 Let us continue our discussion on linear transformations, we discussed one property the other day let us look at some more property some examples then the notion of null space of a linear transformation range space of a linear transformation some examples where we will calculate these subspaces and then probably today the rank nullity dimension theorem okay. Let me recall the result that we discussed in the last lecture what we essentially proved I will rewrite it let Vw be vector spaces over the same field with V finite dimensional and let T from V into W be a linear transformation I will simply call it linear T from V to W is linear. The first result that I discussed is about uniqueness so let us say there are two vectors there are two transformations T1, T be linear maps let script B equal to U1, U2, etc U n I assume that this is a basis script B be a basis of V, V is finite dimension so there is a finite basis for V if T1 of Ui equals T2 of Ui for all i line between 1 and n then we had shown last time that T1 and T2 are the same maps okay this was the first property this theorem was proved in the last lecture. Let us look at certain other results this question was asked last time again V is finite dimensional let us say B as above is a basis let W1, W2, etc Wn belong to W now W need not be finite dimensional so there is no condition on these W's they can be even the same vectors they can all be the 0 vector the question that we addressed last time was is there a linear map T from V into W such that T of Ui equals Wi this question was asked last time let us answer in the affirmative okay we will prove that given any basis B and given any vectors n and number the same as this they can repeat given any set of n vectors there is a unique linear transformation that satisfies this condition okay so the answer is there is a unique T satisfying the above there is a unique T satisfying the above that is what we will show okay so this is another theorem let us prove this theorem okay let us take a general X vector X and V V has script B as a basis these are the vectors so I have a linear combination then there exist scalars alpha 1, alpha 2, etc alpha n such that such that this X can be written as alpha 1 U1 plus alpha 2 U2 plus etc plus alpha n Un this comes from the definition of a basis any a basis is a spanning set so any vector X and V is a linear combination of the basis vectors but what is also important to observe is that these numbers are unique for this vector X the scalars alpha 1, alpha 2, etc alpha n are unique for the X that we started with the scalars alpha 1, alpha 2, etc alpha n are unique for the vector X that we started with what is the meaning of this statement if there exist beta 1, beta 2, etc such that X is beta 1 U1 plus etc plus beta n Un then alpha i equals beta i for all i okay let us prove that quickly again so I am going to demonstrate that this statement is true that is quick suppose that X is also beta 1 U1 plus beta 2 U2, etc beta n Un so I have another if possible let there be another representation for X in terms of the basis vectors U1, U2, etc Un then make use of these two equations to write alpha 1 U1 plus alpha 2 U2, etc plus alpha n Un equals beta 1 U1 plus beta 2 U2, etc plus beta n Un I must have this which I can rewrite this is the same as saying alpha 1 minus beta 1 U1 plus etc alpha n minus beta n Un equals 0 but invoke the fact that U1, U2, etc Un are linearly independent this means alpha 1 equals beta 1, alpha 2 equals beta 2, etc alpha n equals beta n so the representation for any vector in terms of a basis that representation must be unique I fix the basis for a fixed basis there may be several other basis for another basis I have another representation that is a different matter for this basis there is a unique representation okay what is the need for proving this uniqueness we will use this uniqueness to define a mapping T define T from V into W by T of X to be I will use these scalars alpha 1, alpha 2, etc for this X and then use the vectors W1, W2, etc so T of X will be for me alpha 1, W1 plus alpha 2, W2, etc plus alpha n, Wn where remember that these alpha i's are chosen from the representation for X in terms of the basis that we started with that is unique so this is a well defined map the reason why we proved uniqueness is to show that this is a well defined map well definedness here means if X is equal to Y then F X equals F Y if X equals Y then T X equals T Y I am going to leave that as an exercise then T is well defined that is an exercise what is the meaning of this the image for X cannot be two different elements that is well definedness T