 So let us move on to the next topic as I told you yesterday in the last lecture we will discuss the transpose of a linear transformation prove the natural prove the fact that the transpose of a linear the matrix of the transpose of a linear transformation respective to the dual basis is the transpose of the matrix of the original transformation with respect to given basis. This is one of the results that we will prove we will also prove this row rank equals column rank in a much simpler manner using the notion of transpose, okay. So I need to define the transpose of a linear transformation I have Vw vector spaces over the same field over the same field and T is linear be a linear transformation I will define T transpose T transpose is from the dual space of W W star to the dual space of V star define T transpose by T transpose G T transpose domain is W star co-domain is V star. So T transpose G must be of linear functional on V first thing T transpose of something belongs to V star so T transpose of that must be a linear functional on V then this G belongs to W star. So let me say that G belongs to W star and T transpose G G is in W star so I will look at a mapping is defined if it is action on an arbitrary element is defined T transpose G of X T transpose G of X this is a function this is a function this is this must be a functional on W so T transpose G of X what is the definition it is defined as G of T X tell me if this is clear I am defining T transpose G at X as G of T of X remember this X belongs to V T is a linear transformation from V to W so T of X is in W G is a functional linear functional on W so G of this this is in W that we have seen just now. So this right hand side is clear that is the definition of T transpose G this T transpose is called the transpose transpose of the linear transformation is called the transpose of the linear transformation what we have not shown as yet is that T transpose G is linear only if I show that this is linear I can write I can write this it is a linear function on V I must show that I have not yet established so let us do that quickly T transpose G is linear sorry T transpose is linear I am going to prove that T transpose is linear T transpose is linear T transpose of alpha G plus H equals alpha T transpose G plus T transpose H so consider T transpose alpha G plus H this function is known by its action on an arbitrary vector so this of X for X in V this by definition T transpose of something some function f of X is f of T of X alpha G plus H T of X I will use one more bracket but remember that alpha G and H are from W star and W star is a vector space I need to look at the operation in that vector space it is precisely alpha G of T X plus H of T X but go back to this definition this is the same as alpha T transpose G of X plus T transpose H of X I can write this as this is like addition of two functions f plus G of X is f X plus G X point wise so this is alpha into T transpose G plus T transpose H acting on X so what did I start with I started with T transpose of alpha G plus H of X I have shown that this is what I have on the right hand side this is true for all X in V for one thing so what it means is that f X equals G X for all X then f is equal to G so T transpose alpha G plus H is equal to alpha T transpose G plus T transpose H now this equation I know holds for all G H in W star and alpha from the underlying real field okay so it is an easy exercise really provided you understand the definition the definition of the transpose is given by this formula okay let us okay so what is it that we have shown we have shown that T transpose is linear and so T transpose G is a functional on V so T transpose G is a linear functional on V I am sorry yeah T transpose G is a linear functional on V so it belongs to V star okay let us look at some properties of this transpose really one theorem where I list all the properties I remember there I do not take I do not demand that V and W are finite dimension this definition goes through for an arbitrary vector space okay first property is the following one can show that null space of T transpose is the annihilator of range of T this is the first property for properties 2 and 3 to hold we need V and W to be finite dimension let V and W be finite dimensional then condition 2 property 2 that the transpose operation map satisfies is that rank of T is equal to rank of T transpose rank of T is equal to rank of T transpose in order to talk about rank is a dimension of the range of T for instance in order to talk about the dimension of the range of T we must know that W is finite dimensional in order to talk about rank of T transpose you observe that T transpose is a functional on V so I need to know that V is also finite dimension. So V and W have been assumed to be finite dimensional this is this is a formula that will really lead to this row rank equals column rank and property 3 property 3 is range of T transpose equals the annihilator of null space of T range of T transpose equals null space of annihilator of T okay we will prove this and then look at 1 or 2 consequences really. Before proving let us understand this equation see on the right hand side I have annihilator of range of T, T is a linear transformation from V to W so range of T is a subspace of W and annihilator of that means it is a functional on W so anything in range of T annihilator belongs to W star look at what we have on this side null space of T transpose does that also belong to W star that is what is the case T transpose null space of T transpose will be contained in W star okay so it is first meaningful to write down such an equation okay let us prove this you want to show that 2 sets are equal show that each is contained in the other and I observed that these are subspaces of W star so let us take G in W star for G in W star we have the following G belongs to null space of T transpose implies T