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6.1 Orthogonal Vectors and Bases: Problem 8 Previous Problem Problem List Next Problem~10 ~14 and10 point) Find the orthogonal projection of U onto the subspace V o...

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6.1 Orthogonal Vectors and Bases: Problem 8 Previous Problem Problem List Next Problem~10 ~14 and10 point) Find the orthogonal projection of U onto the subspace V of F 4 spanned by -21 ~14 ~8 (Note that these three vectors form an orthogonal set:)10 ~16projv(6)

6.1 Orthogonal Vectors and Bases: Problem 8 Previous Problem Problem List Next Problem ~10 ~14 and 10 point) Find the orthogonal projection of U onto the subspace V of F 4 spanned by -21 ~14 ~8 (Note that these three vectors form an orthogonal set:) 10 ~16 projv(6)



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Let $W$ be the subspace of $\mathbf{R}^{4}$ orthogonal to $u_{1}=(1,1,2,2)$ and $u_{2}=(0,1,2,-1) .$ Find (a) an orthogonal basis for $W$ (b) an orthonormal basis for $W$. (Compare with Problem $7.65 .$ )

Were given a subspace W. Which is the set of all polynomial of degree at most. two P. Two with an inner product inner product of F. And G is the integral from 0 to 1 of fft times. G M T E T. So oh yeah, interested in the projection of the function F f T equals T cute onto the space. W. As a hint, we're told you the orthogonal pollen or meals one To T -1 -1 and 60 squared minus 60 plus one. And then you can So we found that these were orthogonal in exercise 722. Yeah. Now because these are our dog kennel, we can just calculate the fourier coefficients to find a projection of that's under the space. Yeah, these are an orthogonal basis. Mm mm for W. Okay. Yeah. So before a coefficient C one is in a product of t cubed with one Over the inner product of one with itself and Yes. Right. Yeah. So this is the integral from 0 to 1 of t cubed D. T Over the integral from 0 to 1 of one gt. This is uh 1/4 Over one which is 1/4. Yeah the second fourier coefficients. C two is the inner product of t cubed with two t minus one Over the inner product of two T -1 with itself. This is the integral from 0 to 1 of well, t cubed times to t minus one DT over. The integral from 0-1 of to t minus one squared E T. This is equal to the integral from 0 to 1 of two T to the fourth minus t cube pt over the integral from 0 to 1 of four, T squared plus minus 40 plus one B two. Right. And this is equal to Take me up to derivatives and evaluating to 5th drugs. 14 over four thirds minus two plus one. This is to Fix. -14 is 8 20th -5, 20 assists 3 20th We were 4/3 -2 4/3 6/3 is negative. 2/3 plus one plus three thirds is positive. One third which is mm 9/20 right. You know, I like like in the field and finally for a coefficient C3 this is the inner product of she accused, I will be with uh then, I mean New Orleans, these Chicago, the 21st and 22nd, 60 squared -60 plus one. Yeah, pleased by the over the inner product of 60 squared minus 60 plus one with itself. What he said This is the integral from 0 to 1 of t cubed times 60 squared minus 60 plus one. Bt over the integral from 0 to 1 of 60 squared -60 plus one squared E. T. Yeah back I mean mhm. The this is the integral from 0 to 1 of uh 60 to the fifth minus 62. The fourth plus T cubed E. T. over. The integral from 0 to 1 of this is a little tougher. Doctor said 36 T. to the 4th. Mhm. 36 T. to the 4th -72 T Cubed. I do act plus 48 T squared minus 12 T. Mhm. No it's somebody that cares. Plus one. Well he's not taking anti derivative and evaluating we get one minus 6/5 plus 1/4. Yeah. Yeah. Mhm. 36 5th minus. Yeah. Yes. Thinking of buddy 18. Mhm. Yeah. Plus uh 16 minus six plus one. This is uh 40th minus 24/20 is negative. 4 20th Plus 5 20th says positive. 1 20 over. Uh six in Melbourne, Melbourne that 36/5 95th. Yeah, This is 1 5th he looked Which is 5 20th so or 1/4. Mhm. And therefore, oh okay, projection of our function F onto our space W This is going to be C1 times are first function one Plus C, two times their second function to T -1 Plus C. three times our third function 60 squared minus 60 plus one. So, plugging in this is uh 1/4 plus 9/20 Times to T -1 plus 1/4 times 60 squared minus 60 plus one. Mhm. This sympathize too. Three halves, t squared minus 3/5 T plus 1/20.

