56 - Operations on linear maps

I want to remind you that linear maps or linear transformations to start with are just functions right they're functions between two vector spaces that satisfy some additional properties as functions we can do what we usually do with functions or what one of the things we do with functions which is operations on functions we know how to add functions we know how to multiply a function by a scalar we know how to multiply functions okay so we may ask how do what do these notions give for linear linear maps and we'll see that that up to terminology there there's a similarity although things are not precisely the same so here's our first definition so definition let T from V 1 to V 2 and s from V 1 to V to be to linear helps be linear Maps and let I called it alpha be a scalar in the field F everything has to be defined over the same scalar field F so we define T plus s just like we do for functions so what does it mean to define a new linear map we have to say what it does to little V in V 1 ok so what does it do the very most expected thing it's just T of e + SS V so that's how we add linear maps and if we want to ask what is alpha T again a linear map what does it do to a vector V it just takes alpha times T of you okay so that's the definition of these two linear minutes okay so the first thing one has to show as functions we know it's just the same definition for functions but is it true that t plus s is filling your map T was a linear map it satisfied some properties as is a linear map it satisfies some properties is the sum all in your map does it satisfy that T plus s of V Plus u is the same as T of V Plus u plus s of E Plus u and those are very simple things to verify and please do it on your on your own so verify that T plus s and alpha T are linear okay let's do an example so I want to give you two maps both maps are going to be maps from R 2 to R 2 so I'm gonna tell you what T does to a vector of the form a B and what s does to a vector of the form a B so T of a B is going to be defined by multiplying by a matrix we already know that this is a linear map and we had a long discussion that any linear map has a make matrix representation okay so this is what T does to a B in R 2 and s of a B is going to be defined as 0 3 1 negative 1 times a B ok so we know that these are well-defined linear maps from R 2 to R 2 do you agree ok so let's see what T plus s is so T plus s is going to be a linear map on the same domain space with the same range space let's see what it does to a vector a B in r2 so what does it do by definition it does t of a B plus SFA me right that's how we defined the addition okay so what is T of a B I'm just writing what T of a B is and what s of a B is so T of a B is 1 to 0 negative 1 times a B plus s of a B is 0 3 1 negative 1 times a B right that's precisely what these maps TNS do now by rules of matrix multiplication I know we have distributivity so if I have a matrix times a matrix plus a matrix times the same matrix this is the same as the sum of the matrices times a B so this is the same as 1 2 0 negative 1 plus 0 3 1 negative 1 the sum of the matrices times this matrix this is distributivity of matrix multiplication with respect to matrix a ditch addition do you agree ok so what I get is I just add the matrices right so the addition is what 1/5 1 negative 2 times a B okay and note that on root I didn't have to actually calculate this in terms of a B to actually find what the image is what the image looks like as a vector in r2 okay that wasn't necessary okay so at least in this example and that's going to be a theorem a bit later on but at least in this example adding the linear Maps adding the linear maps is corresponds ok works well agrees with adding the matrices that represent them ok but this isn't the case where the maps are indeed represented by matrices ok so we know that's the general case we know we can always do that and we can state this as a general theorem that will follow a bit later okay so matrix matrix addition matrix addition corresponds to map addition to linear transformation addition okay good okay so and and you can verify by the way that it will work the same for for multiplying by a scalar okay you can take a scalar five and calculate what is 5t and then see that 5t corresponds to five times the matrix a that represents d ok so there's a certain compatibility here now things get a bit more interested interesting when we want to define TS not T plus s but rather TS because there's a little twist okay so what's the twist nothing too complicated so the definition is as follows so we're gonna have a map T that goes from a space W to a space U and then a map map s that goes from V to W both of these are linear okay and the picture you should have the picture you should have is the picture of of so let me first write the definition and then write the picture so TS TS this is going to be a new linear map what does it do to a vector V by definition it does T of s of so this is in terms of functions this is what we call composition of functions okay so you first do sov sov sends you into W and then you do t to what you got and that's well defined because T is defined on W and takes you to a vector in okay so this is called you can call it composition of linear maps okay but that the notation is the notation we usually use for products okay we don't have this in in functions usually we put a empty circle here okay and that's not the notation we use here so then this is a slightly confusing the notation for composition of linear maps is