 Okay yesterday we had discussed this result SB linearly independent subset of a vector space V and T be a spanning set also assume that SS finitely many elements as well as T SS U1 U2 etc Um and T is V1 V2 etc Vn then what we had shown yesterday was that m is less than or equal to n okay let us use this result and prove that any two bases for a finite dimensional vector space will have the same number of elements. So let us consider the following I want to prove this theorem that S and T be bases for a vector space V with S equal to U1 U2 etc Um and T equals V1 V2 etc Vn that is I have two finite sets that are bases for a vector space V then they must have the same number of elements that is the conclusion then m is equal to n okay two finite subsets of a vector space V if they satisfy the condition that they are bases for the vector space then they must have the same number of elements okay the proof will use this result that I stated just now. Now S is a basis so S is linearly independent S is a basis so it must be a spanning set as well as a linearly independent set S is linearly independent similarly T I will explore the fact that S is linearly independent and the fact that T is a spanning set by the above result it follows that m is less than or equal to n okay the next part is to reverse the roles of S and T S is a spanning set T is a linearly independent set the number of elements in T must be less than or equal to the number of elements of S. So let me just say similarly n is less than or equal to m and so m is equal to n so the number of elements in any finite basis the number of elements in any two finite bases is the same okay so this leads to the definition of the dimension let V be a vector space with a finite basis the dimension of V is the number of elements in any basis of V. So this is a definition of the dimension of a vector space this number is unique by the previous theorem this number is unique and so this notion is well defined okay if V does not have a finite basis if V does not have a finite basis then V is said to be infinite dimensional if V does not have a finite basis then V is said to be infinite dimensional. Look at the trivial vector space V equal to 0 is 0 dimensional okay the only vector space which is 0 dimensional this is the trivial space single term 0 so remember it cannot be one dimensional because if it is one dimensional then the basis will have precisely one element and this one vector cannot be 0 because it must be linearly independent and so this cannot be one dimension it is 0 dimensional okay. Let us then look at familiar examples of vector spaces that we have seen before and determine their dimensions examples the first one is R n for R n we had written down the standard basis in the last lecture I will call this B, B is E 1, E 2, etc E n where E 1, E 2, etc E n are the vectors that were defined in the last lecture this is a basis this is a basis for R n and it is called the standard basis of R n this is called the standard basis of R n okay. Example 2 let us look at the space of polynomials of degree k for this look at the collection P 0, T 1, etc P k where this was also defined before where P i of T is T to the i 0 less than or equal to i less than or equal to k T is a real variable so this set of polynomials they are k in number this set is a basis and so the dimension of P k is k plus 1 there are k plus 1 vectors in that basis okay. What is the dimension of C 0 1, C 0 1 is a space of all complex valued continuous functions on the interval 0 1 what is the dimension of this space this is infinite dimension what is the argument for that this is infinite dimension see the reason is as follows let us take the polynomials that we have defined earlier define P 0, P 1, etc by P i of T equals T to the i this time T varies in 0 1 define this polynomial that is 1 T, T square, etc now this is an infinite subset consider P 0, P 1, P 2, etc P k, etc this is an infinite subset I am claiming that this infinite subset is linearly independent now we have seen this definition yesterday that an infinite subset is linearly independent if and only if every finite subset of it is linearly independent okay. So we need to show that every finite subset of this is linearly independent but is that something that we have proved before take a finite subset okay let us say I have the polynomials 1 T square T power 100 let us say T power 1000 then can we show that this is linearly independent you look at a linear combination equate that to 0 differentiate as many times as you want show that the coefficients are 0 okay same idea that we used earlier can be used here to show that this is an infinite subset linearly independent subset this is linearly independent this is a linearly independent subset of C 0 1 every polynomial function is continuous so these are first members in C 0 1 now can you get a contradiction if there is a finite basis if there is a finite basis then the previous theorem would be contradicted so this cannot have a finite basis okay the number of if it is finite dimensional then the number of elements in any two basis any two basis will be the same but we have we need to use a lesser known fact which is if you have a linearly independent subset and a spanning set then the number of elements in the spanning set must be greater than or equal to number of number of elements in the spanning set must be greater than or equal to the number of elements in the linearly independent set that is clearly not possible here if it has a finite basis then this is clearly violated okay so there is no finite basis for this vector space okay I have given the reasoning fill up the details here if it has a finite basis then this would violate the inequality m less than or equal to n okay so this is an example of an infinite dimensional vectors