Hello there. So for this exercise we got 54 vectors in R. Five. So these vectors will generate and will span the super space dog of our five. And what we need to do is find the or the basis for the eternal compliment. So um here we got these four vectors and we know that they expand this space. So let's remember the definition for the eternal compliment. Will be any vector in this case in our five that satisfy the condition that X is orthogonal to be for all the B. In this space. The view, this definition of our tonality means that X. V. Is it goes to zero. So the point of this exercise is to first we need to find two pictures that generated space and then check if those are linear in the so to do that. What we need to consider here is a generic vector X. In the compliment. Then this vector we would we only need to to use the generators is enough to use the generators of this based off of you because well all all all the vectors will be greeted as a linear combinations of these four vectors here. So that means that this vector X satisfy the condition that X. B. One. Mhm. Me too X B three X. Before all this R equals to zero. So this translates to solve a system of linear equations. And in the matrix representation of the system, what we have is that the matrix of coefficients is formed by the counts of? Uh huh. The comes of this, this the rows of these matrix of coefficients are made by the vectors B one, B, two, B three. And so he would put the first vector we want on 45 69 three minus two, one for minus one. Mine is 10 minus one minus two minus one, 235 78 times the generic victor. X. So X will be a vector of the form. X one X two X three X four x five. So the set of so you can observe here that we have only four equations for a five dimensional system. That means that we're going to have a farming the ob solutions that will depend on one or two parameters in this case. Uh Yeah, we need to find the if that family depends of more than two parameters and that is a reminder. Free variables. So for that we need to reduce the system to the issue of reform. So reducing the system to the erosion informed, we obtain the following one zero 1 to 1 zero 12 and the next two will be just the risk. So after reducing the matrix the system to the Russian form, we obtain this system of equations. Well it is in your system equals to the zero vector. And here you can observe that we have three free variable. We have X three X four and X five as free variables. So that means that the space the view will be a spun of three vectors of one of two of the first three. So let's find these three generators of the space. So for that we need just to solve this system and the two X x three, X four. Next five or three variables, we can say that X three is equal to T X four physicals to our ex wife is equals two. S. So the general general solution for X, he's equals two. Team minus one minus 11 zero zero plus our minus two minus of one 010 one plus as minus one minus 200 What? So you can observe that the solution for the X. That lies on the compliment of W are generated by three vectors. That means that they are written as a linear combination of just these three vectors. That implies that we complement as it goes to this pan. Of these three factors minus one minute 00 minus 21010 and the last one is minus one minus 2001 So these three factors generate the yeah complement of the of you and we can call them as before of the one is going to be all for two and this is going to be out for three and even more of one of two of three are linearly independent. Therefore they form a basis. So the basis in this case for the compliment of B w will be all for one of two months of three.

Hello there. So for this exercise we got these three factors B one The two and B. three. And these three vectors is a super space in our four in this case. So we need to find a basis for the orthogonal complement of this based off of you. So how to build this whole? Let's recall the definition of the orthogonal complement is equal to the vectors X. On the space. In this case four such that X. Is an internal to the bacteria V. For all the the space of you. And this condition here is they have anticipated. X. V. Is equal to zero for all the. No. So the point is that we can obtain your final compliment. Just be considering the generators of this space. And what I mean is um considering B. One, B. Two and B. Three. So let's be a generic point X on the compliment. Then this implies that these vectors satisfied that x. v. one X. Movie too and six B. Three. All these are equals to zero. Because if this vector X. And any any vector X. In the compliment will be also known to the generators. Actually there is a generic vector of the form X. One. Uh huh. X. Two X three 64 So this becomes a system of linear equations that have matrix representation that corresponded just built the the vectors as columns of this matrix here. So I mean put in here one for 5 2 days be one 2130. This v. two And the 3 -1 three 21 This vector is multiplying X. One X. Two X. Three X. Mhm equals two 00 Okay so we got this system of linear equations and we can, what we need to find is the a solution for digital space orginal space of these interests here. So for that we're going to reduce that metrics to the Russian form the whole system. So we obtain the following we obtain 101 -2 over seven 011 for over seven And 000 zero times X one x two X three X four. And there is the cost to the 000. Okay. So from this you can observe that here we got a free variable that corresponds to explore. And here we got another free variable that response to X three. So what does mean is that the null space he generated by two vectors. Which is equivalent to say that I love you will be generated will be dispatched and of two vectors. Let's call off for one of two. Okay? Um so let's find those vectors. And those vectors will be the will be obtained by calculating the general solution for this system. So the general solution, If X three and X four are free bibles, that means that X three could be close to t the next four. We're going to give the value of our So the general solution X. Is equal to t -1 -110. Lust har And the vector to -40. So so you can observe here that the and any victor in the any vector X. In the compliment Is generated by two vectors by the linear combination of two vectors. These two vectors corresponds to offer one. Mine is a warning one, 10 and alpha two Which in this case is equal to two minus four, zero and seven. So these two vectors expand the whole space dog and even more they are linearly independent. So two vectors selectors that spanned the space. So we know that the space the arsenals compliment will be this pond here of the one of two and Alpha one Alpha 2 are linearly independent. Then the basis for the compliment will be The vectors all for 1/2.