of X cannot be two different vectors the well definedness comes from the unique representation for a vector in terms of a basis okay so T is well defined I am going to leave as an exercise we need to verify that this T is linear and that this T satisfies a required equation the required equations are these we show that T is linear and that T satisfies these equations okay T is linear first let us take a representation X I have taken as before let me take a representation for Y, Y is let us say delta 1 U 1 plus delta 2 U 2 plus etc delta n U n I am taking Y as another vector I want to show T X plus Y is T X plus T Y then if this is my Y then the definition of T Y as above will be delta 1 W 1 plus delta 2 W 2 etc plus delta n W n okay remember you must go back whenever you want to write T of something you must go back to the representation of Y in terms of the basis vectors and then use the linear combination of these scalars along with W n etc W n okay we must show that T of X plus Y is T X plus T Y. So consider X plus Y first X plus Y this happens in the vector space linear combination I can add the coefficients alpha 1 plus delta 1 U 1 etc alpha n plus delta n U n this X plus Y this representation is unique and so T of X plus Y I can now write down that is this scalar alpha 1 plus delta 1 into W 1 plus etc alpha n plus delta n W n this happens in the vector space W alpha 1 W 1 plus alpha 2 W 2 etc alpha n W 1 collecting the first terms from each parenthesis plus collect the second terms delta 1 W 1 delta 2 W 2 etc delta n W n and then go back and see that this is precisely T X this is T Y so this is T of X plus T of Y so we have shown that T is additive T of X plus Y is T X plus T Y T of alpha X T of alpha X alpha is fixed scalar again I must know the representation for alpha X the representation for alpha X will be alpha alpha 1 U 1 plus alpha alpha 2 U 2 etc plus alpha alpha n U n and so T of alpha X will be alpha alpha 1 W 1 alpha alpha 2 W 2 etc alpha alpha n W 1 this comes from the unique representation of alpha X this is alpha times alpha 1 W 1 plus alpha 2 W 2 etc alpha and W n but that is precisely T X so T of alpha X is alpha T X for all alpha and so T is linear okay so T is linear does it satisfy these n equations I must verify that T satisfies these equations okay but look at U 1 for example say I want to know T of U 1 T of U 2 etc T of U n but in order to know U 1 I must know the unique representation of U 1 in terms of the basis vectors but what is the unique representation for U 1? 1 times U 1 plus 0 U 2 etc plus 0 U n for one thing this is a representation for another I know it is unique. Now by my definition T of U 1 is this scalar into W 1 plus this scalar into W 2 etc this scalar into W n so this scalar into W 1 plus 0 W 2 0 W n that is this is just W 1 so I have T of U 1 equals W 1 being satisfied T of U 1 equals W 1 the first equation has been satisfied but what I have done for U 1 can be done for any other vector U 2 U 3 etc for U 2 the representation is 0 times U 1 plus 1 times U 2 etc so use that to conclude that these equations T of U i equals W i these n equations are also satisfied okay. The fact that this T is unique is similar to the previous theorem proof of the previous theorem the fact that T is unique is similar to the proof of the previous theorem that is you want to show T is unique take I will not prove it I will just give a sketch take another linear transformation I will call that S if S from T into W is linear and S of U i equals W i see the claim is there is a unique linear transformation T that satisfies these equations suppose there is another transformation S such that S of U i satisfying these n equations S of U i equals W i yes S is from V to W satisfying these equations then it is easy to see that S is equal to T okay. So this last part is similar to the proof of the last part of the previous theorem I am going to skip this okay so let me emphasize what I said earlier when I wrote down this theorem note W 1 W 2 etc W n this need not form a basis of W first observation this need not be a basis of W in fact W need not be finite dimensional W need not be finite dimensional in fact I can say that W 1 W 2 W n could all be equal could all be equal and the worst case could be 0 also okay there is absolutely no condition on W 1 etc W n that we need in this theorem just arbitrary vectors okay. So these are some of the elementary properties two elementary properties really in connection with existence uniqueness okay let us look at certain subspaces associated with a linear transformation let me first define the null space of a linear transformation I have T