the notation we're used to seeing for multiplication and you might ask huh why would something somebody do such a confusing thing okay and there's a good reason for it which we'll see in a minute in an example but let me just emphasize here that this the picture you should have in mind and sometimes drawing these things help helps so we have via vector space then we have the map s sending us into W and then we have the map t sending us into you right and then we can compose the composition goes directly from V to you okay and the composition is denoted by the way this is something else that is a bit confusing the composition is TS right because we first do s and then we do t okay so this is something you should be already used to seeing with functions that's how we compose functions okay good okay so let me emphasize this TS is the composition of T and s although the notation is TS rather than t composed with this okay so it's the notation we're used to seeing for functions so why is that and let's see an example well maybe before we see an example let's prove that this is indeed linear okay why is this a linear map so let's start with that so let right write it as as a little clean tiny theorem ts is linear proof so how do we usually prove something like this we just follow the definition there's not much much we can do otherwise so let's see what do we need to show we need to show two things we need to calculate TS of a sum U plus V and we need to calculate TS of alpha v right and we need to show that here the Alpha can be pulled out and here that it equals T of the sum P of s of the sum okay so let's see what what we get sorry TS of 1 plus TS of the other I'm confusing myself ok so by definition TS of something is T of s of the something do you agree that's the definition of this product which is actually composition so since s is a linear map we know we can write this as T of s of U plus s so V do you agree ok now since T is a linear map T of a sum is T of 1 plus T of the other so this equals T of s of u plus T of SV good and now by definition T of s of U is nothing else but TS of you plus ts avi so this proves linearity in terms of addition do you agree okay what is TS of alpha v same thing T of s of alpha v alpha can be pulled out of s so this is T of alpha s of V and then it's T of something alpha alpha can be pulled out because T is linear so it's alpha of T of s of e and again by definition this is alpha TS okay very very simple you could even say trivial as long as you understand what you need to prove what you need to show good okay so we know that TS is linear let's do an example let's use exactly the same maps that we had in this example so this is T this is s now we want to see what is T times s okay so I'm going to erase this no the the definition starts with the fact you can only define it for maps that are linear so in the definition rewrote t NS are linear then we define the composition okay you can't add you you can't expect to add a cat in the dog and then show that the the product is linear okay there you're adding linear transformations you can expect maybe the product or the sum or whatever to be linear okay okay so what is TS of a B in this example again okay so we know what it is because we know the definition its T of s of a be its composition right so what is T of s of a be so I'm going to first replace s of a B is this so it's T of 0 3 1 negative 1 times a B do you agree and now I know what T of something is its this so this is 1 2 0 negative 1 times whatever this is 0 3 1 negative 1 a B do you agree but now I these are all matrices now right and for matrices just like before I knew the distributivity distributivity holds i know here that associativity holds for matrix multiplication right so instead of multiplying this times this times this I can multiply these first and then multiply like this so this is the same as 1 2 0 negative 1 times 0 3 1 negative 1 multiply these two first and then multiply by a be do you agree and if you want to know what it is precisely all you have to do is calculate the product of these two matrices so it's 0 plus 2 3 minus 2 0 minus 1 and 0 plus 1 is that correct yep times 8 so what we just discovered is that the product of the linear maps what is at least notation wise looks like the product of linear maps is compatible with the product of the matrices and that's the reason for the notation okay so the precise statement is composition of linear maps is compatible with matrix multiplication okay so composition of linear maps is compatible with matrix multiplication multiplication good and in a sense okay so one way to look at matrix multiplication you may ask why is matrix multiplication defined in this rather weird way okay why is this the way we multiply matrices okay so in fact make matrix multiplication that this is a this statement is an if and only if statement or it that the word compatible says that this is compatible with this but the other is compatible with the first one just as well right so matrix multiplication is compatible with composition of linear maps okay and in a certain sense that's why matrix multiplication is defined the way it is okay and and now you should not think of this notation as because as confusing any longer okay it's confusing because what it really is is the composition but what it really does is the same as matrix multiplication okay good okay so composition of linear maps is compatible with matrix multiplication this is of course not a theorem at least not the way we gave it here it's just a conclusion from an