let us also prove a few more consequences of the previous result let V be a finite dimensional vector space that is V has a finite basis let S be a subset of V then we have the following if S is linearly independent then the number of elements of S cannot exceed the dimension of V if S is linearly independent then the number of elements in S this is the cardinality of S that cannot exceed the dimension of V this stands for the dimension this notation will be used for denoting the dimension that is one property property 2 if dimension of V is less than the number of elements in any linearly independent subset in any subset S then S is linearly dependent okay you will immediately notice that statement B is a counter positive equivalence of statement A so statement B does not really need to be proved condition C if S is property C rather if S is linearly independent if S equals let us say U 1 U 2 etc U n with n denoting the dimension of V if this S is linearly independent that is I take a subset which is which has n elements where n is the dimension of the vector space then if it is linearly independent then it must be a spanning set then S is a spanning set of the vector space B okay a linearly independent subset having the same number of elements as any basis would have has to be a basis a linearly independent subset having the same number of elements as that of a basis of a finite vector space must be a basis okay that is property C okay proof the first one is something we have already proved okay you have to just put the framework there let us take dimension of V to be n this will be useful in part C also and write S as U 1 U 2 etc let us say V 1 V 2 etc V M then we have seen earlier that if S is linearly independent then the number of elements in S that is M that cannot exceed the number of elements in any basis of V M is less than or equal to the number of elements in any basis of V this we have seen before but what is the number of elements in any basis of V that is precisely n because any basis will have n elements since the dimension of V is n okay so this number is n so M is less than or equal to n that is condition A holds property A holds the number of elements in any linearly independent subset cannot exceed the dimension of the vector space B is the contrapositive equivalence so I will simply say B is equivalent to A no proof for that property C S is given to be U 1 U 2 etc U n where n is a dimension of V we must show that S is a spanning set suppose S is not a spanning set we will get a contradiction okay okay if span of S is not equal to V we want to show that S is a spanning set for the vector space V which means you must show that span of S is equal to V if span of S is not equal to V then there exists a vector let us call it X, X is in V such that X does not belong to span of S I have two sets which are not equal and one is contained in the other span of S is always contained in V that is a subspace in fact we have seen so one is a subset of the other but the two sets are not equal which means the superset has an element which does not belong to that subset there exists an X in V such that X does not belong to span S I will look at this set S union X okay include this X define a new set this has these elements U 1 U 2 etc U n and X now since X does not belong to span of S what it means is that X is not a linear combination of the preceding vector X is not a linear combination of U 1 U 2 etc U n so this is a linearly independent set this is linearly independent okay let us recall the argument if this set S union X were linearly dependent then there is at least one vector which is a linear combination of the preceding vectors now that cannot happen for U 1 U 2 etc U n because they are already linearly independent so the only possibility if this set is linearly independent is that this X is a linear combination of U 1 U 2 etc U n we know that is not possible because X does not belong to span V so this set must be linearly independent now this is linearly independent and the number of elements here is n plus 1 so that cannot exceed n but this is a greater number and so you have a contradiction and so this cannot happen that is X there is no X in V which does not belong to span V the number of elements in the set S union X this is n plus 1 this is greater than n which is the dimension of V this contradicts this is a contradiction to property A that we proved just now the number of elements in any linearly independent subset cannot exceed the dimension of the vector space this is a contradiction and so there exists no X in V such that X does not belong to span of S that is S is a spanning set of V that is S is a basis of the vector space V okay so you can this is what we say the dimension is the number of elements in any maximal linearly independent subset or the number of elements in any minimal spanning set okay let us now look at some examples of subspaces and determine their dimensions okay examples of subspaces their dimensions let us first look at R2 okay let us say I start with R3 consider R3 look at the following subspace the set of all X in R3 such that AX1 plus BX2 plus CX3 equals 0 for given constants ABC look at the collection of all X that satisfy this single condition what we know is that this is this is certain plane passing through the origin the constant term is 0 this is a plane passing through the origin we have verified earlier that this is a subspace okay let us determine its dimension let us determine the dimension of this subset that is a plane in R3 intuitively we know that the dimension must be 2 so let us prove that this as a basis considering of precisely 2 vectors okay okay without loss of generality let us assume that A is not 0 at least one of these constant must be non-zero okay this is a non degenerate plane non-trivial