from V into W as a linear map the kernel of T or the null space of T the kernel of T or the null space of T kernel word comes from group theory for instance null space is typical vector space notion null space of T I will call it I will use this notation N of T denoted by this is defined as follows the kernel of T or the null space of T notation is N of T what is the definition N of T is the set of all X and V such that T of X equals 0 set of all X and V such that T of X equals 0 since this is this X is in V this is a subset of V okay so let us observe that this is contained in V so this is a subset of V now this can be shown to be a subspace of V okay let us I will prove that quickly this null space is a subspace of V let us take we want to show that a subset of the subspace then show that it is close with respect to addition and scalar multiplication so let us take two vectors let X, Y belong to null space of T then T of X is equal to 0 and T of Y is 0 I must show that X plus Y belongs to null space of T so start with T of X plus Y I must show that T of X plus Y is 0 but T is linear so T of X plus Y is T of X plus T of Y that is 0 plus 0 that is 0 and so X plus Y belongs to null space of T it is close with respect to addition with respect to scalar much easier T of Alpha X is Alpha T of X that is Alpha into 0 that is 0 so we have shown that X plus Y belongs to null space of T Alpha X belongs to null space of T so this null space is in fact a subspace that is why it is called a null space so this null space is a subspace of V there is another important subspace associated with a linear transformation T some information about the linear transformation T can be derived by looking at the null space okay there is as I mentioned there is another subspace associated with a linear transformation T which also has some information about T that is called the range space of T, T is linear the range of T or the range space of T I will use R of T for that denoted by R of T is defined by R of T is the set of all W in W such that W equals T of X for some X and V that is I collect all those vectors W and W now this is a subset of W this is a subset of W what is a property of this subset this subset as a property that for every element in for every vector in this subset there is a there is at least one pre image in V for every W in range of T there exist some X and V such that W equals T of X every vector in W has a pre image corresponding to T this is called the range space and you observe that this is a subspace of W the other vector space now I am going to leave this as an exercise for you to prove that range of T is a subspace range of T is a subspace of W this is an exercise for you let us now look at some examples and then determine the range space null space to consolidate these notions let us dispose of the trivial cases I want to look at examples let us dispose of the trivial case the first one is the zero map zero from V to W the zero map what is the null space of the zero map null space of zero is the whole of V that is because zero operating on X is zero for all X and V so null space of zero is the whole of V what is the range of zero range of zero is singleton zero this zero coming from the vector space W range of zero singleton zero example to identity linear transformation from V to itself identity linear transformation on V what is the null space of Y singleton zero what is the range of I it works complementary to the zero operator range of I is V okay this is like complementary of the zero operator identity from V to V for identity mapping W must be equal to V so range of I is V okay these are the trivial examples let us look at other examples example 3 let us look at the projections T from let us say R 3 to R 3 defined by T of X equals X 1, 0, X 3 projection on to the so called X Z plane the second coordinate is 0 we have verified that this is a linear transformation what is the null space of T null space of T is a set of all X in R 3 such that T of X is 0 this is a set of all X such that T of X equal to 0 X 1, 0, X 3 this must be the three dimensional zero vector what is the condition that these equations impose on the unknowns X 1, X 2, X 3 that is what you must observe this is a set of all X in R 3 such that X 1 is 0 X 3 is 0 it does not impose any condition on X 2 so it is X 1 equal to 0 X 3 equals 0 X 2 is arbitrary let me just emphasize X 2 is arbitrary can you also give a basis for null space of T now do you agree that this is span of E 1 E 3 sorry just E 2 span of E 2 any multiple of E 2 must be in null space of T and anything in null space of T is a multiple of E 2 because first and