Hi everyone. So today we will talk about invertible map, just like we have invertible matrices. So we have invertible maps. And this is closely connected to the concept of isomorphism, which I was talking about before. Okay, so let me give some definitions. So you may have heard of these terms one-to-one on maps onto map. One-to-one correspondence, also called injection surjection and by deductions. So consider a map T from V to W. I'm not even saying it is linear, just any map. And we end Ws sets if you're like. All right? So T is said to be injective. If different elements of me, if two things are different in elements of V, then the images are also different. If, if mu is not equal to v, then the image must also be different. Right? For example, you would have maps from V to W, which can take two different elements of V back to the same element of w. That's such a map is not injective. If the elements are different, the images must be different. Or equivalently, if the images are the same, if say T1 equals V2, then V1 equals V2. Okay? And tea set to be surjective. People also call it one-to-one injective or one-to-one. T is said to be checked. Her people call it on if the range of T is equal to w. Okay? So here is V, here is W, and here is the map T. So if this maps to the, all of w's, okay? So every point and w is something that is set to be surjective. And T is said to be bijective. If t is injective and subjective, meaning one-to-one and onto. So let me draw a little picture here for the injective part. Well, I think you've seen this concept. Let me just write this here. Show remote is equivalent to T1 equals P2 implies V1 equals V2. If two things have the same image, then the two elements are the same. For all. T is surjective, is equivalent to for every w in W. There is a v in V such that T v is equal to w. All right? So t subject to means you give me any w in w, then it has the image of something. Alright? So both of these concepts are important for us, but let's start with injectivity. And so let me just come back to this injectivity here. Now, injectivity says whenever t1 is equal to T V2 and V1 must be V2. So a special case of this as if t1 is equal to T of 0, then V1 is equal to 0, right? So if T is injective, so okay, here at the proper times, whatever I'm saying, if that's going to be clear. So let T from V to W be a linear map. Then T is injective if and only if the null space of T is 0. So if you wanted to check with our map, the linear map, this is important is only for linear maps. So you want to check whether a linear map is injective. All you have to check is if the null space of T is 0 or not. So let's look at the proof here. Okay, so suppose T is injective. Okay? I have to show that nullspace is 0. If V belongs to the null space of T, then T of v is 0, which implies TV equal to t 0. Oh, I forgot to show this before, okay. That I'm come back that T of 0 for a linear map is always 0. Okay? I'll just write that later. So TV is t. T is injective. So V is equal to 0. Therefore, null space of T must be tough because he set. So this is the easy part. Now we have to figure out the canvas that I've just nullspace of T is 0, then y is injective. Conversely, suppose nullspace of t is just the 0 set. Now have to show that if TB1, TB2 than V1 equals P2, write this definition. To prove injectivity to show that T1 equals V2 implies we've unless you. So if v1 and v2 are in V and TB1, TB2, this implies is 0. So linearity implies. Therefore, V1 minus V2 is in the null space of T. But the nullspace of p is 0. And that proves it. So to check injectivity, you don't have to check it for all v1, v2, we just have a stack of TV. 0 is 0 or not. That's also meaning check the nullspace. But here's the remark which I used. In the proof. We use the following. If T from V to W is linear, is a linear map, then T of 0 is 0. The first nato is in V and W. But p of 0 is 0. And please just do it yourself, okay? Just not complicated. It's basically relying on t 0 plus 0 is 0, etc. Yes. Please try this. Okay? Okay, so now to talk of invertible maps, just like we have invertible matrices. But before we talk of invertibility, talk of one over something. So there has to be something that plays the role of the so-called 1 on the identity. So what is the identity map? So let's define that. If v is a vector space, then the map from V to V, which sends me back to itself, is called the identity map. Okay, so that's my identity, just like you have an identity matrix, this is Martin tMap and you can see why it's called identity does Maps, okay? Note I is linear, right? I v1 plus v2 is V1 plus V2 and so on. Okay? Okay, sometimes we will write this I depends on the vector space. Of course, which vector space? There was an identity map from V to V. And if there's another vector space, there's an identity map from W to W. So to distinguish between those times, when it's not clear from the context, sometimes we will write IV instead of high to specify the vector space involved. And this will be relevant in the next definitions. So what does an invertible map, if you have a linear map from V to W, When is said to be invertible. Well, for matrices venues here, matrix is invertible. Matrix a is invertible if there is a matrix B such that AB is identity, NBA's identity. All right? So same thing here. So a times B. So that's the inverse. So similarly, we can define a linear map and an