 See in the last lecture we were trying to prove this result that the row rank is equal to the column rank okay the proof was left incomplete so first task for today is to complete that and then I will discuss a notion of the matrix of a linear transformation over finite dimensional spaces and then look at some of how to compute the matrix for instance one or two examples of linear transformations where I will tell you how to construct the matrix of the linear transformation and then discuss probably some of the elementary preliminary properties and then continue this discussion further okay. So let me recall that we are our first task is to prove that the row rank is equal to the column rank I will quickly go through what we had done earlier A is an m cross n real matrix and the row rank of A is equal to the column rank of A where what is the row rank so all these will be recalled quickly row rank is equal to the dimension of the row space of A dimension of the row space of A and if R is the row reduced echelon matrix row equivalent to A then this is equal to the number of non-zero rows of R we use the notation small R for that okay let me recall again this capital R is a row reduced echelon matrix row equivalent to A then we had seen that the row rank of A is equal to the number of non-zero rows of the row reduced echelon form of A. What is the column rank of A? Column rank of A is equal to the dimension of the column space of A where what is the column space it is the subspace spanned by the columns of A and we had seen earlier that anything any vector in the column space of A can be written as a linear combination of the columns of A by definition and so if Y is in the column space of A then Y is equal to AX for some X okay so let me write this is the dimension of the subspace Y such that Y equal to set of collect all Y such that Y is equal to AX for some X in RA this is the subspace this is the column space of the matrix A so it is a dimension of this space that is called the column rank the number of non-zero rows of the row reduced echelon form of the matrix A we have seen that is a row rank of A we want to show these two are equal and where we started the proof was to look at the system so I am getting into the proof and the first few steps we have discussed earlier I am recalling these steps S is the solution set of the system AX equal to 0 we are using this notation S is the solution set of the system AX equal to 0 this is a subset of R in this is a subspace of R in and what we know is that this is the set of all X in R in such that from what we have discussed when we looked at the row reduced echelon form we know that these two sets are the same the solution set does not change. Now we are again looking at RX equal to 0 there was one interpretation that we had seen earlier R is the number of non-zero rows I will keep that we have the unknowns X1, X2, etc Xn these are split into two types one type corresponding to the leading non-zero entries of the non-zero rows of R so I have the variables Xc1, Xc2, etc Xcr okay these are the variables corresponding to the leading non-zero entries of the first R non-zero rows of capital R that is i equals 1, 2, 3, etc R that is the n variables are split into two categories one corresponding to the leading non-zero entries the other ones are the rest of the variables we are calling them u1, u2, etc u n minus R the number of variables is n R have been eliminated here the rest of them are n minus R then we had seen earlier that the system RX equal to 0 when expanded using this new notation we will have equations like this Xc1 plus summation j equals 1 to n minus R some constant let us say alpha 1j I am using u, uj this is equal to 0, etc Xcr summation j equal to 1 to n minus R alpha rj uj equal to 0 these are the R equations these are the R non-trivial equations of the system RX equal to 0 the rest of the equations give no information 0 on the left 0 on the right I will now adopt a slightly different notation this is just to recall what we have done earlier the real way labeling of these variables in this manner I will now get back to the original X with the following notation so let me introduce this set j this is 1, 2, 3, etc n where I have deleted c1, c2, etc cr I delete these numbers c1, c2, etc cr from this n then the cardinality of j number of elements in j is n minus R number of elements in j equals n minus R these equations I will rewrite Xc1 plus summation j element of j alpha 1j I go back to the variable Xj sorry this is yeah this is fine right Xc1 is a number so I must use the subscript here so not the superscript there so let me go back and correct this this is alpha 1j uj subscripts see these are numbers each on the right I have 0 this is a number this is also a number and this is a number that comes from this u1, u2, etc u n minus R so this is uj similarly these equations will be alpha 1j Xj equals 0, etc Xcr plus summation j element of j alpha Rj Xj equal to 0 all that I have done is to relabel the variables u1, u2, etc u n minus R as Xj for j element of j u1, u2, etc u n minus R are now Xj for j element of j is that okay if it is in j then it does not correspond to c1, c2, etc cr those are written separately okay what is the advantage of this okay again get back to what we have studied earlier how do you get the set of all solutions of this system look at the so called free variables that is look at the variables that correspond to look