third coordinates are 0 okay that is null space of T what about range of T range of T is the set of all this time I will use Y in R 3 such that T of X equals Y set of all Y in R 3 such that T of X equal to Y X belongs to R 3 for some X in R 3 let me write on this side this is the set of all Y in R 3 such that X 1, 0, X 3 equals Y 1, Y 2, Y 3 some X in R 3 I have just used the definition of T of X T of X is X 1, 0, X 3 X 1, 0, X 3 equals Y 1, Y 2, Y 3 now you see that this really imposes a condition on Y 2 Y 1 and Y 3 are arbitrary so Y 2 is 0 that is the only condition okay so can I write okay so range of T for this example is the set of all Y in R 3 such that Y 2 is 0 okay now can I give a basis similar to what we did for the null space can I give a basis for range of T span of Y 2 is 0 so look at E 1 and E 2 E 3 both these vectors Y 2 is 0 E 1 and E 3 and you can verify that these two vectors satisfy this condition. Now in this example you observe that the dimension of null space of T plus the dimension of the range of T the dimension of the null space of T null space is one dimensional that is one dimension of range of T it is two dimensional 2 this is equal to 3 that is the dimension of the domain space of T, T is from R 3 to R 3 the domain space is three dimensional forget about this the domain space is three dimensional we observe that dimension of null space plus dimension of range of T is the dimension of the domain space this is part of a general result the rank nullity dimension theorem okay let us look at two more examples before I discuss these two numerical examples let me also give you the notion of injective and surjective linear transformation injective and surjective linear transformation let me give the definition a linear map T from V to W T said to be injective or 1 to 1 T said to be injective or 1 to 1 if as a function it is injective that is if T of X equals T of Y implies X is equal to Y distinct elements have distinct images X not equal to Y implies T X not equal to T Y that is the same as saying T X equal to T Y implies X is equal to Y distinct elements have distinct images that is injectivity the notion of injectivity of a linear transformation surjectivity so I will call this first part second part T from V to W is called it is linear I will not emphasize that again all the objects in this course will be linear transformations T from V to W is called surjective if for every W vector W in capital W there exists at least 1 vector X in V such that T of X equals W T said to be surjective if every element in the codomain has a preimage every element in the codomain has a preimage there exists X such that W is equal to T of X there exists at least 1 X such that W equals T of X this is surjectivity of a linear transformation before I look at those 2 or 3 examples let me also get the connection between these notions and the subspaces null space of T and range of T let me first make the following observation observation 1 T is linear T from V to W is injective if and only if null space of T is the 0 subspace T is injective if and only if null space of T is a 0 subspace observation 2 T from V to W is surjective if and only if the range of T is the whole space W T is surjective if and only if range of T is the whole space W of these 2 observations the second one is straight forward that comes from the definition T must be an on to map for example just as injectivity is 1 to 1 surjectivity is on to so in some cases I will use this notion on to we will refer to T as an on to map if it is surjective okay so from the definition of an on to map it is clear that range of T is W let us quickly prove that first statement is true T is injective if and only if null space of T is single term 0 T is 1 1 if and only if null space of T is single term 0 proof of this let us take this implies this let T be 1 1 I will use this notation for 1 1 injectivity let T be 1 1 I must show that null space of T is single term 0 let us take X in null space of T I must show that this X is 0 okay then I have T X equal to 0 but T is a linear transformation so T of 0 is 0 we know this property basic property so T X equals T 0 but this is like T X equal to T Y since T is 1 1 it follows that X is equal to 0 okay 0 belongs null space of T there is no problem null space of T is a subspace so it has to have at least 0 vector but what happens in this case when T is injective is that 0 is the only vector in the null space of T that is what we have shown we have started with an arbitrary X in null space of T we have shown that X is 0 so this proves one way if T is 1 1 then null space of T is 0 conversely suppose that null