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Let Pn be the vector space of polynomials of degree atmOST " In the problems below , let V = Nul(E), where E P R? is E(p) Let $ = {1,4,02,08,1'} be the st...

Question

Let Pn be the vector space of polynomials of degree atmOST " In the problems below , let V = Nul(E), where E P R? is E(p) Let $ = {1,4,02,08,1'} be the standard hasis fo1 P; and let Ps R' _ P be the 'linear combinatins o[ S' linear transformation Let [Js + Rihe its inverse the cvorinates with respect to $ linAr trasformatiomFindl the 2 x 5 matrix A ; R' VP ER?,2. Fiudl a hasis for Nul(A).Apply Ps to the vectors our hasis for Nul(A) to show that 8 = [2 _ 2t+4,7'

Let Pn be the vector space of polynomials of degree atmOST " In the problems below , let V = Nul(E), where E P R? is E(p) Let $ = {1,4,02,08,1'} be the standard hasis fo1 P; and let Ps R' _ P be the 'linear combinatins o[ S' linear transformation Let [Js + Rihe its inverse the cvorinates with respect to $ linAr trasformatiom Findl the 2 x 5 matrix A ; R' VP ER?, 2. Fiudl a hasis for Nul(A). Apply Ps to the vectors our hasis for Nul(A) to show that 8 = [2 _ 2t+4,7' _ 3t+2,/' _ 4/+3] hasis for V, (Recall that V = Nul(E))



Answers

Let $T : \mathbb{P}_{2} \rightarrow \mathbb{P}_{4}$ be the transformation that maps a polynomial $\mathbf{p}(t)$ into the polynomial $\mathbf{p}(t)+t^{2} \mathbf{p}(t)$
a. Find the image of $\mathbf{p}(t)=2-t+t^{2}$
b. Show that $T$ is a linear transformation.
c. Find the matrix for $T$ relative to the bases $\left\{1, t, t^{2}\right\}$ and $\left\{1, t, t^{2}, t^{3}, t^{4}\right\}$

Were given a linear transformation. Well, we're giving a transformation, I should say from space of second degree polynomial is to the space of third degree polynomial that maps a polynomial p t into the polynomial t plus five times PT in part a were asked to find the image of a polynomial of degree to So we're given that the panem you'll degree to pft is to minus t plus t squared. So we have that the image of this polynomial pft is going to be the image of the polynomial tu minus T plus T squared which according to our transformation, this is the polynomial T plus five times two minus T plus t squared and foiling So yet to teen, then minus t squared and then plus T cube. So the tea cubed here we also have 10. I guess the best way to start with the constant. So we have a 10 and then we have to t we have a negative five t so negative three t and five t squared minus two squared is plus fourty squared and plus t cute. So we get 10 minus three t plus four t squared plus t cute in Part B were asked to show that this transformation is, in fact, a linear transformation. As I alluded to earlier Do this. We're going to let p of tea Q of t the polynomial of degree to and we're going to let Alfa and data the scale, er's assuming these are point no meals over the real numbers. Really, if it could be any elements of a ring, in fact, actually, I'll make this simpler instead of trying to prove everything at once. Why don't we just prove one step at a time so we'll just take one one scaler? See, then we have that. The transformation of pft plus q of t the image of this polynomial well, we have this since P and Q are both polynomial of degree to it follows that there. Some is also a polynomial of degree to since the degree of P plus Q is at most two. Therefore, it follows that by the transformation, this is going to be T plus five times the polynomial pft plus Q of T and then using the distributive property we get. This is equal to T plus five times pft plus T plus five times Q of t. And we see that this is the same as the image of PFT, plus the image of Q of T and therefore we've shown part of linearity. Now we have also the t of C times pft Since C is a scaler, it follows that C times pft is still a second degree polynomial. All right, Paula, no emulation say of a degree at most two, since we're not multiplying by a factor of X and so the image of this polynomial is going to be the polynomial t plus five times see times pft and then using communicative ity of scale Ear's. This is the same as C Times T plus five times p of tea, which is the same as C times, since PFT is a polynomial of degree at most two, the image of pft and therefore it follows that t is a linear transformation. In Part C were asked to find The Matrix 40 relative to the bases, won t t squared, repeat two and won t t squared cubed four p three. So I guess one way to do this is staffed by finding the images of the bases so we have that t of one. Well, this is a polynomial which is t plus five times ones. This simply t plus five. And therefore it follows that the vector t one sub. And I guess I was called this he for this basis of p two, this is going to be the column vector five one zero. And because the basis for P three has four basis specters, I've had another zero here. Likewise, we have that t of t. This is going to be t plus five times T, which is the same as T squared plus five t and therefore t of T subi is the column vector zero five one zero and finally t of t squared. This is going to be t plus five times t squared, which is the same as it cute plus five t Square. And so we have that t of t squared. Sophie is the column vector with entries 00 five one. And therefore we have that The matrix for T relative to these two bases is going to be the matrix. Who's calling? Victors are t of one. So be t of tea said Be and t of t squared. So be I. I really instead of using be it probably could have used to see if it doesn't really matter here. And so our matrix is the Matrix with entries 5100 Columns 5100 0510 0051

