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Problem 4 Let f : V _ V be a linear transformation on a vector space V with dimension n. Prove that for any scalar a € F, det(af) = a" det(f)....

Question

Problem 4 Let f : V _ V be a linear transformation on a vector space V with dimension n. Prove that for any scalar a € F, det(af) = a" det(f).

Problem 4 Let f : V _ V be a linear transformation on a vector space V with dimension n. Prove that for any scalar a € F, det(af) = a" det(f).



Answers

Prove Theorem 5.2: Let $V$ and $U$ be vector spaces over a field $K$. Let $\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}$ be a basis of $V$ and let $u_{1}, u_{2}, \ldots, u_{n}$ be any vectors in $U .$ Then there exists a unique linear mapping $F: V \rightarrow U$ such that $F\left(v_{1}\right)=u_{1}, F\left(v_{2}\right)=u_{2}, \ldots, F\left(v_{n}\right)=u_{n}$

Hello there. Okay, So for this exercise, we need to prove one of the most important theorems in algebra because it's widely used in case of finite dimensional spaces. So here we're going to assume that the V is finite dimensional, and we got a linear map f that goes from B to you. So the theorem say that the dimension of B it's going to be equal to the dimension of the kernel, plus it and mention of the image of the mapping. So I wanna give you some intuition behind this theorem. Of course, withdraw winds are not precisely approved for for this theory. But it's important because it gives you some intuition, as I mentioned of why this can be true, at least, and then you can figure out how to prove it. So, basically, let's build. And here you got the the set V. And over here you got this set. You okay? So you got a map f that goes from V to you. And what this map does is that it takes elements there. It takes some elements from the some subset, and it maps to another subset of youth. It can also map to the whole set of you. But let's consider first. This is Escenario here. Okay? So basically, you're taking some elements from V and map it into you. And this is the so called image of F. Yeah, but you need to map all the elements from the so the rest of the elements from B are going to be mapped to the no element here in you and these elements here on the set, we are called the Colonel of F. So basically, what this said is that you either map to some non zero element on you or to the zero element. And if you put together all this, I mean the kernel of F and the per image of the image If you take the image of the image of F basically, you obtain this set this upset on the or it could be, uh, So let's call this, um w it's going to be, uh, subset, either proper subset or it can be the whole set V. That is my to to you. So the pre image of the image is going to corresponds to this set here. And if you put together this upset, w and the colonel, You obtained the holes at the so that have implications on how is the dimension going to behave Because you're you're going to have some, uh, set of elements on the that are spawned by some basis, that, uh, span, actually the Colonel and some span for this set here in red. And if you put together those I mentioned, you should obtain the dimension of the holes had been. But this is only if v is finite dimensional. So this is just to give you some intuition behind this theory and before starting with the formal proof. So, having said that, yeah, let's start with the real proof. So for the real proof, we're going to consider an analog of this. An implication of the dimension is that if the dimension of B, I'm going to raise this, so I'm going to start with the Colonel. So if the dimension of the colonel of this map is equal to R, for example, that implies that the basis of the Colonel of the Map is a set of our victors. I mean, I I equals to you equals to one up to our, um these are elements on the and the same happened for the image. If the dimension of the image of the mapping is equals two s, that implies that the basis for the image of the of the map is generated by s elements I'm going to call them, um, I'm going to call this differently. I'm going to call them. Uh, I and the basis for the image are going to call you. I I equals to one up to us, and those are elements on you. So that's why I chose this small You Okay, so we