example but it is going to be a theorem and we're going to prove it a bit later okay okay I want to show you one more example in this situation why the composition of the other way around is defined as well right because this is a map from both these maps are from the same V to the same V they're both maps from R 2 to R 2 so you can either do first s and then to the result duty that's what we did here but just as well you can do first E and then to the result do s that's just as well defined ok so let's see what s times T is for this example let's write it here so what is s times T of a B for that precise example so by definition it's s of T of a B right T of a B so we can we can skip steps we know what it is we need to multiply the matrices in the other direction and then operate on a B so let's write the matrices the matrix for s was 0 3 1 negative 1 the matrix for T was 1 2 0 negative 1 and then that acts on a B right and the product of the matrices in this order we would be surprised if it would be the same as the product India in the reverse order right we know that matrix multiplication is not commutative right so what do we get we get 0 plus 0 so we get a 0 here 0 minus 3 1 & 2 plus 1 3 so that's the matrix representation of s T okay and so this is the matrix for SC the matrix 40s we saw in the previous board was 2 1 negative 1 1 and they're different okay so in general in general if T s is defined there's no reason for s T to even be defined let's look at this board a minute right if saying that TS is defined means that the looking good means that the composition works in this direction okay the the range of s is included in the domain of T that's when we can define this for it to be defined the other way the we need we need U and V to be the same right we need to be able to compose the other way around and in general that doesn't hold right that's that's a special phenomenon but even when it does hold even when both are defined for example what example when we're working with linear operators from a space to itself they're usually not equal okay so let's write that in general in general when TS is defined when TS is defined st may not be defined right it's not necessarily defined okay but even if it is even if both are usually st is not the same as TS okay and this usually is precisely the same statement as matrix multiplication is not commutative okay okay I want to write down two more examples and from them move on to some more definitions so do we want to erase this let's erase here so here's a couple more examples so let's look at em two of our and take a map we're gonna call it s that sends us to our four and then take a map which we're going to call T that sends us to our three of X and the maps are gonna be the following what s does to a two by two matrix is send it to a plus B 0 C plus D 0 that's a vector in R 4 right and you can see that it's going to be a linear map because each entry here is a linear combination of the entries here we already know that ok and what T does T of a general vector in R 4 let's call it Alpha Beta Gamma Delta right these are Greek ABCD what it does is send it to the polynomial beta plus Delta x squared plus beta plus Delta X cubed ok and again you can see that each of the coefficients here is a linear combination of these therefore this is a linear map as well ok and we can compose them write the domain of T is the range of s we can compose them so what is T s TS is something that operates on M 2 R so I have to see what it does to a matrix of the form ABCD and what does it do its T of s of the matrix so it's T of a plus B 0 C plus D 0 do you agree and T of this thing is the polynomial right that takes beta what is beta 0 what is Delta 0 so what's this polynomial going to be 0 this is going to be the 0 polynomial ok so when we take this linear map which is obviously not the zero map and we take this linear map which is obviously not the zero map the composition is the zero map okay and that shouldn't be too weird because we know that we can multiply matrices that are not zero and get the zero matrix right there are what we call zero divisors in matrices okay so that's one so let's write that ts is the zero map what we defined is the zero map in terms of linear transformations it sends everything in every two by two matrix it says to the it sends to the zero polynomial okay by the way what is s T in this case it's not defined doesn't make sense at all right s T would be take a vector here send it to a polynomial then do s to it but s doesn't have a clue what to do with polynomials S operates on two by two matrices okay so let's maybe write that note that s T is not defined in this case okay let's write another example yet one more example each each of these examples demonstrates some phenomenon that you can encounter that's why I'm writing several examples so here's another example let's take the map going from R 3 to R 3 and this is going to be the map s and then there's gonna be a map T that again sends us to R 3 and how does it work so I have to tell you what s does to a vector X Y Z right and what it does is it sends it to X y 0 do you remember what we called this map a projection remember this is a projection it takes a point in three-dimensional space and projects it onto the XY plane okay doesn't care what it's Z coordinate was every point above the point XY is going to be sent to XY 0 okay so this is a projection and then T T again operates on guys like this so T takes an X Y Z in general and sends it to minus X Y Z what did we call a map like this a mirror is