Mario is. Yeah. He's just trying to see some. Are you were given the vector space V. Of polynomial over our Of degree less than or equal to two. I tried to like I can't I cannot tell you one. I have no idea what happens he said to to you here. No, I don't. Dude samba music. You refer over our with an inner product. Find by the inner product of two functions. F. M. G. Is the integral from 0 to 1 of the 50 chinese G. O T B. T. Website. Whereas defend the basis of the suspects. W. Orthogonal to the function ancient too. Did you see it was to T. Plus one beaches. Dude rob thomas number one song. Well that's the that's the song. We should. First of all We know that affects here. I'll call F. zero will be orthogonal to each. If the inner product of that zero and H equals 01 other words zero is equal to he simply a final. Yeah. Any time we have to google swell show really? I'll let F. Zero T. V. Equal to 80 squared plus B. T. Plus C. And therefore the inner product of S. Zero T. Do a speech. You are sneak. Is any girl from 0 to 1 of a. T squared plus D. T. Plus E. Times. Yeah To T. Plus one. Security. Yeah. I went orgy was you want this to be equal to zero? Well let's solve this integral. Mhm. Thanks. Come on. This is this game shit Adams this is the integral from here or the one uh to a T. Cube. It was because plus en plus two B. T. Squared. Plus he plus you've seen how risky pussy time be thine. Yes. Well this is equal to getting the derivatives. Uh A. T. To the fourth over to bless State was to be over three and 2 kids plus B. Plus to see over +22 squared plus two times two continue called 0-1 which is over two plus A plus to be over three plus B. Plus to see over to plus C. And this is a favor to plus A with the room. The idea of which is yes. 56 oh plus uh Seventh shit. E. To see you then want us to be equal to zero. Let's speak now for example what one king goo one. Yeah uh refuse on earth knows what C. equal to zero and take be equal to ah negative five. Then it follows a. Is equal to uh positive seven. And so we get that that zero of tea is 70 squared minus five to how's your by Giannis? Because that shit looks like an Iranian Russians. Now we know that there's one more victory. No thesis. But it's so spruce To find it. You know that this vector f. one of T. Which are called A. T. Squared plus B. T. Plus C. Well it should just be port. Mhm. And what's your we want the inner product of once again F. Zero and H. Zero. So once again we want 56 A. Aliens get home washing plus seven six B. Plus. To see equal zero. We also want in a product of F. One. H. T. V. Zero To be in a project F. one, mate. Well sorry that's okay. This is they're rubens were records. It's the person that's what actually instead to find another vector. First four star general to H. And also nearly independent the F. Two or F. Zero. Well he's good. Let's take uh These are the zero and let's take mhm. A. to b. 12. And if I was that C equals -5. Have since equals negative five In F zero and equals zero and Uh one opposed F one and a 200. The independence FT is given by 12 to swear minus me minus five. As have you ever been, There are always like and therefore F zero and F. one. Yeah. From a basis for or some space. Help me.

