 We are discussing the notion of vector spaces we have seen several examples let us now look at a few properties of that hold in a vector space before that let us go back to the axioms the existence of negative element okay that is axiom corresponding to the existence of the negative element for every U element of V I mentioned that there exists V element of V such that U plus V equals 0 okay. Now this V such a V is unique that can be proved easily I want to introduce a notation I want to introduce notation this notation is just to appeal to our intuition it is an additive inverse so we will call it the negative element we will denote it by minus U see this is just a notation presently we do not know what this means okay this what we know is that this is a vector in V which satisfies this equation so minus U is that unique vector V that satisfies this equation okay. I want to discuss four properties that hold in a vector space for elementary properties which we will be using on many occasions let V be a vector space then the following hold the following properties hold first property alpha into 0 equals 0 this is true for all alpha element of F here after I will write R for F this these properties hold for C also okay so unless I specify the underlying field will be the field of real numbers for all alpha element of R let us observe that this is scalar multiplication okay scalar vector so this 0 is a 0 vector the right hand side vector is also the 0 vector okay property 2 0 times U equals 0 for all U element of V this time you need to differentiate the 0 on the left is a scalar 0 that comes from the field the additive identity the 0 on the right is a vector 0 this holds for all U property 3 if alpha into U equal to 0 then either alpha equal to 0 or U equal to 0 okay again we are using the same 0 for scalar as well as vector so let us understand that this 0 is the vector 0 this is scalar 0 this is again vector 0 that is property 3 property 4 it is a only with regard to property 4 that I wanted to use this notation so what property 4 says is minus 1 times U is minus U for all U element of V so minus U is precisely minus 1 times U minus 1 into U okay these are elementary properties quick proof for proving 1 and 2 I will use the following result the equation X plus X equals X for X and V has so can you say something about X X must be 0 the equation X plus X equal to X for X and V has X equal to 0 as the only solution as X equal to 0 as the only solution okay let us prove this once we have this properties 1 and 2 will follow immediately okay let us prove this quickly what is the reason so I have X plus X equals X and I will add minus X to it I know that minus X in the vector space that is minus X is a negative of X just now I introduce that notation so I had minus X plus X plus X that is minus X plus X minus X plus X I know this is a negative element so this is 0 I know the addition is associative so minus X plus X plus X again use the fact that this is 0 0 plus X equals 0 but I know that 0 plus X is X so X is equal to 0 so if this holds for X then X must be 0 and clearly 0 satisfies this equation okay so 0 is the only solution of this equation let us now consider the proof of 1 I will set X to be alpha into 0 okay then alpha into 0 is alpha into 0 vector which can be written as 0 plus 0 that is distributivity alpha into 0 plus alpha into 0 that is X is equals X plus X just from what we have seen now it follows that X is 0 okay that is the first one I hope this is clear it follows essentially from the fact that the equation X plus X equals X has X equal to 0 as the only solution so that is first one proof of the second I will again set X as 0 into U then previously we used the vector 0 now we will use a scalar 0 and use the property that alpha plus beta times U is alpha U plus beta U so look at X that is 0 into U that is 0 plus 0 this is a scalar 0 into U that is 0 into U plus 0 into U which is X plus X and so X is equal to 0 so that is property 2 exactly quite elementary okay property 3 we need to show that I am given that alpha into U is 0 we must show that either alpha is 0 or U is 0 or both okay if alpha is 0 then there is nothing to prove let us take the case that alpha is not 0 remember that alpha comes from R alpha is a real number non-zero every non-zero real number it is a field so as a multiplicative inverse that is 1 by 1 by alpha so let us look at this equation I pre multiply this equation by 1 by alpha 1 by alpha into alpha into U is 1 by alpha into 0 1 by alpha into 0 is 0 1 by alpha into alpha I am using axiom 7 alpha into beta U is alpha beta into U 1 by alpha into alpha into U 1 by alpha is the inverse of alpha so this is 1 1 times U condition axiom 8 1 into U is U U is 0 okay that is that is proving 3 prove a 4 okay 0 times U I start with this 0 into U that is I can write this as 1 plus minus 1 into U this happens in the field 0 is a scalar here this happens in the field 1 plus minus 1 is 0 in the field this is 1 into U plus minus 1 into U that is U plus minus 1 into U I will add minus U on both sides 0 into U I know as 0 okay so minus U plus 0 on the left minus U plus U plus minus 1 into U minus U on the left minus addition is associative this is 0 plus minus 1 into U that is just minus 1 into U so we have shown that minus 1 into U is minus U so the negative element that we denoted by minus U is minus 1 into U okay okay so this is quite elementary okay let us proceed we had looked at several examples yesterday several examples of vector spaces I will pick up one of them we used this notation pk I think I used pn yesterday but I want to look at pk pk is the vector space the real vector space of all polynomials of a degree less than or equal to k for a fixed positive integer n for a fixed positive integer k this is a real vector space by which I mean the coefficients coming from the field of real numbers it is easy to see that for L less than or equal to k take the non-trivial case it is easy to see that p L is contained in pk strictly and that