plane so take A to be non-zero then I can write X1 as minus B by AX2 minus C by AX3 remember I started with a single equation AX1 plus BX2 plus CX3 is 0 single equation in 3 unknowns X1, X2, X3 ABC are given constants so I can fix one of them and determine I can fix two of them determine one in terms of the other two so I have fixed X2 and X3 determining X1 in terms of X2 and X3 then any vector X in W can be written as X is X1, X2, X3 that is X1 is minus B by AX2 minus C by AX3, X2, X3 are arbitrary this can be rewritten as minus let us say X2 into minus B by A1, 0 plus X3 into minus C by A0, 1 so all that I have done is to write X as a linear combination of these two vectors let us call this as V1 this as V2 set V1 to be minus B by A1, 0 this time I am writing the vectors as row vectors set V1 to be this and V2 to be minus C by A0, 1 okay then what is the claim? The claim is that these two vectors form a basis for W so this must these two vectors must be linearly independent and they span W okay the fact that these two vectors span W has been demonstrated here any X and W can be written as a linear combination of V1 and V2 is that okay that is in this notation this is equal to X2 V1 plus X3 V2 I have written any X and W as a linear combination of V1 and V2 and so this is a spanning set these two span W is clear linear independence are they linearly independent one is not a multiple of the other so these are obviously linearly independent clearly V1, V2 are linearly independent and so I have a basis V1, V2 is a basis of the subspace W which is a plane passing through the origin so the dimension of any plane passing through the origin is 2, so dimension of W equals 2 okay we also prove the converse the converse is true if the dimension of a subspace is 2 then it must be a certain plane passing through the origin okay so let us prove that that is we are determining subspaces of dimension 2 precisely okay we are saying that in R3 if a subspace has dimension 2 then it must be a plane passing through the origin and what we have just now shown is that if it is a plane passing through the origin then its dimension is 2 okay so this is complete understanding of two dimensional subspaces of R3 okay so the converse is what I want to show next Firstly let us show that if W is a subspace of dimension 2, W is a plane passing through the origin okay so let us look at W and call a basis B of W consisting of elements u this be a basis of W I know that there are two elements because the dimension is 2 I must show that this W is a plane passing through the origin okay let us take X and W I will take an arbitrary point in W I will show that this arbitrary point satisfies the equation of a plane okay I will show that this point satisfies the equation of a plane so it follows that any point in W must lie on the plane and it will then follow that these two are the same okay let us look at okay before that let me write down X as a linear combination X is let us say alpha times u plus beta times v let me use coordinates u equal to u 1 u 2 u 3 v equals v 1 v 2 v 3 and X is equal to X 1 X 2 X 3 okay so I have X 1 X 2 X 3 equal to alpha times u 1 alpha u 2 alpha u 3 plus beta v 1 beta v 2 beta v 3 this is my X that is X 1 equals alpha u 1 plus beta v 1 X 2 equals alpha u 2 plus beta v 2 X 3 is alpha u 3 plus beta v 3 any X in V is a linear combination any X in W is a linear combination of the vectors u and v I have written down the expanded form of the coordinates of X of the components of X. Let me also define a new vector I will call that as a vector z let me define a new vector z to be the cross product of u and v z is a cross product of u and v remember that u and v are basis vectors so none of them is 0 none of them is 0 one is not a multiple of the other and so u cross v must be perpendicular to both u and v okay vector calculus z is perpendicular the vector z that we have defined is perpendicular to u and v then perpendicularity is what a dot b equal to 0 and for three dimensional vectors a dot b is the component wise multiplication and then addition a 1 b 1 plus a 2 b 2 plus a 3 b 3 so what I have is z dot u is 0 as well as z dot v that is z 1 u 1 plus z 2 u 2 plus z 3 u 3 this is 0 as well as the corresponding equation for z dot v equal to 0 z 1 v 1 plus z 2 v 2 plus z 3 v 3 this is 0 okay this z is perpendicular to u and v now u and v are plane vectors z is perpendicular to u and v we are in three dimensions we will show that X belongs to the plane generated by u and v so we will show that X is perpendicular to z we will show that X is perpendicular to z it would then follow that X must lie on the plane generated by u and v I will write down the equation of the plane explicit so the last step is to consider X dot z I will show that this is 0 X dot z is X 1 z 1 plus X 2 z 2 plus X 3 z 3 use the formulas for X 1 X 2 X 3 so let me write z 1 first z 1 into alpha u 1 plus beta v 1 alpha u 2 plus beta v 2 plus z 3 into alpha u 3 plus beta v 3 I have written down the formulas for X 1 X 2 X 3 in terms of the coordinates of components of u and v so this is take alpha outside z 1 u 1 plus alpha outside z 2 u 2 the last term alpha has been taken outside z 3 u 3 plus the other set of terms beta into z 1 v 1 plus z 2 v 2 plus z 3 v 3 but I have just now observed that these two numbers are 0 so this is 0 so X dot z is 0 that is expand the expanded form gives me X 1 z 1 plus X 2 z 2 plus X 3 z 3 equal to 0 where z is a fixed vector z 1 z 2 z 3 are fixed numbers they are fixed numbers because they come z is u cross v u and v are fixed the coordinates components of u and v are fixed so z is fixed that is the coordinates z 1 z 2 z 3 are fixed numbers so this is like A X 1 plus B X 2 plus C X 3 is 0 so X lies on a plane passing through the origin I have taken X as an arbitrary element here so any