at the unknowns that correspond to those entries in j look at the entries look at the free variables u1, u2, etc u n minus R earlier look at the variables Xj for j in j you assign arbitrary values to them come back and substitute determine Xc1, etc Xcr that will give you one set of values take another set of values for Xj j in j another set of values you compute all the solutions using this okay among these solutions let us give a new notation for the following variables how do I I am introducing Sj for j in j I am introducing the vector this is a vector I will use subscripts for real number scalars and superscripts or vectors Sj j in j each Sj let me emphasize belongs to Rn to avoid ambiguity each Sj belongs to Rn how many Sj's are there there are n minus R Sj's what is the definition of Sj Sj is that vector which has the value let me write Sj is that vector whose jth coordinate is 1 that is I am looking at one specific assignment for the free variables I will take a j in j and then look at the following specific assignment of the free variables I will take j in j and then put Xj equals 1 that is the coordinate corresponding to the first j that I have chosen from capital J that is equal to 1 X i equals 0 for all i element of j i not equal to j I am only looking at the free variables now I am only looking at the free variables there are n minus R free variables u 1, u 2, etc u n minus R earlier Xj j in j now these are the n minus R variables I will look at a very specific assignment now of values for these free variables pick a j in j let us say 1 belongs to j then I will write down a vector I will call that S1 first coordinate is 1 there are some other elements in j assign 0 for all those values substitute in this equation you will determine Xc1, etc etc Xcr give to those values fill up that vector that is my S1 is that clear okay for any particular j in j assign the value Xj equal to 1 that is a free variable j capital J corresponds to free variables, capital J corresponds to indices which are the free variables so I assign the value Xj equal to 1 all the other coordinates corresponding to j again only those are free the rest of them these are not free these you have to go back to these equations substitute and then determine this what I am doing is fixing one of these equations fix that variable Xj equal to 1 assign the others to be 0 and then determine Xc1, etc Xcr I know some values 1 minus 1 e pi, etc come back fill it up I have Sj I do this for each j in j there are n minus r Sj's we obtain n minus r such Sj's okay each of the Sj is a solution of Rx equal to 0 each of the Sj is a solution of Rx equal to 0 so each Sj belongs to S each Sj is a solution so each Sj belongs to S can you see that these vectors are linearly independent immediately what is the reason the reason is very similar to the standard basis vector vectors being linearly independent this entry is 1 how do I this entry is 1 for Sj let us say I look at Sj plus 1 that entry will have Sj plus thus j plus 1th entry 1 all other free variables will have the value 0 forget about the very values that these variables take there is a 0 it is like a column it is like a column there is a 1 and 1 entry all other entries are 0 these vectors have to be independent okay so for one thing these Sj's are linearly independent the argument as I said is very similar to the argument that the standard basis is a basis the standard basis vectors are linearly independent so Sj's are linearly independent okay let me tell you what I am trying to prove I am trying to prove that the solution set the solution space the solution subspace is of dimension n minus r I want to show that the solution space is of dimension n minus r I want to show that it has a basis consisting of n minus r elements I am explicitly determining such a basis the first step is to prove that these vectors are independent second set is second step is to prove they are they form a spanning set that I will do next if I had done this then it follows that the subspace S the subspace S is of dimension n minus r use rank nullity dimension theorem we will get row rank equals column rank okay so the next step is to show that these vectors span any solution that is if I take some x star as a solution some x star that satisfies A x star equal to 0 I must show that this x star is a linear combination of these n minus r vectors okay okay I am giving the argument you can fill up the details how do we get a solution of A x equal to 0 there is only one procedure that we know what is the procedure there are free variables give some values to the free variables come back and determine these variables that is a solution this is the general method okay I have n minus r free variables take one of them I give arbitrary values to the n minus r variables let us let us say I take the j th coordinate I look at the entry corresponding to x j I look at the entry corresponding to x j that entry is alpha let us say then I can write that as alpha times 1 then look at the next entry corresponding to j in j some x j plus 1 x j plus 2 some entry that is some beta that is beta times 1 I keep doing this for each j in j I look at x j that entry and then I write that as if it is 0 I leave it as it is if it is not 0 it is some constant alpha then I write it as alpha times 1 leave it and do this for all x j for j in j I will have n minus r vectors can you see then that the x star that I started with is a linear combination