this holds also p L is a vector space in its own right p L is a vector space in its own right okay since I have included equal to I will include equal to here also it is a vector space in its own right with respect to the same operations let us observe this with respect to the same operations as you have in pk same addition and the scalar multiplication okay so this motivates the definition of a subspace okay which is what I want to discuss next from this we will look at the notion of a subspace it makes sense to say that p L is a subspace of pk whenever L is less than or equal to k p L is a subspace of pk whenever L is less than or equal to k what is a subspace a subset W of a vector space V is called a subspace of V if W itself is a vector space with respect to the same operations by which I mean vector addition and scalar multiplication from V with respect to the same operations in V such a subset this is a special subset which is also as vector space such a subset is called a subspace of V okay we have already seen an example p L is contained in pk p L is a subspace of pk whenever L is less than or equal to k let us look at other examples but before that to verify that these examples are indeed examples of subspaces we need a device so let us prove this result which will be useful in the examples. So we have the following theorem for a subspace for those of you who have studied group theory this will not come as a surprise a subset W of a vector space V is a subspace of V if and only if the following two conditions hold first condition x, y in W implies x plus y is in W it must be close with respect to the vector addition where this plus is the same as the addition in the vector space V only to emphasize x element of W alpha element of R implies alpha into x belongs to W this is closer with respect to scalar multiplication so one needs to verify just these two conditions to show that certain subset is a subspace okay just these two closer axioms really okay let us prove this first and then look at examples of sort of prove this theorem there are two parts it is if and only if there is a necessity part there is a sufficiency part if W is a subspace that is the easy part if W is a subspace then these condition holds that is the easy part the W be a subspace of V then obviously W is a vector space so 1 and 2 hold trivially because it is a vector space it is a vector space in its own right so one part is easy if W is a subspace then these two condition hold okay it is a converse that is interesting conversely suppose that W is a subset of V such that conditions 1 and 2 hold we must show that W is a subspace let us quickly write down the axioms of a vector space okay this is also useful in recalling the definition of a vector space what are the axioms first we must verify that it is closed okay so X Y in W implies X plus Y is in W then second condition alpha in R X in W implies alpha X is in W these two are the closer axioms these two axioms hold because conditions 1 and 2 have been assumed then associativity X plus Y plus Z equals X plus Y plus Z for all X Y Z this holds because these are elements of V and in V associativity holds so this equation holds in W commutativity X plus Y equals Y plus X that is I am picking elements X and X Y and Z see to emphasize this is true for all X Y Z in W I am writing down the axioms for W being a vector space X and Y are from W but it does not matter I am looking at X plus Y in V so these two are the same so I have associativity I have commutativity so let me say this is there this is there this is there this is there all these four let me write down the other conditions the other act for each X element of W there exists Y element of W such that X plus Y is 0 this we do not have presently we will prove this okay presently we do not have this we will prove this but what what we have is X is in W so X is in V so I know there exists Y in V such that X plus Y is 0 why should this Y belong to W we will show that this Y will indeed be in W if those two conditions hold okay but this right now we do not have the next one is also not there well even before this look at this axiom there exists there exists 0 in W such that X plus 0 equals X for all X in W we need to verify this even before verifying this what I know is that there exists 0 in V why should that 0 we will actually prove that 0 is in W also okay so this we do not have presently so we will prove that this the 0 in V belongs to W and then prove this existence of additive inverse what are the other axioms then with respect to scalar multiplication I have written down axiom 1 axiom 2 is what distributivity alpha times X plus Y this is alpha X plus alpha Y for all alpha in R XY in W again this is true because this holds in V and so this is true condition with regard to distributivity over the fields alpha plus beta into X that is alpha X plus beta X alpha beta in R X in W this again is true what else do we have alpha into beta X this must be alpha beta into X alpha beta in R X in W this again is true because this happens in the vector space V finally 1 into X equals X for all X element of V for all X element of W we want this to hold in W this is also true because if it is in W it is in V and for elements in V this holds so this is also true we need to only verify these two conditions first 0 belongs to W and that for every X in W there is a negative element okay yes yeah W is a subset in which the conditions 1 and 2 hold we must show that W is a subspace by definition we must show that W is a vector space I have written down all the axioms of a vector space and the tics correspond to those which we do not have to prove the first two tics we already have as part of the conditions of the theorem the others hold for any subset not just a subspace the other axioms hold for any subset so we do not have to prove the others they are already there for you we need to only prove these two I will prove this first and this I will prove that 0 belongs to W where the 0 comes from V okay so