Hello. And uh today we're gonna be solving a problem in linear algebra up. Excuse me. So let's get started here. So we've been given a factor space P. Of all the polling um meals a degree and over the reels. Um And it's also beginning to subsets of P. Ah They were stated to be the even and er palma meals and people. And those polling males are defined by these two qualities here. And as you can see these are definitions of even in odd functions. Um So one thing to point out here as you might see his little bracket with in the definition of even polynomial. Um And they can ignore this here. Just trying to be fancy if you right you write this this is true for any polynomial for all X. Products and the reels. It's it's true. Um So we don't really need to worry about that bracket. We know this is true right here. So don't worry about that. Um Those who have been given, we've been given the vector space The polynomial degree and we've given the two subsets. One of the even Paul Mills one of the Odd. Now we're asked to prove that even in the odd palma meals are sub spaces of people. So to do that we just have to look at with sub spaces and the definition. And then just prove that that holds I'm and uh subspace has to be closed under addition and scalar multiplication as pretty much the definition of subspace. It's just subset that is closed under addition and scalar multiplication. So that means that if you have any two elements in the subspace or is the kind of subspace, you can multiply them by a real number and you'll also get another element in the same subspace. Or you could add any two elements together. Um In subspace we'll get another element in subspace. So you're stuck. You can't get out you with addition and scale modification. You can't leave. That's basically what it's saying. Um So how do we prove that? Well we just 20 years that we've been given. It's always good with these problems to look back at your the definitions or what they've given you. Um And of course they use some some term like subspace. You could always look up, make sure you know exactly what that means. But we're going to use what they have given us. Let's just take two functions F. N. G. Um So let's say F and G. If you add them together, you get a judge, F N. G. Or even paul no meals. Um Now let's try to prove that that H. Of X is also an even polynomial. Um As you can see if we do this, then we have actually proved that that the even poll numbers are closed under vector edition. Because we added to even polynomial. And we got another one. Okay so let's try to prove that prove that we just use the definition they gave us and work from there. All right. So you see what I did there? I just took it's negative X. Redefined it as we did here. Um And then I redefined these terms using the fact that they're even paul no meals. You can write them also in like this. Um Because H. R. Because we have negative X. Is gonna equal Activex if it's even. And then we're right back where we started and we did 80 bucks. So um That's good to remember. We we didn't have to work too hard for this but we just used definitions. So it was a good thing to remember when you're doing these kind of problems. Now let's do scalar multiplication. We write this little more clearly as you can see. You want to make sure when you're proving something that you define everything pretty clearly to even be constant or define that. Right? So we just did basically the same thing we did before. Um Just redefine get back, get back to where we to find it before appeared. You can read it like that online knowing that Fx says even we can rewrite F. Negative except for X. And and right back where we started. So now we know that even palmas are closed under vector addition and scalar multiplication. Therefore even um even subset upon our meals is also a subspace of P. I am. All right. So let's just anything with the odds. I'm just start doing this because it's pretty much the same thing as before. Birds can apply the odd rule instead. All right. So I hope you so I just did there the same thing we did before. Did you redefine your, we started as what we we stated it was the ancient negative X. Is a combination in summation of F N. G. So we just put that back and and we use the rule for odd Paula. No meals were given before and then we just see the associative property. And um we factored out negative side basically. And yeah, we have shown that um that H of negative ax equals negative H. Of X. The do you sit there and scale of implication here might seem tedious, but it's yes we got to do when you have to prove something can be thorough. So this is definitely using the associative property. There you go. Mm. So we have shown that adding to odd Paula, no meals. We get another Ottawa polynomial. Because this equality is true. And we've all shown that scaling. Yeah polynomial. You will also get there are polynomial because this quality true. There we go. We've showing that both even and odd polynomial. Czar closed under vector addition and scalar multiplication. Therefore God and even Poland is our subspace is of pete. So now we have their second for our question and they are the complementary to subspace complementary. Well we're given well I've provided a definition of what that means. Treaty said basically be complementary. So one is that their intersection is just zero vector. And the other thing is that if you add the two set of spaces together, you will get entire vector space they came from. So another way to think about it is a linear combination. Further. Any element to be in the vector space can be found. There can be reached through a linear combination of now, then one to face the element of the other subspace through. A complimentary. Better way to say is if you add the two subspace together, you get the entire vector space. I think that's the best way to say it. So let's prove it now. So the first thing you want to do, we have taken intersection to even the odd functions and see what we get. Um Well this is actually kind of interesting. But if you want to have an even an odd function, just go and look at our definitions. We just need a function that satisfies both. Use at the same time, let's see if we can just write it out and see if that works. Like if we can think of anything. All right. So because it turns out this is what equality we get. We want to have all of them. If you want to have