transpose G equals 0 G belongs null space of T transpose T transpose G remember is in V star so it is a functional it is a functional it is a function a function is 0 means its action on any element is 0 so this implies what we also know that it is a functional on V so T transpose G of X this is 0 for all X in V a function F is identically 0 if F of X equal to 0 for all X in the domain T transpose G X let us use the definition of T transpose this means G of T of X is 0 for all X in V definition of T transpose G of T of X is 0 for all X in V now when X varies in V this T X is in W and this T X is in the range of T so from this can I write G of W equal to 0 for all W in range of T I want to show G of W equal to 0 for all W in range of T let us take W in range of T then W can be written as T X question is G of T of X is that 0 yes G of T of X is 0 for all X the thing you need to observe is that T X belongs to range of T so G on any element in the range of T must be 0 this is precisely what we mean by saying G belongs to range of T annihilator G belongs to range of T annihilator range of T is a subspace if what is the notation I am using W let us say if Z is a subspace of a vector space V then Z 0 is a set of all functionals F that satisfy the condition that F of Z equal to 0 for all Z in Z that is precisely what we have done so G belongs to annihilator of range of T can we retrace the steps what have we proved null space of T transpose is contained in range of T annihilator arbitrary G in null space of T transpose we have shown G must belong to range of T annihilator to show the we also need to show that range of T annihilator is contained null space of T transpose but can we retrace the steps here suppose G belongs to range of T annihilator can I conclude G of W equal to 0 for all W in range of T that is the definition so from here to here I am allowed to go can I go from here to here G of W 0 for all W in range of T can I conclude G of T of X is 0 yes that is the definition G of W equal to 0 implies G of T of X is 0 from here it is the definition of the transpose so from here to here to here I have gone just now from here to here definition of transpose if I have a function F such that F of X equal to 0 for all X then F is a 0 function so from here to here I can go and if T G is 0 that is the same as saying G belongs null space of T transpose so these two sets are the same is that clear so this proves the first formula so null space of T transpose T transpose is the annihilator of range of T let us prove the second formula when the spaces are finite dimension proof of the second formula recall that if Z is a subspace of V, V is finite dimensional then this was shown in the last lecture or the previous last lecture then dimension of Z plus dimension Z is 0 this is equal to dimension of V so I will make use of this formula let me start with the range of T rank of T is a dimension of the range of T I will use R for that R is a rank of T that is a dimension of the range of T I will apply this formula for the subspace range of T that happens in W let me now assume that dimension W is M dimension V is M now look at rank plus dimension of range of T annihilator this is a subspace of W so this is equal to M but we have just now shown that range of T annihilator is null space of T transpose R equals M minus dimension null space of T transpose. Now I will apply rank nullity dimension theorem for the operator T transpose rank nullity dimension theorem for the operator T transpose I must know that the domain of T transpose is finite dimensional domain of T transpose is W star W is finite dimension so W star is finite So I can write this as M minus rank plus nullity is a dimension rank of T transpose plus nullity of T transpose equals dimension of what? Dimension of W star that is M rank plus nullity of T transpose so I want nullity of T transpose nullity of T transpose is M minus R I am sorry not R see I am using this equation may be I will write it on this side somewhere I am using this equation rank of T transpose plus nullity of T transpose equals dimension of the domain of T transpose that is W star that is M from this I will take this number nullity of T transpose is this number dimension of null space of T transpose nullity of T transpose nullity of T transpose is M minus rank of T sorry rank of T transpose nullity of T transpose is M minus rank of T transpose cancel M rank of T transpose so I have shown that R is equal to rank of T transpose but R is by the definition rank of T okay so rank nullity damage theorem plus this formula connecting the dimension of a subspace and the dimension of it is annihilator this is okay finally the last formula let me write down we are seeking to prove that range of T transpose equals null space of T annihilator let me first prove that this left hand side subspace is contained in the right hand side subspace and prove that their dimensions are the same then it will follow that they are the same okay we have seen this kind of an argument before. So let us take G in range of T transpose I am seeking to show that G belongs to null space of T annihilator that is I want to show that G of W is 0 whenever W belongs to null space of T okay let G belong to range of T transpose an element is in the range of a transformation so this G can be written as G can be written as T transpose F for some F in W star yeah that is correct T transpose is a function from W star to V star so G is T transpose F that is what is the meaning of this G of X is equal to T transpose F of X which by definition of the transpose is F of T of X belongs to V G must G yeah G is a functional on V so G belongs to V star so this X belongs to V okay. Now I look at let us say W suppose W belongs to null space of T I am seeking to show that G of W is 0 suppose I show G of W is 0 then it means G belongs to null space of T annihilator but that is straight forward