they share the same additive identity that must happen but that is easy look at condition 2 from condition 2 which states that for all alpha in R and X in W I have okay just choose alpha to be 0 then 0 into X belongs to W for all X in W but the property first property that we proved 0 into X is 0 and so 0 belongs to W okay that is easy take the scalar to be 0 so this also holds so let me remove I am sorry this holds so let me remove this we have proved that 0 belongs to W and this is easy it is enough just to prove that 0 is in W this holds in the entire V we need to show existence of additive inverse okay how do you do that take alpha equal to minus 1 consider the case alpha equal to minus 1 I am again looking at condition 2 that is it is close with respect to scalar multiplication take alpha equal to minus 1 then minus 1 into X belongs to W whenever X belongs to W property 4 that we proved last time minus 1 into X is minus X minus X belongs to W for all X in W so that was easy okay so existence of additive identity that is existence of 0 and existence of negative element and so W is a vector space in its own right with respect to the same operations of V and so W is a subspace by definition okay so you need to only verify that to verify that a certain subset is a subspace you need to only verify that the closure axioms are satisfied okay so let us now look at examples quite a few of them examples of subspaces I have already given one let us look at one coming from linear equations homogeneous linear equations so I want to list a few subspaces I will take this as example 2 look at V as R n by which I mean a real vector space underlying field is R I am given a fixed matrix of order m by n m rows n columns I will define the subspace WA I will define the subset W which depends on A as the set of all X and R n such that ax is equal to 0 this example was given as an example of a vector space yesterday's class I hope remember this is a subspace this is a subspace of R n remember this is contained in R n okay A is m cross n X is in R n so the product is m cross 1 so this belongs to R m this is a 0 vector of R m okay to write this in detail this is 0 0 et cetera 0 that has m coordinates then this WA is a subspace let me prove this very quickly by using the previous theorem let us take we only need to prove that it is close with respect to addition and scalar multiplication so let us take 2 vectors x y in WA then we have ax is equal to 0 and ay equal to 0 solutions of the homogeneous equation I must show that x plus y belongs to WA as well as alpha x consider z as x plus y I will call x plus y as z I must show that az is 0 so let us look at az, az is ax plus y but I know that matrix multiplication is distributed this kind of a thing can be split ax plus ay that is 0 plus 0 that is 0 so az is 0 that is we have shown that z belongs to WA and so it is close with respect to addition scalar multiplication still easy suppose ax equal to 0 and alpha is a real number let us call w w is not a good idea z prime as alpha into x look at az prime that is a of alpha x but I know from matrix multiplication alpha is a scalar this is alpha into ax alpha into 0 that is 0 so I have shown that z prime belongs to WA okay I am treating this as the first example of a subspace and proving it in detail that it is indeed a subspace from the next example onwards I will just wave and it means it is an exercise for you okay so this is a subspace example 3 you have any questions example 3 V is R n cos n this time the space of square matrices with real entries let me define w as the set of all matrices A that satisfy this condition okay let me write like this the notation probably I could include that also here what is aij for me aij is the i throw jth entry of A just to confirm this notation okay i throw jth element is a jth throw i th element these matrices are called symmetric matrices A equals A transpose okay set of all matrices that satisfies the condition set of all matrices A that satisfies the condition A equals A transpose set of all symmetric matrices all A such that A transpose equals A if you are familiar with this operation of taking transpose the rows become columns as a result the columns become rows look at this subset okay obviously a subset this is a subspace w is a subspace that is to verify that w is a subspace you must ask whether this question has an affirmative answer if A and B are symmetric matrices then A is A plus B symmetric A plus B is symmetric obviously you can verify that even if you do not know this equation if A is symmetric alpha scalar alpha times A is that symmetric yes okay so by the previous theorem these two conditions are enough to verify so this is a subspace let us try and do a similar thing for complex matrices okay this time the underlying field is the field of complex numbers so B is C n cross n set of all n cross n matrices whose entries are complex numbers the underlying field is understood to be understood to be C in this let me define w as a set of all A such that I will give a similar definition A ij is A j i bar this is called the conjugate transpose okay for all one less than or equal to i, j less than or equal to n that is A star equals A where A star is the conjugate transpose A bar transpose to write down an equation similar to A transpose equal to A I have A star equals A that is A star is you take the complex conjugates of the matrix A first and then take the transpose that is A star such matrices are called Hermitian matrices okay A is a Hermitian matrix sometimes we also use the word self adjoint A is a Hermitian matrix is this a subspace of intuitively we would say yes okay the answer is no this is not a subspace of V the reason is the following I will not give the proof but you try to fill up the gaps you take a Hermitian matrix the diagonal entries of a Hermitian matrix must be real numbers okay let me just write down a 2 by 2 Hermitian matrix this is how it will look like where this is easy to see a 2 by 2 Hermitian matrix just to