invertible linear map. Linear map T from V to W, is said to be invertible. If there is a linear map S. Now, you cannot multiply maps. You have to do competition. So for the composition to be defined, S has to go from W to B. Okay? So linear map T from V to W is invertible if there is a linear map S going the opposite way. Remember this map S has to be linear and such that now saved to do composition. So as composed with T as the identity of V and T composed with us, or tf is the WTO. So you will see that not every, just like not every matrix is invertible or not every map from V to W is invertible. Only certain maps are invertible and test. Like for matrices. If a matrix has an inverse, it has unique same bitmaps, right? So let's just prove this remark. If T is invertible, then T has. Let's start then with homily one map from W to V. Such that this is true, I pick cannot be two maps to inverse x. Okay? And why? Proof? Suppose S1, S2 a maps from w to v such that t is the identity on V, S to T. And T W, T S2 is okay, then a happy show. S1 is equal to S2. Okay? So. Ok, and there are different ways of showing it, but essentially is this. So firstly, let's look at this. S1 and S2. Okay, now, I did not prove this, but composition is associative. So let's look at it this way. That's one of t is the identity on v. Compose with S2. Now the identity matrix composed with anything you're mapping v to v and then applying S2, S1, S2. I want you to check it please. Whenever you compose the identity map with anything, you just get back the same map. But this is equal to just from associativity of the product is the same thing as S1, S2. Now ts, T2 is I tablets. So it says when compose with i w, i up compose with anyone is say, map. And these two are equal. So therefore, if you go back and look in your book on linear algebra, this is essentially how you also prove that if a matrix has an inverse, it has a unique inverse. Okay? So this remark here, V to W is invertible, can be unique map from w to v is called the inverse of T and denoted by t to the power minus one. Okay? So if you see T-inverse, what I mean by that is the map as such that as composed with T and decompose with this, alright, entities. Okay? So let us go back here. This will remind you a linear map T from V to W is said to be invertible if there is a linear map S from W to V. So firstly notice then is this the so-called inverse S has to be linear by definition. Say there is a linear map as from publicly so that the compositions are identities. So the next proposition is the following, right? Let's just, this is very subtle point and I want you to pay attention to this linear map T from V to W is said to be invertible. If there is a linear map S from W to V such that T. Suppose I found a map S from W to V, such that these two things will hold. But I don't know whether it's a linear map or not. Okay? So I'm given a linear map T from V to W. And suppose I am able to find a linear map. Suppose I'm able to find just some map as from W to V, such that these two things as composed with T that can decompose with this is identity. Then I cannot say T is invertible. Because for the inverse to be, for it to be invertible, map as you find, not only must have these two properties, but must also be a linear map. So the next proposition says, the linearity is not needed to be checked. If you have a map T from V to W, and you have a map S from W to V. So these two things are identity. Then S will automatically turn out to be linear. So if you want to check something is invertible, all you need to check is to find a map as such that these two things are true. Okay, That's the first part. The second part actually is this, that if you've studied mappings and stuff, not just for linear maps, okay? The fact that there is a map F from S compose and decompose f is the identity. This is equivalent to the State fact that T is a bijection. This is just true for maps. If T is a bijection, it has this composition inverse. And conversely, if there is a composition inverse, t is a bijection. Okay, So putting all of this, whatever I've been talking about, putting it all together. This is the proposition that I want to show. The linear map T from V to W is invertible if and only if T is a projections. So if you have a linear map and you want to check invertibility, meaning the existence of the S which is linear and that's composed decompose. But that's all you need to check is is T objections as T1 to T12. So let's do the proof. Okay. It's not very complicated or anything. Notice that this will be a bill not depend on any basis. I'm not saying B is a finite dimensional vector space or not. So let's do it this way. Suppose T from V to W is invertible. Okay? Hence, there is a linear map S from W to V such that T, S and P and W st is the identity on. Okay? Now if you've studied mappings and stuff, you know that these two, these two things automatically imply t0 is a by dx. Let us prove it. Okay? We claim this implies T is a byte actions. Okay? So why? So injections. If T1 is equal to TV2, then this implies you apply S to both sides. Okay? But Estes identity I, V1 is V1. So if T1 equals V2, then this is, this is for all V1, V2 and V1 is equal to t v2. That applies to both sides as TV, but SDS identity. So you get HIV and TB maps just V12, V1, and V2 to be right and surjectivity. Okay, I'll have to show that every element of w is T of