of these that is immediate almost it is again very similar to the standard basis in a standard basis there is only one entry one all other entries are 0 this is more or less like the standard basis because of the reason that these values x e 1 etcetera x e r are not in your control they can take other values nonzero values they are not in your control but look at the rest of them the rest of them are free and we are we have looked at a very particular choice in constructing x j this 1 0 0 type and so please fill up the details what follows is that each solution so let me write let y belong to s then it can be seen that y is a linear combination of these s j's for j in j an argument almost duplicating the or imitating the argument for the standard basis can be applied here to show that these vectors are not only independent they span the solution set s so what is the meaning then so dimension of the subspace s is n minus r okay I have given an explicit basis for this subspace let us go back and use this okay so let us go back and see what s is s is set of all I will now use this set of all x such that A x equal to 0 okay the dimension of this subspace is n minus r now let me define a linear transformation T from R n to R m by T of x equals A x matrix multiplication x n R n what is A the matrix that I started with A is the matrix that I started with I am defining a linear transformation this is only to this is a via media to apply rank nullity dimension theorem rank nullity dimension theorem is applicable for a linear transformation I am going to practically apply it to a matrix to apply it to a matrix I want this step T x equal to A x then we know that this T is linear okay rank nullity dimension theorem can be applied to the linear map T by rank nullity dimension theorem rank of T plus nullity of T equals dimension of the core domain I am sorry domain that is n rank plus nullity is dimension of the domain core domain could be even infinite dimensional okay rank of T this time in terms of the languages in terms of linear transformations earlier it is row space column space rank T what is rank T rank of T is the dimension of the range space okay so rank of T is a dimension of range space range space of T that is dimension of range space of T let us write the complete definition it is a set of all y n R n such that y can be written as A into x for some x n R n I am sorry y equals T x I am writing down the range of T okay but T x is A x so this is the dimension of the subspace y such that y equals A x but we have encountered the subspace before we have encountered the subspace before column space dimension of the column space column rank column rank of A so rank of T is a column rank of A what is nullity of T dimension of the null space of A dimension of n of T that is what dimension of the subspace of all x such that T x equal to 0 that is a dimension of all x that satisfy A x equal to 0 because T x equal to A x this is what we have determined as n minus R dimension S is n minus R S is the set of all x such that A x equal to 0 so this number is n minus R nullity of T is n minus R so go back to this equation n is equal to rank of T that is column rank of A plus rank of T plus nullity of T column rank of A plus n minus R cancel n column rank of A equals R what is R now number of non-zero rows of capital R that is the row rank of R that is a row rank of A that is a row rank of A now this is a proof that involves lots of calculations primarily using rank nullity dimension theorem but there is a more sophisticated way of proving this using transposes we will look at this proof quite later okay but this is using all the calculations that you do for a row reduce echolon matrix homogeneous equations non-homogeneous equations etc okay so this is one of the most important results in linear algebra that is why I have made it a point to prove this completely okay so let us move on to the next topic the matrix of a linear transformation okay so I want to discuss the notion of the matrix of a linear transformation this is another fundamental notion I hope you remember the statement that I made some time ago when we discussed examples of linear transformations I will that is something we have done even now given a matrix so recall this given a matrix A in A with real entries m cross n the mapping T from R n to R m defined by T of x equals A x this is linear given a matrix there is a natural linear transformation that can be associated with this matrix and I also made this statement there is a certain converse which is true now what is the converse we will I will make this statement of the converse precise and also show how to construct how to prove this statement in a constructive manner the converse statement is that given a linear transformation there is a matrix associated with this linear transformation which behaves in precisely this manner given a linear transformation between finite dimensional spaces there is precisely one matrix corresponding to a corresponding to bases two bases such that the matrix will do what you have here so this is more or like the defining equation of a linear transformation between finite dimensional spaces okay so that is the statement that we are going to prove so how do you go about it I have still not made the statement precise so I will simply say a certain converse is true okay so let us go back and look at finite dimensional vector spaces so I have let V W I have two finite