got this, and we need to notice something. This said you will. Actually, these are not elements on you, but are elements on the image of F to be more precise. And this fact here implies that there exists some vectors, the eye on the such that if we apply the map to these vectors, we obtain each element of the basis. Okay, so we need to take into account these three things here. So did I mention imply that we got some finite, uh, set of elements on the basis of the colonel and image in particular for the image that implies that we got some set of elements on the set V and we got something extra that say the therm and is that the I mentioned of be need to be equal to R plus s, which occurs the correspondent and mentioned for the Colonel and the image. So in this sense, we can say that the basis for the said V is generated by the elements on the basis of the Colonel and the image of the image of the of the of the mapping. That means this V I s So we want the to the V s. So this could be a basis for this, said V. However, we need to prove it that this is a basis, because if this is a basis here, we got the first S elements that come from the colonel basis and these elements here that are to hear our elements are elements from the Colonel and here in green, we got s elements that comes from the pro image of the image of the map. And if this is a basis for being one of the basis for the that means that the dimension of B is going to be equal to R plus s and the term will be proved. So we need to prove that this set here that we define as a V of B is a basis. Okay, So to prove that we need to first show that it expands that we have be the basis for B this possible basis for B response. So that means taking presentation here. The span of the V returns us the the set, the initial set. Okay, so we need to show this. That means that any vector on the set V can be written as a linear combination. Uh, the elements in this case from equals to one up to our A i alpha I plus decimation from I equals to one up to S of B. The. So what we got here is a linear combination of elements of from the basis VfB. Okay, so that is the meaning of his party. So let's show that this happened, actually, So let's consider this vector on the city, and we know that this if we apply the map to this vector, this is going to be on the image of the mapping. And this implies that f f B can be written as a linear combination from one to s of elements. V I You are here. Okay. Why? Because the elements you I are the basis for the image. And this is an element of the image of the mapping. So this element can be written as a linear combination of the elements of the basis. Then we can consider a better we had That's going to be greeting us. This mission from I equals to one up to s of a I the I the I minus. This is the vector V that we have the fine previously. Okay, so we got this and again, we're going to apply them up to this vector b had and we obtain the following we obtain here the summation. So first, the information from I equals to one up too. S of B I f of the i minus f b. And we know what this expression for, um, for f applied to be is right here. And moreover, we know that f b I is equals to you. I so from this we obtain here. Sorry. So then we have here this information from I equals to one up too. s of B I u I minus f f b. And this is equal to zero. Okay, but what are the implications of this? Well, basically, we have shown that f of jihad is equals to zero. That implies that we had is an element of the kernel of the mapping. And we're going to use this fact because if this is part of the colonel, then we had can be written as a linear combination of elements from one to our Alpha I A i where a I are part of the basis of the kernel of the mapping. Okay, so these are the elements on the on the basis of the colonel and this vector V is part of the colonel. So we can write this vector V as a new combination of the basic elements of the Colonel. So that's exactly what is written here. And we know the whole expression for be hot. We know that we had is equal two decimation from I equals to one up to s of V i v i minus V. So if we replace this into this equation, we obtain the following decimation from I equals to one up to S B I V I minus V is equals two