space of T is single term 0 we must show that T is injective let us start with T of X equals T of Y we must show that X is equal to Y that is injectivity then T of X minus T of Y equals 0 T is linear T of X minus Y equal to 0 which means X minus Y belongs to the null space of T which I know has only the 0 vector and so what I have shown is that X minus Y is a 0 vector so X is equal to Y so I started with T X equal to T Y I have shown that X is equal to Y so T is injective so a linear transformation is 1 1 and only if the null space is single term 0 it is on to if and only if the range space is the entire codomane W okay. Let us now look at 3 examples an example of a linear transformation which is 1 1 but not on to another example of a linear transformation which is on to but not 1 1 third example which is both 1 1 and on to so I want to continue with the examples this is probably example 4 let me define T from R 3 to R 2 by T of X R 3 there are 3 coordinates R 2 let us say X 1 plus X 2 X 1 minus X 2 for each X in R 3 let me define T of X in this manner then this is a linear transformation that does not need any proof we have seen this several times T is linear what is the null space of T let us calculate that null space of T is a set of all X in R 3 such that T of X is 0 X 1 plus X 2 X 1 minus X 2 that is X 1 plus X is 0 X 1 minus X 2 is 0 what can you conclude from these 2 equations X 1 equals X 2 is 0 the set of all X in R 3 such that X 1 equals X 2 equal to 0 is T 1 1 is T a 1 1 map it is not because X 3 is arbitrary so let me emphasize X 3 is arbitrary so the null space is actually one dimensional it is spanned by E 3 but what about range of T what is range of T okay let us calculate see this T is not 1 1 see this T is not 1 1 because null space consists of a non-zero vector at least one non-zero vector range of T set of all Y in R 2 such that Y equals T of X again it is X 1 plus X 2 X 1 minus X 2 X is in R 2 R 3 that is my range space this is a set of all Y in R 2 such that Y Y 1 equals X 1 plus X 2 Y 2 equals X 1 minus X 2 for X in R 3 the question is what is the condition that these equations impose on Y in R 2 if at all there is a condition we want to determine range of range of this linear transformation T the question that arises is do these equations impose any conditions on the left hand side numbers Y 1 and Y 2 the answer is no let me give an argument for that let us take see this is a subset of subspace of R 2 let us take the two basis standard basis vectors for R 2 1 0 0 1 suppose I show that the standard basis vectors have pre-images does it follow that any vector will have pre-images does it then follow that range is the whole space it would then follow that T is on to okay so the question is look at these equations this is my question do there exist X 1 X 2 of course X 3 such that such that this time X 1 plus X 2 right so let me write like this X 1 plus X 2 equals 1 X 1 minus X 2 equals 0 I want to answer this question I want to ask a similar question do there exist numbers that X 1 plus X 2 is 0 X 1 minus X 2 is 1 this is one question this is another question is it clear that these two systems have solutions look at the first case X 1 equals X 2 equals half in fact that is a unique solution second equation is trivially satisfied half minus half there is no X 3 X 3 can be taken to be arbitrary second set of equation second system X 1 equals X 2 equals X 1 is half X 2 is minus half X 1 is half X 2 is minus half this is 0 this is 1 so both these systems have a solution but what is the advantage this is the vector E 1 this gives rise to E 1 that is 1 0 this gives rise to E 2 the second standard basis vector 0 1 these two vectors can be if these two vectors can be written as a linear combination of certain vectors in R 3 that is what this means see these equations do not include X 3 so X 3 is arbitrary really X 1 and X 2 only satisfy certain conditions X 3 can be taken to be arbitrary so I can write so the summary is the following all the time saying is E 1 E 2 belong to range of T what we have seen just now is that E 1 E 2 belong to range of T since E 1 E 2 okay what this means is that range of T is two dimensional but range of T is a subspace of R 2 which is two dimensional so the spaces must be the same so range of T is the whole of the space W that is R 3 R 2 in this example okay so this is an example of a linear transformation which is not 1 1 but on to range of T is the whole of R 2 I have shown this by showing that I have picked probably the simplest basis of R 2 and then shown that any basis vector in R 2 can be written any basis vector of R 2 has an image in R 3 T of X