ones were the ones that were unnecessary which ones did I throw out right I could start tracing but this could be a process of 17 intermediate steps and it could be very hard to trace I don't need to okay so be u+ w which is the set of those four matrices 1 0 0 1 0 1 negative 1 1 0 0 2 negative 1 and 0 0 0 negative 1 is a basis for u plus W the dimension of u plus W equals 4 and one further conclusion u plus W is everything right because of the second part of the theorem we had before if you have a subspace with dimension the same as the ambient space it's everything you plus W equals V do you agree everybody ok one more subspace that we want to study is the intersection of U and W that's also a subspace let's do that so for that I'm gonna resort to this theorem which is here ok so the intersection I can before doing anything I can say what how big the intersection is going to be right one one-dimensional right because u plus W was four dimensional right so before I erase it let's let's okay let's do this let's do this let's this ok so now we're looking we're continuing the example and we're looking at you intersection W so the dimension of you intersection W is 1 since this we know by the theorem and in our example the dimension of u plus W was four right that's what we just saw the dimension of you was to the dimension of W was three therefore this has to be one do you agree this is a weird way of writing an equation let's just put the one here but good so we know that that I mention of the intersection is going to be one but that doesn't tell us what the what a basis for the intersection is what is the intersection okay by the way is is this a situation of a direct sum no right there is a non-trivial intersection okay there are elements which are both in U and in W simultaneously okay so it's not a direct sum so we know that we're expecting something one-dimensional how do we find the basis so for this we can for example look at what is an element that satisfies being both in U and in W that's usually easier than satisfying being in U Plus being in W okay that sorry being a sum of something in U plus something in W being in the intersection is usually easier to observe so look at this board for a second being at the intersection means it's simultaneously of this form and simultaneously of this for that's what it needs to be in the intersection it's both here and here okay so this puts only one restriction the the the inverse diagonal elements should be equal but this says that they're not equal there one is the minus of the other and that doesn't leave my choice B has to be 0 do you agree and there's no restriction on a and C has to be of the form a plus B but B is 0 so do you agree that elements in the end section are merely what we call scalar matrices a zero zero eight does everybody see that so let's write it here indeed elements in the intersection in U intersection W are of the form a 0 0 a and a basis for u intersection W is merely the set 1 0 0 1 is a basis good ok so this is an example of it's not just an example it's in fact a classical example it's a very standard form of question that you would you will see a lot even in exams and homeworks and you kind of get a subspace get another subspace and are asked about the sum the intersection is it a direct sum find the basis for the intersection find the basis for the the psalm and so on and so forth ok so this is really a very standard sort of question in this context ok questions everybody good ok so this this wraps up our discussion of bases and dimensions for subspaces coming up next is a further step towards the idea that we already are really getting a feeling for that sometimes two seemingly different spaces are actually the same space in disguise ok and that's coming up next in in in in terms of the notion of a coordinate vector next

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Length: 52min 16sec (3136 seconds)

Published: Thu Nov 26 2015