look at G of W G of W is equal to anything in W I have this formula F of T of W but W belongs null space of T so T W is 0 so this is F of 0 F is linear so 0 is 0 so if W belongs null space of T then we have shown G W is 0 so G is a functional that must belong to the annihilator of null space of T so G belongs to annihilator of null space of T I started with G in range of T I have shown this so it follows that range of T transpose is a subset of subspace of null space of T annihilator I will next show that the dimensions coincide so it follows that these two subspaces are the same look at dimension of the annihilator of the null space of T I am appealing to this formula dimension of annihilator of null space of T this happens in V there is a dimension of the vector space V minus dimension of null space of T I will apply ranked on the T dimension theorem N minus dimension of null space of T I will go back to this formula rank of T plus nullity of T equals dimension of V that is N what I want is dimension of null space of T nullity of T that is N minus rank of T N minus N minus rank of T that is rank of T but just now we have shown the rank of T is rank of T transpose rank of a linear transformation is the dimension of the range space of the transformation which is what we wanted to show you agree we have shown that range of T transpose is contained in null space of T annihilator we have also shown that these two subspaces have the same dimension so these two subspaces coincide that proves the theorem so again you see that the essential formulas are ranked nullity dimension theorem and the fact that dimension Z plus dimension Z 0 is a dimension of the space where Z is a subspace okay we will now look at derive the two consequences okay the first one is not really a consequence I will use this notation let B V comma B W denote basis for V comma W respectively let B V star B W star denote the corresponding dual basis of V star and W star so I am following this notation B basis B star dual basis I also need to index using the spaces V and W let A be the matrix of T is a linear transformation let T be linear and the T be linear A be the matrix of the linear transformation relative to the basis that I started with B V B W and B be the matrix of T transpose the natural the natural basis that one must consider for T transpose or the dual basis for these two but T transpose remember is a function from W star to V star so T transpose is B W star to B V star T is from V to W so the matrix of T will be B V B W T transpose on the other hand is from W star to V star so B star W to B V star these are the basis this is the order in which we need to take of course these are ordered basis then I told you that these two matrices are related by the formula that one is a transpose of the other remember that if the dimension of V is n and the dimension of W is m this matrix is of order m cross n this is of order m cross n this is of order n cross m this is of order n cross m so please remember this I want to show that A transpose is B I have introduced a notation what is A transpose? A transpose is the matrix okay let us say if what is the definition of A transpose A transpose equal to C means C i j equals A j i C i j equals A j i so if A is of order m cross n C must be of order n cross m and you know the operation of taking transposes this is the definition so we will have to show that these two matrices are the same okay now you will see that it is essentially the formula for writing down the matrix of a linear transformation relative to certain basis so for preciseness I need to write down basis so I want to prove this result let us say B V is n dimensional I will have U 1 U 2 etc U n I will write B star V the dual basis of this I will use the usual notation F 1 F 2 etc F n let me use B W the notation will be slightly different from this let us say I have W 1 W 2 etc W m I am assuming that the dimension of W is m the dual basis denoted by B W star that will also have m elements this time let us say I use G 1 G 2 etc G n all that I will do is write down the matrix of the linear transformation T relative to these two basis the matrix of the linear transformation T transpose relative to these two basis and just compare their actions so by definition this is something we have used earlier T of U j what is the j th column of the matrix A T of U j equals the number I have written summation i equals 1 to j m A i j W i you need to verify quickly and tell me okay now there is an i and a j so the running index this is something we have used before instead of this I will use a running index to be k okay so let me change this to k k is the running index j is the free index that is T U j okay so this is the formula for T U j this U j as many U j's I have there are n U j's this is the definition of the matrix of T relative to these two relative to these two basis B U B V any element of U action under T that must be a linear combination of W n etc W n what is the similarly the formula for T transpose T transpose is the function W star to V star for W star I have this for V star I have this so I must look at T transpose of I will write similarly G j that must be a linear combination of these vectors these functionals I will write a similar formula summation k equals 1 to n this time because there are n functionals here k equal to 1 to or may be l equal to 1 to n for T transpose for T the matrix is A for T transpose the notation is B so I must write this in terms of B so this is B l j instead of W I have f f l do you agree with this this is the same as this A has been replaced by B W has been replaced by f U has been replaced by G T has been replaced by T transpose matrix of a linear transformation relative to two basis this is what we have this time j runs from 1 to n how many G j's are there I am sorry this must be G subscript j this is also f subscript n the notation that I have been