decimation from I equals to one up to our or of a i l i and we can boot along this term that is the vector V and we obtain the following that the is actually equal two. The summation from I equals to one up to s of the I The I minus decimation from I equals to one up to our A i of I These terms here are the basic elements of our basis. So the eye and alpha I are part of or set v of b So this equation here means that any victory we on the set capital V can be written as a linear combination of the basic elements of or set. We cannot say that as a basis yet proposal for for a for A for a basis of these set capital V. Okay, So this part here implies that the spon of b we generate the set V So the first property for a basis has improved. Now we need to continue. The next thing that we need to show is that this said V is linearly independent. So here I'm going to regret the the basis that we have considered here. And it is formed by Alfa one of two up two of our and here B one B two up to the S. So this is our possible basis. We need to show that these elements are linear combination are linearly independent. So that means to be linearly independent is equivalent Say that if we take decimation from I equals to one up to our of x i alpha I plus the summation Plus this mission from I equals to one up to s of why I the I This is equals to zero if and only if all the x I on the why I r equals two here. Okay, these are coefficients of the is linear combination and we need to show that the only possibility to this equation to be equal to zero is that all these coefficients R equals to zero That immediately immediately implies that the basic elements these elements of the set are linearly independent. So let's show this. We're going to apply this map to this summation here to this expression. So we got this. Okay, Mhm from I equals to one up to our of X I al I blast the summation from I equals to one up too s of why I the i is equals two f of zero The no vector will map to the new vector so there is equal to zero and this map is because it is a linear map. We got this summation from I equals to one up to our of x i f applied to each element of I blast the summation from I equals to one up to s of y I an f applied to each element v I and this expression should be equal to zero. So here you need to notice something. Alpha I These elements are part of the kernel of F that implies that for for every alpha I on the colonel sorry for yes, not on the colonel, but on the basis of the colonel. If we apply the map to this elements, the result will be the no. The new element on the on the image or just equals to zero basically. So this whole part vanished 20 and we end with the expression here and we know that f apply to V i r equals two ui which are elements of the basis of the image of this map. So we got the summation from I equals to one up to s of why I you I and this is it goes to zero. We know that these ui are part of the basis of the image so that immediately say that these vectors are linearly independent. And this implies that for every I why I is equal to zero mhm. Okay, so we got that All these coefficients r equals all these coefficients Why I r equals to zero so or equation up here it's called Triangle. We got all this. Why I r zero So this element here vanished and we end just with this summation equals to zero. So I'm going to write that here below. So this implies that the equation triangle becomes the summation from I equals to one up to our of x i l I equals to zero. And as I mentioned before, Alpha are elements of the basis of the kernel of the mapping. So these elements are linearly independent, and that implies that for all i x i r equals to zero Okay, So we have shown that based on these two results here, that the only way for this linear combination of elements of or set the I b V plus this nation from I equals to one up too s of why I mhm the I is equal to zero if and only if x I and y i r equals to zero for all I Okay, so we got this and this implies that or said v is linearly independent. So we got that we we on one hand is linearly independent and on the second it expands the set V. So this means that the is a basis for V And we know the dimension of this, the dimension of the it's going to be our plus s as we expected.