vector in W must lie on the plane passing through the origin okay so we have determined all subspaces of dimension 2 in R 3 okay so also look at the following result what are the possible the following result is motivated by this question what are the possible dimensions of any subspace of a finite dimensional vector space all integers lying between 0 excluded till the dimension of V okay intuitively that is clear let us prove it quickly a quick proof of this fact let W be a subspace of a finite dimensional vector space V then dimension W cannot exceed dimension of V dimension W cannot exceed dimension of V okay intuitively this is clear a subset cannot have a greater dimension than the superset proof is as follows let us take B to be a basis of W then B is linearly independent in W any basis must be a linearly independent spanning subset so this is linearly independent W but linear independence in W is the same as linear independence in V because it does not depend on anything that has got to do with W the only thing that I know is that the vectors are in W so this means that B is script B is a linearly independent in V but what we know is that any the number of elements in any linearly independent subset cannot exceed the dimension of the space okay so the number of elements in B cannot exceed the number of the cannot exceed the dimension of the space V but this number is the dimension of W that has dimension W is less than or equal to dimension V okay so that is an easy consequence okay let us now look at R3 and then list all the subspaces and their dimensions we will use this result consider the vector space R3 the dimension of R3 is 3 let W be a subspace of R3 then by the previous theorem what we know is that dimension of W equal to 0 1 or 2 it can be 3 also so let us say it is include that and then see what happens if it is equal to 3 dimension W is 0 1 2 or 3 dimension W equal to 0 this means W is singleton 0 0 dimensional 0 is the only vector space look at the other extreme dimension W is 3 okay in this case W equal to span of B where B has 3 elements okay let us say X Y Z if W has dimension 3 then it has 3 elements in any basis but now look at the result that we proved today if I have a linearly independent subset that has the same number of elements as dimension of the vector space then that must be a basis for that vector space if I have a linearly independent subset having the same number of elements as the dimension of the vector space then this must be a basis for that vector space okay using that result it follows that B is a basis for R3 follows that B is a basis for R3 okay but this means W is equal to R3 is that okay in this case W must be the whole of R3 yes 0 1 2 3 these are the four possibilities the dimension of the previous result says the dimension W does not exceed dimension B it can be equal so these are the yes 0 1 2 3 these are the four possibilities B has dimension 3 dimension is a non-negative number dimension is 0 1 2 etc up to dimension B dimension of R3 is 3 so it is from 0 to 3 0 is really trivial because the only vector space which has 0 as its dimension is a 0 space so that is anyway trivial but we must include it here it is a non-negative number it is a number of elements in a basis so it can be 0 so 0 should also be included even though it is trivial okay so these are the two extremes if the dimension is 0 then the subspace is a trivial subspace if the dimension is 3 the dimension of the original space then the subspace must be equal to R3 so we are left with these two possibilities dimension is 1 dimension is 2 dimension 2 we have disposed of dimension W equal to 2 we have disposed this means W is a plane a certain plane passing through the origin this we proved just now if dimension W is 2 then W is a certain plane passing through the origin so we are left with dimension W equal to 1 so what is your guess in that case it must be a certain line passing through the origin okay so let us dispose that off by giving a quick proof so we want to consider the case dimension W equals 1 let us first observe that let us first observe that if let us call something else let us call h let h be the set of all x in R3 such that let me first write down the set of all points lying on a certain line passing through the origin set of all x in R3 such that x1 is a times t x2 is b times t x3 is c times t where abc are given constants t varies in R abc are given constants t varies in R we have seen this before we have shown that this is a subspace okay now if x belongs to h then x1 by a equals x2 by b equals x3 by c that is this is a symmetric form of a line in R3 that is x lies on a certain line passing through passing through the origin that direction ratios are abc okay x lies on a line through the origin what is also clear is that if x is in h then x can be written as x1 x2 x3 that is equal to use these equations I will take t outside t times abc I will call it t times u where u is the vector whose components are abc okay now this is a non-trivial line so there are 3 constants abc at least one of them is not 0 so the vector u is not 0 a single vector non-zero linearly independent and what we have shown is that any x in h is a multiple of this any x is a linear combination of this and so this is a basis for h so dimension of h is 1 what we have shown is that if the collection of all points lying on any straight line passing through the origin that is a subspace of dimension 1 converse maybe I will leave it as an exercise if h is a subspace of R3 of dimension 1 so that h is precisely the set of all points lying on a certain line passing through the origin okay that is going to be an exercise for you it then follows that we have understood all one dimensional subspaces of R3 they are precisely lines passing through the origin okay let me stop here.