following is that for functionals I will use subscripts for coordinates I use subscripts for different vectors I use superscripts different vectors I use superscripts functional subscripts coordinates will not appear here I think okay then I must show that A is equal to A transpose equals B so let me look at consider T transpose G j of U i consider T transpose G j of U i this is by definition G j of T of U i by the definition of the transpose operation T transpose G j U i G j of T of U i for T of U i I will use this formula T of U i I use this formula so this is G j of summation k equals 1 to m instead of j I must use i A k i W k instead of j I must use i I am looking at T of U i now G is G j is linear so this is equal to summation k equals 1 to m A k i G j of W k summation k equals 1 to m A k i G j of W k but G j and W are related they are dual G this is the dual basis corresponding to this B W corresponding to this we know how these are related they are related by the delta j delta i j G i of W j is delta i j let us remember that this j is fixed please understand this for a fixed j and for a fixed i I am doing all this this summation k varies from 1 to m for a fixed j among those values that k takes k will take the value j k will take the value j so for that j G j W j is 1 all the other terms it is 0 there is only one number then this with k equals j A j i directly there is only one term because G and W are related by the formula for the dual basis so what I have done is to show that T transpose G j of U i I have shown that this is equal to A j i I will look at a different formula for the left hand side show that it is equal to B i j then I am through I have used the formula of the I have used the definition of the T transpose to get A j i I will use another formula for T transfer for the left hand side show that it is equal to B i j okay so let us consider this again also T transpose G j of U i I will use this formula the second formula T transpose G j acting on U i so I will come back to this this will be summation L equals 1 to n B L j F L acting on U i T transpose G j formula is this that acts on U i so remember T transpose G j is a linear functional on V it is a linear function on V okay now again what happens L is a running index i and j have been fixed in particular i is fixed when L takes a value i this is 1 all other terms are 0 when L takes a value i this is B i j all other terms are 0 that is a proof so this is equal to B i j okay so there is nothing complicated about this proof write on the formula and simply apply the formula formula for the matrix of a linear transformation relative to certain basis okay so from this it follows that A transpose is B so I have not used the previous theorem this is only using the definition of the matrix of a linear transformation you have to ask the right question what is the matrix of the transpose relative to the dual basis that is the right question okay the final result the final result is given a matrix E its row rank and the column rank coincide we have proved this before by using complicated calculations we will use notion of transposes and the theorem connecting the dimension of certain ranges and get a conceptual proof row rank of A equals the column rank of A okay the proof is we have to use the language of linear transformations so let me define T define T from R n to R m by T of x equals A into x matrix multiplication x is in R n what is the matrix of T relative to the standard basis the standard basis I will use just this B domain is R n so B n B m I must write B R n B R m but that is complicated so I will just write B n B m B n is the standard basis of R n B m is the standard basis of R m so what is the matrix of T relative to the standard basis T x equals A x what is A E 1 what is T E 1 A E 1 first column of A x so the matrix of T relative to the standard basis that is equal to A okay okay let us look at I will also use this notation B n star and B m star for the dual basis corresponding to B n B m respectively this is a notation for the dual basis for B n and B m okay what is a column rank column rank of A is equal to I will start with the range of T what is the rank of T rank of T equals dimension of range of T rank of T is dimension of range of T this is dimension of range of T what is range of T range of T is the subspace set of all y in R m such that y is T x for some x in R n this is a range space but T x is A x okay so this is dimension of set of all y such that y equals A x what is the subspace A x for any x is in the column space so this is a column space of A dimension of the column space of A A x is a linear combination of the columns of A we have seen this before that is precisely the column rank so what did we start with we started with rank of T we have shown that it is a column rank of A where T is a linear transformation which is defined through A by the formula T x equals A x what is rank of T transpose rank of T transpose similarly is the column rank of the matrix of T transpose related to the dual basis which we have just now shown is A transpose we have shown that the matrix of T transpose related to the dual basis is the transpose of the matrix of the transformation relative to the initial basis that we started with so this is a column rank of A transpose column rank of A transpose is precisely the row rank of A but we have also shown rank of T equals rank of T transpose this is finite dimension the spaces are finite dimension we have also shown rank of T is rank of T transpose so since rank of T is the same as rank of T transpose it follows that row rank of a matrix is equal to the column rank of that matrix. So we have done two different proofs this is much simpler but remember much simpler more conceptual but you need the notion of the transpose of a linear transformation okay hence forth we will call this number as a rank of the matrix okay so let me stop here.