slang what did we call it more formally a reflection okay so T reflects with respect to the Y Z plane okay okay so what does TS do T s what does TS do so TS of XY z TS of XYZ is T of s of X Y Z right s of X Y Z is XY 0 right and then T of that so it's T of X y 0 and what is T of that minus XY 0 do you agree ok so the so let's write that this is a this is a projection and this is a reflection ok so in this case the composition is defined in this order by the way what is s T in this case what is s T of X Y Z its s of T of X Y Z so it's s of minus X y 0 sorry minus X Y Z right we first reflect and then we project what do we get the same thing so in this example both maps are defined both are operators so it would make sense that they're defined and in fact they're equal right so in this example in this example st and TS coincide good ok so here's another definition here's another definition so we know what what it means to compose linear maps we know what it means to add them and multiply one by a scalar division is is probably not gonna make sense right just like we don't have division of of matrices right it's probably not gonna make sense and but what could make sense is at least division in some sense which is the inverse right so this is the following definition definition so let T from V to V now it's an operator it goes from the space to itself be linear one-to-one and on to okay a linear map that's one-to-one and onto as a function what do we know if a function is one-to-one and onto it has an inverse so as a function t has an inverse right and we're going to denote the inverse just like we do for functions denoted T to the minus 1 that's how we denote the inverse function f to the minus 1 and what T to the minus 1 goes from V to V goes backwards but they're the same space so inverse is still a linear operator right and what does the inverse satisfy the property of being an inverse satisfying that if you take T and then do T inverse or if you take T inverse so this is first doing T inverse and then doing T but you could also do T inverse on top of T first duty and then do T inverse and both are going back and forth and we'll get you where you started that's the property of being one-to-one and on - right that's the invertibility so both of these are equal and they're both equal the identity map I remember I I is the map that sends every little V to itself okay so what's missing from this something is missing here everything here holds because these are functions what's missing is why is T inverse a linear map that's not in the definition that again requires proof ok so again a claim claim proposition T inverse which is the inverse function of T as a function is not just an inverse function it's all in your map we need to prove that ok so again how do we how do we prove this maybe let's take a new board how do we I'm gonna regret erasing stuff but I don't have much of a choice so how do we prove this we have to show that if we take V Plus u and do T inverse to it we get T inverse of V Plus T inverse of U right that's and likewise for a scalar product ok so let's do it so prove we want to take T inverse of a sum v1 plus v2 and show that this equals to T inverse of V 1 plus T inverse of V 2 that's what we need to show so let's denote denote remember that T inverse is the inverse of T that's how it's defined so V 1 V 1 is T of something right every vector is T of something because the map is on - so I'm gonna denote V 1 is gonna be P of W 1 and G 2 is gonna be T of some W 2 and not just some a unique element ok it's one-to-one and on to so let's so W 1 and W 2 exist and are unique because the map is invertible because it's a one-to-one and onto map ok so T inverse of V 1 plus V 2 equals T inverse of T of W 1 plus T of W 2 right and ok what is can I write that this equals T inverse of T of W 1 plus T inverse of T of W 2 nope that's what I'm trying to prove I'm trying to prove the t + versus linear right Tinh the next the next step to be T inverse of this + T inverse of this is using the fact that it's linear but I don't know that that's what I'm trying to prove okay what I know is linear is T that I do know so I can go the other way and say this equals T inverse of T of W 1 plus W 2 that's because T is linear good ok and now T inverse of T of something is the identity his W 1 plus W 2 right that's the property of being an inverse map if you do T and then do T inverse to what you got you get to where you started and what was W 1 plus W 2 W 1 plus W 2 well T of W 1 is V 1 so W 1 is T inverse of V 1 and W 2 is T inverse of V 2 good clear and that's what we wanted to prove okay so this proves the T inverse respects addition and then you need to prove the same thing for so I'll leave that as an exercise use others similar arguments to show that T inverse of alpha v equals alpha t inverse a V don't be tempted to use them to use this fact and showing it right you can only use the linearity of T along the way and the fact that every vector is T of something okay because it's a it's a invertible length okay okay let's look at some examples let's look at some examples let's start with looking at this example here okay so is the map t here invertible does it have an inverse t sense X Y Z 2 - X Y Z can I go back the answer is yes and what's the inverse map of T it's T itself T and T inverse are the same map you take 1 2 3 you send it to negative 1 2 3 right so all you do is add a minus sign to the first coordinate and that's precisely what you do if you want to go the other way right so in this example T and T inverse both