both those equalities satisfied and have to satisfy this equality here, the polynomial would. Um Well, if you think about it, the only poll numbers that can satisfy All three of these conditions is zero polynomial. Um if you multiply negative one for anything, you're not gonna get the same thing back there no longer to be equal that right away. Should give you a hint that zero is being the shelling answer. Um And that's that's another way you can look at it too. Sleeping it Dimitri frogs around the X equals Y axis. But but for the evens that's that's around the Y axis. So the only you really function that is symmetric around both. That's actually just zero function. So there we go. We've shown that The intersection of the even the Odd Palma Mills is in fact zero. And if you do do some research, he'll realise that zero is even and odd. It's the only even and odd function. I think that's what it could be. Even believe in our polynomial. We're gonna put paul and we're now because because in the vector space, asking the polynomial inherits that from the vector space. So 2nd thing must prove that every polynomial mp can be made up of an even an odd vulnerable. Okay. Um well to prove this, we just have noted polynomial means it's basically the summation of one or more mono meals, whatever I know meals there are constants times variables, two uh to non negative powers. So that's that's that's the definition of polynomial non negative. The key word here is also negative into german hands. So knowing this definition, we can just take a polynomial. I'm gonna just use an example. Just help explain myself a little better but take any polynomial. Um Every polynomial is going to have um Variable terms that you might think. Well what about plus 1? Well this trick here Plus one does have a variable time. Um At least you can write it that way if you want. Uh So now what do you think? What you might ask? Why is this important to know? We're in definitions. This doesn't this is math. Come on. Well, okay so logically speaking every term, every Manno meal and a polynomial must have either an even power for an odd power. Um And you can think about and yeah it's true. Every minimal and a polynomial must have even or an odd power. And what does this mean for us? So, well it means that we can take any polynomial And split it into two. You split it into other polynomial and they if you add them together they will get the original. So you can rewrite any polynomial as a sum of two is a sum of an even and odd polynomial. And does that sound familiar? It should because that is our definition here for any vector in V, w and Z and W W prime respectively. Out together you get the well, same thing. We're all pulling on meals. You can find an even and odd polynomial just by breaking the terms apart. Um If you have them together you will get the polynomial mystery for every polynomial. Just because of this definition. And real quick you might wonder like what about like X squared? There's only even term there. Well, trick question. There's never there's never real. There's a there's always a little trick going on some of these things. But anyway, so for ones like this, you could just say they have experts even and the zero factor is the odd. Um Then yeah, that would be your case where you still haven't uh even an odd function added together to get a polynomial. Uh M. P. I'm saying functions but I met even and odd Paula mill to try to be clear. All right. So, we've proven this. Remember? It didn't ask as they proved that was one point. Didn't say prove. You don't have to be as rigorous as what we did up here, but still keep it. You know, we want to make it logical still. Um But let's go to the last question. So these sub spaces in varying under the differential transformation. So, there are some words we might have to understand here. Uh So differential transformation was that that is a differentiate. Yeah, assuming everyone here knows what should be what differentiating is. Um Yeah, that's differentiating his team derivative. Um It's called the transformation here because as you can see, we took and polynomial where we took it as a technically a polynomial, but there's a polynomial and we took it and we got another polynomial back. So it is a linear transport. Okay, then I could define that, but that's a little out of scope of this. This problem. We already know just from what they told us that the differential transformation is the linear transformation uh because they're they're using it in this context with the vector space. Um But if you didn't want to go prove that, you can go look another video on what differential transformation or where the linear transformation is. You'll see what I mean by that this still in your transformation, but that doesn't really matter for this. We just know from the problem, but this is linear transformation. Um But we want to show that the subspace is in variant um under this transformation. And then you know what that means. So basically I've given the definition here. Um But you can summarize this by saying that if you take an element of a subspace and you apply the transformation, you will also get you'll get back an element of the subspace. So using that transformation you will not leave the subspace. So like we talked about earlier player was being closed under better addition and scalar multiplication. Um And for this problem we we got lucky. It's pretty easy because as you can see what I just did, I just did. We took an even polynomial and we use the differential transformation. And we actually got in Ottawa back. So we left the subspace, we left E. He was left we jumped out so to speak. So is not is not variant under the differential transformation about odds. And I guess they're not going to be either. It didn't pass three there. You see you took there we took a odd all no meal and we got back even. They're both are not in variant right? So you have any questions or any anything you wanted to clarify especially with differential transformation. What that is in terms of like what a linear transformation is. This is a differential transformation taking the derivative. But you don't know what linear transformation is. I'd encourage you just look at other videos on that on that topic. Um Anything else in here? But I hope you were able to see how these problems are able to be solved and hope we learn something. So thank you.