Get her home. Bye. Dukie Brown sound. I like trick. Yeah. I'm saying that there is a sense were given to inner products F of U. V. And G of UV. Mhm. No voices. Yeah. And these are inner products. It'll say on a vector space. V over the real numbers. It's Oh, we're asked to prove some statements. So in part A were asked to prove that these some of F plus G is an inner product on V. Also where F plus G of UV is defined using point wise edition as F of U V plus G of UV. Well, to prove this, we're going to verify the Inner Product Axioms for F plus G as well. 10. Well, first want to prove the inner product axiom. I won. So we'll let A and B. B. Real numbers are scholars and U and V be vectors. In fact, let's do you one, U. two and v. Than the inner product F. Well, the function F plus G. Of a. U. one plus a the U. Two. The a definition this is F of a U. One plus bu to the plus G. Of au one plus B. U. Two. The oh And now this is equal to eight times f of U. one D plus B times F of U. two v plus eight times G of You. one. The plus B times G of you too. The good stuff. Since both functions F. And G. These inner products satisfy property. One of inter products. Yeah. And this is intern equal to eight times f of U. one V Plus GF. U. one v. Right? Plus B times F. Of U. Two V plus G. Of U. Two V. And then using the definition of addition for these products, Once again, this is eight times F plus G. Of you won the plus B times F plus G of U. two v. Therefore it follows that I didn't like to come when we got on back Property I one holds for F Plus G. Really good chemistry. Let's see if the other properties holds so for property too well, F plus G of you. The by definition this is F. Of U. V plus G of UV. Which is equal to F. V. You plus gov. You since both F and G satisfy property I to foreigner products. And this in turn, is equal to by definition F plus G of the you. Therefore it follows the F plus G also satisfies property I to foreigner products. Finally, we want to prove the positive definite this property. Well, it is suppose that U. Is a non zero vector, then F Plus G. Of you. You a definition, This is F. Of you, you plus G. Of you, you and these are both going to be greater than zero. So it's greater than zero plus zero, which is zero. Since the inner products F and G satisfied Property I. three foreigner products grab not have more of a way of life. Right? You should. And therefore it follows that the function F plus G satisfies well, not quite. It satisfies part of I. three, but we aren't finished yet. On the other hand, we have F plus G. Of 00 This is by definition f of 00 plus G of +00 which is equal to zero plus zero Is equal to zero. So Since both F&G once again satisfy property i. three. Foreigner products. Therefore, since F plus G of you, you is always non negative and is equal to zero. If and only if you is zero. It follows that F plus G also satisfies in a product property I three. Since it satisfies I have one through I three. It follows that F plus G is also an inner product started pot on V. Over the real numbers. All Patreon podcast. So in part B. Whereas to prove the scalar product K F for K A positive scalar is also an inner product on the Well, All right. I know. I mean, we're not fired. This is where K times F of U. V is defined to be K times F of U V. I didn't do anything fucking wrong. Just a meeting. But once again, to prove this, I'm gonna have to prove the three inner product axioms for KF. So, again, we're going to suppose that F is an inner product on V over the reels. Yes. Well, let A and B. B real numbers the scholars and then U one, U. Two and V be vectors in B. Then it follows that K times F of Yeah, AU one plus b. U two v. By definition, this is K times F of a U one plus B. U two V. And then this is equal to K times eight times F of U one plus B times F F. You to be Sorry. This should be f of U one d. No, this is since F satisfies property, I won of any products and this is equal to right? Yeah. eight times k times f of U one. The Plus B Times K Times FF. U two. V always. And if you're distributing now, we can use the definition to rewrite this as A times K times F of U one V plus B times K times F of you too. The Yes. Therefore it follows that the function KF also satisfies property. I one foreigner products show it satisfies property I to consider K times F of U V. By definition, this is K times F of U V. And then this is equal to K times FV you since F satisfies in a product property I too symmetry. It's a good room. And then my definition, this is K times F of V. There is yes, you. And therefore it follows the function K F satisfies the symmetry property. I too. For inner products, there's one more property to prove it's the positive definite miss property. This is the hardest part. So, to prove this, first suppose that you as a non zero vector, then we have that K times F of you. You by definition is K times F of you. You Yeah. Which we know is going to be Since K is positive. And well, this is going to be greater than zero since f of you you Is greater than zero. And this is because FC is an inner product and so satisfies property I three. Yeah, that's part of it. On the other hand, suppose that U. is equal to zero. What is K F of zero? Zero? By definition, this is K times F of 00 which is K times zero, which is zero. Once again, since F of 00 is zero. And this is again since F satisfies uh huh. Property I three of any products. This shows that function KF also satisfies Property I three of inter products. Since it satisfies properties I won through I three. It follows the KF itself is another inner product slot on the vector space? The over the reels. Yeah. Is he a better troll than?