exist and they're the same do you agree do you see that ok what about s s doesn't have an inverse right what s does s projects it kills the Z coordinate makes it there's no way to go back right so what's hidden here the reason you can't go back is that s is not on to right and not one-to-one right it sends everybody doesn't matter what Z they have to zero right another way of saying it is that s as a colonel right okay so s is not invertible so T equals T inverse whereas s inverse does not exist that's in this example do you agree okay so note that we're saying it in various ways just as a function this function is not one-to-one and onto right but in terms of being a linear map we another way of saying the same thing is that s has a kernel a non-trivial kernel okay and we know that having a kernel is the same as not being one-to-one okay so everything with x and y being 0 you can put whatever Z you want and it's going to send you to the zero vector it's gonna project you onto the origin okay so it has a non trivial a one-dimensional kernel therefore it's not one-to-one therefore it's not going to be invertible good ok what about for example the first example that we had the one we saw with the matrices the one with the matrices so let's it's no longer on the board uh let's recall what it was quickly so the first example that we had we had three examples so far and the first one the first one at least T was T of V was a V where a was the matrix 1 2 zero negative one right so T of V was 1/2 T of a B was 1/2 0 negative 1 times a B that was T of the first example there was also an S but let's look at T first ok so the claim is that t is invertible okay and can you guess what the inverse is gonna be right T inverse is going to be represented by the matrix a inverse okay and again that's going to be a general theorem which requires proof okay so let's let's see that T inverse is represented represented by a inverse so first we need to find a inverse how do we find an inverse matrix we have various ways of doing it right we put the matrix a 1 2 0 negative 1 then we put I here and then we do row operations until we reach the canonical form here hmm I write but if it's invertible the canonical form is gonna be I write and here we're gonna find a inverse that was a theory okay so what are we gonna do the first thing is I want to make this a 1 so I'm gonna multiply our two is gonna become negative R 2 so I get 1 2 1 0 and then 0 1 0 negative 1 right and then I want to make this 2 a 0 so I'm gonna subtract from R 1 R 1 is gonna become R 1 minus 2 R 2 and what I get is 1 0 1 minus 2 right plus two minus 2r2 and then here I'm gonna get zero what well this I'm not touching in fact do you agree and that's it he just became I so I just became a inverse okay so this is a inverse so a inverse in this case is the matrix 1 2 0 negative 1 and just the coincidence a coincidence but note that in this case just like in the previous example a inverse equals to a in this case so which happens to be a okay that's not usually gonna be the case it just happened in this example okay so this is a inverse what we want to see is that a inverse is exactly the matrix representation of T inverse ok so what do we need to show what do we need to show we want to see is it true is it true so let's recall what T does T of a be T of a be is a times a B so it's 1 2 0 negative 1 times a B right and what that is is a plus 2 B that's the first entry and then 0 minus B that's what T does to a B right so if T inverse is really presented by a inverse if I multiply this matrix by this I should get back to a B that's what it means to be the inverse a map that sends this back to a B right so let's let's see that T inverse of a plus 2 B minus B the claim is that this is precisely a inverse which happens to be the same e times the vector a plus 2b minus B and if we're right this is going to be precisely a B and it is right it's a plus 2b minus 2b which is a and then it's 0 times this and therefore is so in this example t inverse we just said that T inverse is represented by a inverse okay good so again that's going to be a general theorem and I'm gonna leave it free for you for practice so find s inverse there was an s in this example two was given by a different matrix find the inverse map of that it's a linear operator and it is invertible okay we have tons of ways of deciding whether a matrix is invertible or not remember that was a theorem with many many parts you can check the determinant is 0 or not you can check that when you do row operations you get to I you can check that for any be ax equals B has a solution remember we had a lot of characterizations for being invertible so s is invertible as well find s inverse and show that it's given by the inverse matrix of the matrix that gives s okay okay so i want to i want to write a couple more facts so first of all here's a theorem well both of these both of these facts are going to be theorems so if T from V to V is a linear operator remember that an operator is just a map from a space to itself then T is invertible if and only if as a function when can you say that a function is invertible if and only if it's one-to-one and onto right but as a linear map there's a stronger statement it's invertible if and only if T is on - and if and only if T is one-to-one you only need to check one of them and why is that true because of a theorem that we had previously okay so let's write the proof the proof is gonna be a one-liner the proof is gonna stem from the fact that for a linear map these two facts are if and only