equals E 1 we have solved T of X equal to E 2 we have solved so these basis vectors have pre-images in R 3 and so range is the entire space so T is on to so this is an example of a linear transformation which is not injective but surjective see this X comes from R 2 I will write the same coordinates X 1 plus X 2 X 1 minus X 2 0 T is from R 2 to R 3 this time then T is linear that is easy to see my claim is T is injective let us prove it quickly let us start with T X equal to 0 I want to show that X is equal to 0 now T X equal to 0 implies X 1 plus X 2 equal to 0 X 1 minus X 2 equal to 0 0 equals 0 the row reduced echelon form the last row is 0 this is not the row reduced echelon form but you can see that this has 2 non-zero rows and the last row is 0 now look at these 2 equations these 2 imply immediately X 1 equals X 2 equal to 0 that is the same as saying the vector X is 0 so I have started with T X equal to 0 I have shown that X must be 0 so T is injective T is injective T is not surjective can you give me one reason T is not surjective which means I must exhibit a vector in R 3 which does not have a pre-image which one 0 0 1 E 3 does not belong to range of T E 3 does not belong to range of T because E 3 has a third coordinate 1 but if it is in the range of T then the third coordinate must be 0 so E 3 does not belong to range of T so T is not surjective so T is not on to okay this is an example of linear transformation which is injective but not surjective one final example before I conclude T from R 2 to R 2 is defined by T of X equals X 1 plus X 2 X 1 minus X 2 okay this linear transformation has a property that it is both injective and surjective. Injective T of X equal to 0 implies X 1 plus X 2 is 0 X 1 minus X 2 is 0 as before X is 0. Now surjectivity again one can show that the right hand side vector if I take E 1 as a right hand side vector then the system X 1 plus X 2 equal to 1 X 1 minus X 2 equal to 0 has a solution change the right hand side vector to E 2 X 1 plus X 2 equal to 0 X 1 minus X 2 equal to 1 has a solution that is one way of looking at it the other way of solving this problem is to link this with a particular matrix we are actually solving solving system of linear equations link this with a particular matrix that matrix comes from the definition of T look at the matrix whose entries are 1 1 1 minus 1 okay. Now what we have shown is that this matrix has a property we have shown T is 1 1 which is the same as saying this matrix has a property that the system A X equal to 0 has 0 as the only solution then from an equivalence condition that we have proved before it follows that A X equal to B has a solution for all right hand side vectors B that is the reason why A X equal to E 1 as well as A X equal to E 2 have solutions but if A X equal to E 1 A X equal to E 2 have solutions it means T is on 2. So please fill up the details this T is on 2 also link this with what we have learnt before okay the second argument to show that T is on 2 relies basically on if you go back to that example you will realize it relies basically on solving two systems of linear equations two systems of linear equations A X equal to E 1 A X equal to E 2 E 1 E 2 are the standard basis vectors in R 2. So I would like to ask does A X equal to B have a solution for all B if I know the answer if I know S is the answer for this question then E 1 E 2 any number E n they will have a solution. So my question is to find out whether A X equal to B has a solution for any B I am saying in this example the answer is S because we have shown that T is 1 1 which is the same as saying that the homogeneous equation A X equal to 0 has 0 as the only solution the homogeneous equation A X equal to 0 where A is the matrix which comes naturally from the transformation T this A X equal to 0 has 0 as the only solution we know that this is square matrix homogeneous system has 0 as the only solution then we know that this matrix is invertible which is the same as saying that the system A X equal to B for any right hand side vector B has a solution which is what we wanted. So A X equal to E 1 has a solution A X equal to E 2 has a solution I am not interested in knowing the solutions I want to know that the solutions exist A X equal to E 1 A X equal to E 2 both have solution. So standard basis vectors E 1 and E 2 have been written as a linear combination have been written as a linear combination of certain vectors in R 2 they are in range of T that is really what we want to conclude E 1 E 2 both are in range of T it follows T is on to as before so this is an example of a linear transformation both 1 1 and on to so let me stop here.