Problem. We're given age, which is son and dimensional. Subspace, subspace off. Hey, which is also and dimensional. And where will we want to show ages people to be so first to Christine to notice that that stuff space age as the basics with the in vectors and those in vectors has to be the new independent. So already down here. So that is Asia has a basis. Oh, and electors, because age it's in dimensional and more over those vectors has to be leaner. Independent, no. Are we nearly independent? Being dependent now spelled it wrong. Independent Dent. Okay, so Oh, I forgot to say that these factors has to be, you know, the independent in B because H e's a subspace off. All right, so now we can apply bassist zero, you know, a textbook, these this serum. So it says by our basic faces, the room. So they, the vectors will constitute well constitute a basis for yes. Well, so factors constituted he two, right? So that means issue. Because why? Because Asian each and b r o r both spend by the same set of vectors. So there's no other way to find a dis joint Asian V to make this happen. So this is our conclusion. We just have a vectors, that kind that span, uh, both H and visa. These two spaces has to be his two spaces have to be.

Okay, so consider a finite dimensional inner product space V. Along with a linear functional fight. Okay, the dimension of this inner product space fee is going to be end. Now, the result that we want to show is a classical result. Um We want to show that there exists a unique little V. Inside this inner product space with the following property. So for all vectors, you inside feet. Okay, when we apply the linear functional onto you, so any you um this is always equal to the inner products of you envy. Okay, so this holds for any you that you can pick inside the inner product space fee. So how do we go about showing a result like this? Well, note that there are two parts to the argument. You need to show that there exists such a fee. And then the second part, we need to show that it is unique. So let's do existence first. Okay, Yeah. Existence of therapy. Okay, so the key fact that we're going to use is the fact that because the inner product space V is finite dimensional, there exists a basis that we can use. So let e one dot dot dot up to bn be an Ortho normal basis of your choice. Okay. And because this is an Ortho normal basis, we can write any vector you in the following way. So we can take the inner product of you along uh with, you know, like each of the basis factor E one. And that's the coefficient. And we'll just multiply this coefficient along the direction in which each of the basis factors is pointing to. And then you can just sum them all up. Okay, so E n uh right. So I've just done, you know, like a basis expansion of you. And now note that all of these coefficients are in our products, which means that they're scholars. And so when we apply a linear functional fly onto you, by linearity, we can pull all these uh scalar factors out, and also by linearity. We can expand it throughout the addition. So what that means is we can write this in the following way. Yeah. Okay. So I've done very little here. I'm just using the linearity property to break up the addition and to pull out the scalar factors. Now, what we want to use is another property of the inner products, which is the homogeneity property. And so because of homogeneity, we can pull in uh this five factor, this five E one all the way to five p.m. We can pull these into the second argument of the inner product. Okay, As long as we take the conjugate, so we can write this in the following way. E one bar. Five E one plus dot dot dot plus E N five E n bar. And so there's a bar on top of all the files because we have to take the congee. It that's the conjugate homogeneity property of the inner product. And we are putting it into one big inner product, just violent charity. Right? Because this is in addition. And the argument, the first argument is always just you. So we can just lump this into a big inner product. Okay. And at this point we are done. Why is that? It's because we can just pick RV to be the second argument over here. Right? So no note that whenever you apply five, you that's always equal to in a product of you and this fee. Right? And so we're done. We have shown the existence of fee just by using some basic properties of linearity and the inner product. Okay, so now we need to show uniqueness. So let me refresh my screen and let's talk about uniqueness of the So typically how uniqueness arguments go is as follows. So you suppose that there are two candidates. So let's say one is called V. And the other one is called the star. And you're supposed that both of these candidates satisfy the property that you desire. So five of you in this case is equal to uh you in a product V. But that's also equal to you in a product the star. Right? So you're supposing that there are two candidates, both of which satisfy the desire property. Now, what you want to then show is that if both of these candidates satisfy the desire property, then these two candidates must be the same one. If you can show that, then we have shown uniqueness. Okay, so that's how these arguments go. So how do we go about doing that? Well, let's do some algebraic manipulation of the inner products. So this implies that you v minus you in a product V star. That's equal to zero. I've done nothing here, I'm just, you know, doing some algebra. And so I can then use the linearity property of inner products to lump them together into one in a product. Now, at this point, let us recall a basic fact of inner products, which is the definite this property. So suppose you have some vector X in a product with itself and that's equal to zero then by the definite this property of inner products. We know that X itself must be the zero factor. Okay, so now how do we use this fact on the right hand side? Well, note that you can be any vector right inside this inner product space, So we can pick you equal to v minus V Star. So this is a choice that we have made for this proof. Okay? We can pick any of you, but we have chosen this one to make our proof go through. Okay, So now we have that B minus V Star. V minus V Star in a product that's equal to zero. Ha ha! At this point we can recall the definite this property of inner products on the left hand side and conclude that the minus V. Star must be the zero vector and therefore V must be equal to the star. Okay? And Sarah, we have shown the uniqueness property because we have shown that any two candidates V or V star that satisfies the inner product property listed above here must be the same factor in the end. So therefore it must be unique. That's it for this video. Thank you very much.


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