if a map is on - if and only if it's one-to-one providing one thing that the dimensions of the range in the domain are the same that was the theorem that we had and if it's a linear operator there of course the same cuz they're the same space okay so the proof is we had a theorem previously that if the dimension of V 1 equals the dimension of V 2 and that theorem followed again with one line from the rank nullity theorem okay you can look back and see how we proved it that if this holds then T is one-to-one if and only if T is on - and that completes the proof if it's one it's the other and then it's both if and only if it's invertible that's a general fact about functions I'm not going to prove it here okay a general fact about functions a function is invertible if it if and only if it's one-to-one okay functions between sets not not linear function sir okay good okay and one last theorem for this clip for this lesson so if TNS are invertible then TS is invertible remember that being invertible is defined for linear operators okay so both of them go from V to V therefore the composition makes sense okay we can compose them but the composition is going to be invertible as well just by writing this it should make complete sense that it's true if you do T and then do s and you can go back on both because both are invertible then you can go back on the composition right and what's going back gonna be and you tell me what is TS inverse going to be you can even say precisely what this is going to be right it's gonna be s inverse T inverse in the other direction and the reason is so we'll prove it formally in a minute but the reason is the following picture so intuition we have V we have so we did TS so s goes from V to V and then T goes on back to V these are both operators so TS is this map right so what's the map going backwards gonna be so here there's a map going backwards T inverse here there is a map going backwards s inverse that's because both are invertible and this map going backwards on the one on the one hand it's going to be TS inverse right it goes backwards but on the other hand it's the composition of these two backwards arrows and the way we we write the multiplication is we first do T inverse to V and to that we do s inverse so just like we write composition right we first do T inverse and then s inverse and we write it like this do you agree okay so that's the intuition and and the this intuition is is in fact the proof okay the the line between being intuition and being proof is very very blurry here so the formal proof would just be to write it a bit really a bit more formally so what do we need to show we need to show what does it mean to be the inverse what it means to be the inverse is that if we take TS and multiply it by TS inverse we should get the identity that's that's how we defined the inverse map right so that's what we need to show we need to show that if we take TS and multiply it by the inverse we get the identity right so if we if we're claiming that this is the inverse tell me if you're following the logic whatever TS inverse is when I put it here I should get the identity that's the definition of being an inverse my claim is that this is it this is the inverse so in order to prove it I should put this here multiply and see that I get the identity do you agree that's what it means for this to be in fact the inverse good is the logic clear so how do I show that the this product is the identity I have to take a little V do this to it and see that I get V that's what it means to be the identity okay so everybody following the logic I didn't write anything I didn't write a single equal sign yet I'm just understanding what it is that I need to show and then showing it would be what will end here right it's gonna be very easy but you have to first understand what you need to show good everybody with me okay so what is this this is by definition the way we we we do a product of maps is do this and to that do this so this is T s of s inverse T inverse of V right that's the definition of composition right and what is T s of something so this equals this equals T of s of that's what TS is of s inverse of T inverse of V how many do I have to close for right and s of s inverse of whatever is like doing nothing so this equals T of T inverse of e because whatever T inverse of V was I did s inverse to it and then I did s to it I got back to where I started and likewise here this equals I of V which is just okay so we proved that indeed if we take this map and compose it with this we get the identity map and therefore this is the inverse okay and I recommend so this is this is abstract I recommend for practice take example one for example the the one with the matrix what was it 1 2 0 negative 1 and then there was a matrix for s and show it find T s I think we already did that multiply them in the other direction and show that you in get indeed get the inverse meant that it does that the inverse operation whatever T s does s inverse T inverse sends you back ok so practice it on an example just to get the feeling that it's the right thing okay Oh so this completes this discussion of operations on linear maps okay what I want to do in the next lesson is to argue formally to write it as theorems that indeed there's a compatibility between the operations on the linear maps and the corresponding operations on the matrices that represent them okay so that's what we're going to prove next

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Length: 58min 21sec (3501 seconds)

Published: Mon Nov 30 2015