give you a feel of the fact that this is not a subspace a 2 by 2 Hermitian matrix has this form alpha, beta, gamma, delta are real numbers this will be the complex conjugate of this the diagonal entries if you go back and check this A I equals A I bar so the diagonal entries will be real numbers. So if A is Hermitian then the diagonal entries are real numbers I this is a complex vector space so the scalar comes from C look at I times A I into A the diagonal entries are not real so it is not a subspace this is not a subspace of C but this is a subspace of R this time the underlying field is R then this is a subspace okay so what we have done is to write down a formula similar to the real case but we observe that this is not a subspace but it is a real subspace okay that is example 4 look at I introduce this notation yesterday V is okay let me introduce W first and then V let me say V is C 1, 0, 1 C 1, 0, 1 this is the vector space I have not included this yesterday but you can easily verify that this is a vector space and what are the elements the vector space of all differentiable of all once at least once differentiable functions whose first derivatives are continuous similar to C 0, 1 C 0, 1 is a vector space of all continuous complex valued continuous functions this is a complex valued functions which satisfies the property they are at least once differentiable and that the first derivatives are continuous you can verify this is a vector space okay basically two axioms to be verified if F and G are functions that belong to V then F plus G belongs to V and alpha times F belongs to V essentially that is what you need to verify so this is a vector space we also know from calculus that a function is differentiable then it must be continuous and so this is a subset V prime V is C prime, 0, 1 it is not a very good notation let us say this is V 1 for me and this is V 2, V 2 I will use C 0, 1 which I introduced yesterday this is obviously true because every differentiable function is continuous this is a subspace V 1 is a subspace of V 2 and V 1 is not equal to V 2 because there are continuous functions which are not differentiable okay that is example 5. I want to go back to example 2 and then specialize it for R 2 and R 3 the reason why I do this is these examples will be useful a little later when we discuss a notion of linear independence and dimension okay so let us go back to example 2 example 2 is set of all solutions of the homogeneous system okay I am looking at a particular case I want to look at this example 6 which is W is the set of all X in R 2 the plane such that X 2 equals a fixed constant C X 1 C is fixed constant real constant okay this from a monetary geometry of two dimensions we know that this is like Y equal to M X plus C the constant term is 0 so this is the set of all points lying on a certain line whose slope is C lying on a certain line passing through the origin passing through the origin is important okay W is a subspace of R 2 can you see that W is not the entire R 2 and also W is not singleton 0 example 7 W is a set of all X in R 3 such that see here I am following the notation that if X is in R 2 then X is written as X 1, X 2 okay that is a notation that I am using always X in R 3 now X 3 equals a times T X 2 equals B okay let us say X 1 X 2 equals B times T X 3 equals C times T a, b, c are fixed constants T is in R look at the set of all vectors that satisfy this condition so can you tell me what this subset of R 3 looks like is it similar is it not similar to the previous example this is also the set of all points lying on a straight line this is set of all points lying on a straight line passing through the origin if you want you can write down the symmetric form of the line X 1 by A equals X 2 by B equals X 3 by C okay please verify that this is a subspace set of all points lying on a certain straight line passing through the origin passing through the origin W is a subspace of R 3 I will call this W 1 I will define W 2 next W 2 is a set of all X in R 3 such that let me use different constants now C 1 X 1 plus C 2 X 2 plus C 3 X 3 equals 0 for fixed constants C 1 C 2 C 3 I collect all X that satisfy this condition C 1 C 2 C 3 are fixed numbers collect all X that satisfy this condition what is the geometric interpretation this is a plane passing through the origin this is a set of all points lying on a certain plane passing through the origin please verify that this is also a subspace okay W 2 is a subspace W 1 or W 2 they are not the entire R 3 they are not 0 okay there are non-zero vectors in each of these let me include this and conclude today's lecture W equal to 0 is a subspace of any vector space and this will be called the trivial subspace just 0 that is a minimum you need to have for a subspace okay for instance if you want to show that a subset is not a subspace you show that 0 does not belong to that this may work out in certain examples. So this is a subspace because 0 plus 0 is in W then whatever be alpha into 0 is in W so this is a subspace the trivial subspace V itself is a subspace we will be interested in subspaces which are neither V nor singleton 0 those are called proper subspaces any subspace W such that W is not singleton 0 and not V will be called a proper subspace this is called a proper subspace if you go back to the previous examples you will you have examples of proper subspaces example 4 this is a proper subspace not every matrix is a Hermitian matrix there are continuous functions whose first derivative there are continuous functions which are not once differentiable there are points in R2 which do not belong which do not lie on a certain line passing through the origin similar here this is a plane passing through the origin there are points in R3 which do not lie on this particular plane passing through the origin these are all proper subspace okay I will stop here tomorrow's lecture we will look at the notion of dimension that is the notion of linear independence dimension and then the notion of basis.