dimensional vector spaces over the same field let us say the field of real numbers let us say dimension of V is n and dimension of W is n let me also write down two bases explicitly let me use the following notation script B V will be a basis for V B V let us say the entries are U 1 U 2 etc U n this is a basis of V B W will be a basis for W let me call it V 1 V 2 etc V m a basis of W so I start with two given bases then given any vector x and v it can be written as a linear combination of these vectors let us say alpha 1 U 1 plus alpha 2 U 2 etc plus alpha n U n alpha 1 alpha etc alpha n are real numbers they are unique for the x that I start with these real numbers are unique for the x that I start with what I will do then is define the matrix of x matrix of a vector matrix of x relative to the basis that I started with relative to the basis B V define the matrix of x relative to the basis B V by the notation for me will be x B V with these parenthesis this matrix if it is a vector it is a column so what is that column it is alpha 1 alpha 2 etc alpha n this is equal to this this is actually a function in order for this to be a function the right hand side must be unique whenever x is unique but we know that it is unique because the representation is given in terms of a basis okay so this is that then the definition of the matrix of a vector relative to a fixed basis okay let us let us observe one thing immediately if I take the same basis B V instead of U 1 U 2 etc U n I take U 2 U 1 etc U n then this matrix will change this will be alpha 2 alpha 1 etc okay so there is really an ordered basis that I am dealing with okay we must deal with ordered basis but here after when I write down U 1 U 2 etc U n then it follows that U 1 is the first vector in that basis U 2 is a second vector etc U n this order I should always remember when I write down the matrix of a vector okay so the notion of ordered basis needs to be introduced but I will just keep it over ordered basis means a basis whose vectors form a finite sequence sequence means that is the first element of the sequence second element of the sequence etc so it is just an ordered set of vectors which also forms a basis why is it important when you write down the matrix it is important because you say there is a first coordinate there is a second coordinate etc alpha 1 is the first coordinate of the now I am writing a vector in an abstract finite dimensional vector space using numbers I am writing down the I am giving a representation to a vector in an abstract finite dimensional space using real numbers alpha 1 etc alpha n by this so there is a coordinate first coordinate of the vector second coordinate of the vector etc in order to make sense that it is a first coordinate second coordinate etc you need to have a fixed basis where the elements are taken in only that order okay that is an ordered basis. Let us make one simple calculation an elementary example just to consolidate let us take R 3 for instance for R 3 there is a standard basis let me call that B 1 the standard basis E 1, E 2, E 3 given the standard basis in this order this is always implicit the order is implicit what is the matrix of x relative to B 1 given any x in R 3 what is the matrix of x relative to this this is the simplest basis you can write down the matrix immediately but let us do it from the scratch this x can be written as okay I will start with the following x is the vector which has 3 coordinates so I have x equal to x 1, x 2, x 3 I am intentionally writing this as a row vector then this x can be written as x 1, E 1 plus x 2, E 2 plus x 3, E 3 this representation is unique and so what is the matrix of x relative to the standard basis I am calling that B 1 that is a column vector according to the definition x 1, x 2, x 3 okay it is as easy as just taking a row and then putting it this order this is the simplest basis suppose I have another basis let us make a calculation corresponding to that in particular if let us say x is E pi 0 then the matrix of x relative to B 1 is E pi 0 okay let us say I have another basis B 2 consisting of this time let me call them U 1, U 2, U 3 where U 1 let me say is the vector 1, 1, 0 U 2 is the vector 1 minus 1, 0 U 3 is the third standard basis 0, 0, 1 I have taken another basis just to illustrate that the matrix corresponding to this basis will be different from the original one corresponding to the standard basis. I want to calculate x relative to this B 2 okay instead of doing the general vector let me take this particular vector and do the calculations here I want to determine these numbers let me now call them alpha 1, U 1 plus alpha 2, U 2 plus alpha 3, U 3 first I must write this I have just written x and then I need to determine the matrix of this whole thing relative to B 2 please verify the calculations it will turn out to be see this is 1, 1, 1 minus 1 so it will turn out to be the following for this x that I started E pi 0 you can verify that it is E plus pi by 2 E minus pi by 2 0 this is the column vector this is the matrix of the vector E pi 0 corresponding to the second basis which is obviously different from the matrix of the vector corresponding to the first basis the standard basis. So when I change the vectors rather when I change the basis the representation of the vectors will obviously change okay but there is a relationship between them we will be able to demonstrate that there is a relationship between them okay.