Hello there. So for this exercise we have two sets in this case and call it. I only got one. And then you get to so we have two sets of two vectors each one. And we need to show that this pan of these vectors generate the same space started. So for that we need to show that actually if we pick for example some vector I'm going to call it exit one that lies on the span of only got one. Then this vector actually corresponds to a vector x. two defined in this pan of omega two. Get basically we need to show that these two vectors cars. So to do that let's see what happened for the first case. So this sub space. Uh huh. W. one will be The one generated by Omega one. So just to use some notation. So if I pick the vector here that means that X. Will be written as a linear combination of the vectors of Formula one. That means see candy. And here the vector one 01 and the vector 01 zero. So a generic generic form of a victory in W one will look like C. D. See for any constants C and D. On the reel. Okay, so keep in mind the structure of how the constants are located. So now let's speak a vector in W. Two. And here remember that these for any effect for any constant CNT. So in this case We have that X2 lies in the span of Omega two. That means that the vector will be written as a linear combination of these two vectors here. So X two will be equal two. Let's go alpha and beat the Corresponding coefficient 1, 1, 1 and one minus. Okay, so there is a generic form of a factor that lies in this space W. two. That is a span by this Said uh only got two. So if you some express this vector X two, that means that this will look like alpha. Let's be to oh fuck minus peter. An alpha was big. So you may say, well they are not the same. You you observe them and you say oh that they are not the same. They're not the same. They are not the same vectors. Therefore they don't correspond. But actually if you observe what you need to observe is the structure. So this vector here and these constants here and here are the same. So we can rename them. So let's go see prime and see prime. And this is different from the other two. So let's call the prime. That means that a picture next to is written as C. Prime The prime. Right? For every for any constant C. Andy this means that these two vectors here are the same. Therefore the Earth spanned by the same. They lie in the same subspace. Therefore we can say that this pan oh me! That one is equal to this pun of omega two. That is equals to a subspace off. That means that the space of you is generated by. Okay, this uh the space that generates these two vectors is the same. Okay, so that's the first part. And now we need to actually we're going to visualize that. So if this this to set some bigger one and then you get to generate the same space. That means that if you pick, so that's a part, you pick a vector. Any vector in this case, I'm going to choose the vector minus 31 and seven. And you calculate the projection of you Using the set Amiga one. That will be the same. The projection of you on the set Omega two, that will be equal to the projection you on the subspace of you. Okay, so basically this means that we just need to focus on this. The projection of you on the set Omega one is equals to the projection of your own this economic. So just to remind you what artist back these sets. So omega one is just set Pictures, one, and the vector 01 zero. And when you get to is this set 111 one minus two. What? So we need to calculate the projections. So I'm going to give some names to these vectors. So the first vector will be you one and the second will be here too. And for the case of Omega two, I'm going to call them gamma one and Gamma too. Right. So let's start by calculating the projection on Amiga one to the projection of you On the Middle one means the inner product of you with you want times you want by the discourse of the north. Yes. Being the product of you with you too. Times you too divided the square often love you. So this is equal 2 -3 plus zero plus seven over two times a vector you want that is one 01 plus zero one human. And the final result of this is the factory too. One two. And let's see what happened with the They said only got two. So the projection of you on America to in this case is the financing their product of you would got my one one divided by the square of the norm plus the in the product of you with mama to mama to hear divided by the square of the north. So this becomes minus three plus one. Was seven over three 111 plus minus three plus two plus seven over six Times a vector one minus 21 and the result in vector Is 212. You can see that these two vectors are the same, so that means that the projection of you in this pan of Omega one is equal to the projection of you on this pan of omega two. With this act, which is actually the projection of you on the subspace dollar. Because the subspace W can be generated by omega two or omega one.


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