Okay. So in this exercise we have a transformation from the space of one of the great too Into these faithful in the Mills of Degrees one. The idea of these exercises that we need to find the transformation matrix mm relative to the following basis that are the usual or canonical basis for each space. Okay, so the transformation is as follows you need zero was a 1 -2. A one lost 3. 8 2. Okay, so in order to construct this transformation matrix, we need to map each of the elements the basis of the domain onto the basis and right into the basis of the image. So let's see what happens if we apply this to the basis elements one in this case you can observe that we're going to map this into one. Then we have the basis element X. And we're going to map these to one of the will t X. And also to -2. Okay, okay. You can observe that. They want you here and here here we have coefficient one and here we have coefficient minus do. And if we have access square using this transformation, what we're going to obtain is -3. Okay, so this is the correspondent uh vectors associated to the transformation of the basis elements in data may but we can represent of course these factors. So this will be the picture in this basis. Of course will be one zero. Like this interface is deprived. The first component corresponds to one and the second component is multiplied by The 2nd element in the basis B Will be one and -2 in the base price. And finally this element here will be the vector zero minus on the basis. Of course the price. So from this we can construct the transformation matrix and the transformation matrix between the to be prime will be the if column that we're picking the correspondent basic element from B. We're going to put the transformative factor in the basis of the prime. So here we should 10 here, one -2 and finally 0 -3. And there's the transformation Yeah, no, we need to verify for the second part, Dimitri satisfy the following condition. We pick any arbitrary vector in the two we're going to obtain is equal to take to transform this vector and this will be in the basis. Right? Okay. These two operations are thank So let's very fine. We have the parametric here. This matrix was 10, 1 -2 and 0 -3. What is a generic vector? Institute? Well, a denying that truth can be too equals two. The zero lost ft. one plus two. Thanks to square. So we need to multiply the matrix by the vector. zero, see one. And this will be should be opposed to this part. We're going to check this later. So first what happened here is that what we obtained is The zero was he wants and here minus two times 1 -3 times. This is in the basis of the price. Okay, so we can put this into the form of a picture in people and will be the the deploying on you zero plus, you want plus minus To find the one plus, declined to act. Now what happens is we need to verify what happened with this part. It corresponds to the left hand side. Rich. Now let's see what happened with the right hand side, on the right hand side. We need to just apply to transformation two the point on your ex or directory. And this transformation would have using the matrices mm Is equal to zero. Lost the one minus two times he wants blood. three times X. Just flow into. And you can verify that these two expressions are the same. So these equations


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