something. Okay? So given w and w, we have to, well, you have to show there is a v and v such that w is TV, right? Okay. Now you are given that T is invertible. So DFSs I didn't, T of w. Therefore, for any w in W, W is okay. Therefore, we can write this as T of w is w. So I've managed to write w as T of V, where V is equal to B. Therefore, T is surjective. Okay? Actually what I've done here, I have not used linearity of S at all in all of this. I did not at all use the fact that the linear map S, which is the inverse of T, which is TSS iSCSI. I never use the fact that it's linear. The argument I gave it, this is very standard argument in algebra. If you've studied groups, you've seen the same argument about bijections Or when you started mappings. Okay? So if T is invertible, implies T is not to go the other way. Suppose T is a bijection. Okay? Of course we are also given T is linear. To begin with. Let us get rid of the statement. The linear map T. So the linear map T is already there. Okay? T is already linear to begin with. Now you're talking of is this linear map invertible or not? So the linear map T from B to a is invertible. If not, we also know T is linear. Ok? So if T is a bijection, we can define the map. Let me write this again. We have to show T is invertible, P is invertible. I have to build that map S, which is linear. So you have to construct a map from W to V such that number one is linear. And to st is the identity. And ts and t. That's what we have to do. Okay? So firstly, what I'm going to do is I'm going to build a map S, which has this property too. Both of these. And then I will show that the map as I build is linear. Ok. Now if you've studied mappings, then bisection automatically implies two. Actually, this is very well known. And what is new here, It's actually get to automate. You want to also prove that this is linear. But I'm going to do 12 both. Alright? But first of course we have to construct the map S. Okay? So I have to tell you, given a w, which, which element of V does it go to? So given, sorry, given W and W, since t is my direction, which means T is onto, there is a v in V such that T, because T is a bidirectional. But also since T is injective, bijective, there can be two different V such that T v w. There is exactly one B&B such that TB is yes. So we can define f of w is phi. Right? Let me go over this again. I have to construct a map S which reverses t. So you give me any w, v and w and w. Since T subjective w is the image of some subjective 12, now T is injective, then they cannot be two different V's which go to the same w is a unique one. So you give me a W, there is a unique V such that TV is W. So my inverse map is send this tablet back to that tree. Hence, we have created a map from W to V. So that, you know, if you have v, you get TB and then sends it back to v. Ok. And if you start with the W than it is some TV. So add sends it back to v and then T sends it to TB. So the first one says composed with T is the identity on V and T composed with us. So if you define your map S, Thus the way I described it, given a W, T of V, some v and a unique wreath or defined S sub w to be v, sends it backwards. Okay, here is TV S and V and then sends it forward, or V TV sends it back. Which is exactly these tools. So I've constructed the map S such that these two things are true. So I'm almost there to prove that, Almost done that T is invertible. Okay? Too. Prove that T is invertible. The only thing remaining is written. There are two things you need to do to prove T is invertible. You need to build this map as from W to V, such that S versus t. The second they have to show that S must be linear. So that to prove T is invertible. It remains to show. All right, Now how do I show f is linear? So let's say given W1 and W2, W. Okay? So consider these two things. I have S W1, W2, and I have this other object S of W1, W2. Okay? I have to show these two things are the same. Okay? So how do I show this? Now? We have to go back to the construction, but actually there is a simpler way. I know. D is injective. So if d of this is equal to T of that, then these two things must be itself because T is injective, right? So that's all we need to do. And then, then we apply the fact that T is the inverse of S. Let's call this a and B. To prove a note, T of S W1, W2, right? Remember T is linear. So it's TFS W1, W2. And this is by the definition of the identity. So this is identity w1, w2 is w1, W2. Okay? So that's one side. But then apply T to the other side. So this is ts applied to W1, W2 and TSS identity. This is w1, w2. So these two quantities are the same. So therefore the injectivity of t implies if two things are the same and T is injective, then similarly to this, okay? You can show they are equal by applying the t to both sides. Who can be done. Okay? So please go back and watch this video maybe 23 times. So make sure you understand all pieces here. Okay? That seems like a long proof, but actually it's fairly elementary. Okay? So the linear map T from V to W is invertible if and only if T is a bijection. If T is a bijection and T is linear, then automatically can build the S such that S T compose and decompose as her identity. All right, S will also be linear. So please go over this again. Okay? Tsunami come to this notion of what is called isomorphisms. Okay? Two vector spaces are set to be. I mean, intuitively, they are isomorphic. If really we're just labeling these sets V and W differently, but really they're identical. For example, right? Let's start with this. Okay? So we have this P4. Okay? So these are all, Let's just make it simpler. Let's make it P3s have to write this is x plus a2 x. Let's make it P2 is 0, a1 and a2 and right, and then there is R3. We turn all of these, let's call it B1, B2, B3. Let's call it Texas. Okay? Now, as far as vector space structure is concerned, P2 and R3 are really the same vector space. Sorry, it's not a good name. Let's call it B1, B2, B3. Okay? Any polynomial, you can say it's determined by a0, a1, a2, a3 coefficient. And here are the three numbers, b1, b2, b3. So we can actually define this map T from P2 to R3. Let me just send this polynomial to just the vector a1, a2. Okay? So t also does preserve the structure. Meaning if you add two polynomials, okay? You get that. You add here like that. And same with scalar multiplication. Really, the way you add polynomials is exactly the way you add vectors. Scalar multiplication is also the same. So really for all practical purposes, as far as the vector space structure is concerned, P2 and archaea are really the same. Vector space, does different ways of representing the vector. Okay? So that is sort of the notion of isomorphism. Say you're studying a vector space, you're setting R2, you might as well just study P2 or vice versa. Okay, but let's make this notion little clearer and precise. Two vector spaces, V and W over F are said to be isomorphic. If there is an invertible a linear map T from V to W, yes. Okay? So, so that means that v and w have the same additive structure. So let's use the example, right? So P2 is isomorphic to R3, has seen next, right? So first you have to define the map, which I have already done before. So it is 0 plus a1x plus a2 x square. Two is 0 plus a1 and a2. You can easily check that T is linear. All right? You add two polynomials, same thing as adding vectors, so on. Okay, So that doesn't prove invertibility though. We claim is invertible. Right? So to proving whether it be from that proposition, all we need to do is check the direction. And that is almost trivial, right? So firstly is injectivity. Remember the theorem, pre-check a linear map is injective. You just have to check the nullspace is heat. All right? So if that's easier, 0 vector, then A1, a2 is the 0 vector, therefore is a CDO. And therefore the polynomial to the 0 polynomial. Okay, and who buys the subjective? That is also, are you up? Yes. Right. Y is t onto. You. Give me anything here. There's always a polynomial going with it. Write R3. What does the polynomial which is mapped to it. So it's T B1 plus B2. B3 x square is okay. So before t is invertible. So P2 is isomorphic to R3. So that's one of the vector space structure is concerned. These two are identical, right? Remark. So you know, sometimes I'll write this. So V is isomorphic. Okay? Pause the recording in the middle and I forgot to restart it. I'm not sure where I stopped, so I may be repeating some things, please excuse me. So we will write V isomorphic for W means we'll write it this way. We will use this notation to say v, That's morphic to w. Okay? So I want to prove this proposition that to find out dimensional vector spaces V and W are isomorphic if and only if they have the same dimensions. Okay. Forgive me if I said all this and it's already in the recordings. Okay. I just got mixed up. So let me make sure I'm still recording. Yeah. So suppose V and W are isomorphic. So there is a linear invertible linear map T from V to W. So there is a linear map T from V to W and d is a by ejections. Now bisection means it's injective. So from the proposition the nullspace is 0. And by action mean the subjective, so the rain spaces the whole thing. Okay, by the way, no, 0 if it's injective is only true for linear maps. Okay? So we have a linear map T from V to W whose nullspace is 0 and range of T is w. Now from the rank-nullity theorem, we know the dimension of V is the dimension of the null space of T plus the dimension of the range of T. But nullspace of p is 0. So its dimension is zeros and range of T is all of that. So we get dimension V is equal to dimension W. Not have to prove the converse. Suppose you have two vector spaces which are finite dimensional and have the same dimension. I have to show they are isomorphic. Okay? So I have to construct this linear map from V to W, T from V to W, which is what you have to do. So how did we define that? Okay? Since dimension V is dimension W, any basis for V and W will have the same number of elements for w0 and w1 through wn. Okay? Now remember that if you have a basis for V, then you can add some elements of w. You can always build a map proposition. We know. We can build a linear map. T from V to W with V is equal to Ws. Not the way you do it did in fact, TF any alpha1, v1 through vn is actually alpha and W. And that's how you defined right here. So I will leave it to you. Show T is injective, showing novelties, hero is surjective. Okay, I'll just leave it for you to do it. Okay. It's not very hard that the surjectivity is clear. I mean, it's w1 through wn are in the range t and so on. So it just, I will leave it for you, please do that. And I will stop here.
Rakesh Rakesh's Personal Meeting Room
From Rakesh Rakesh October 06, 2021
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