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Consider R1 ,R2 and R3 as a 1*t digit, 2nd Digit and 3rd Digit respectively of vours registration number and R= vours registration number For example:For registrati...

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Consider R1 ,R2 and R3 as a 1*t digit, 2nd Digit and 3rd Digit respectively of vours registration number and R= vours registration number For example:For registration number642; then use as:R= 642RI: =6,RZ. = 4R3. =2OR For registration number 34; then use as:R= 034:RI-0,R2-3,R3-4Question No. 1: Let, 1,a2 and a3 are the basis for R"[CLO(Marks 8+6+6)a1 =H a2 = l a3 = Find the following:Y =Orthogonal basis for R? and Express Linear combination of orthogona basis_ QR Factorization The minimum d

Consider R1 ,R2 and R3 as a 1*t digit, 2nd Digit and 3rd Digit respectively of vours registration number and R= vours registration number For example: For registration number 642; then use as: R= 642 RI: =6, RZ. = 4 R3. =2 OR For registration number 34; then use as: R= 034: RI-0, R2-3, R3-4 Question No. 1: Let, 1,a2 and a3 are the basis for R" [CLO (Marks 8+6+6) a1 = H a2 = l a3 = Find the following: Y = Orthogonal basis for R? and Express Linear combination of orthogona basis_ QR Factorization The minimum distance from Y to H The subspace H is formed by the linear combination orthogona basis. (or Mathematically H = WV1 v3h)



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Find a basis of the subspace $W$ of $\mathbf{R}^{4}$ orthogonal to $u_{1}=(1,-2,3,4)$ and $u_{2}=(3,-5,7,8)$

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 start with a subspace that is respond by these three vectors. We want B. two and b. three. Okay. And we need to find an Ortho normal basis for the space. So the first step is to, well the point is that we know a procedure that takes a set factors. Yeah two mm. And then it returns after playing the punishment procedure. It returns a set of orphan yeah sets the Northern Normal said. In terms also. So you start with a set of pictures and you obtain an Ortho normal set. But to apply this procedure we need to know that these vectors are linearly. Mhm. So before starting to calculate and or applying the garnishment procedure for this set of B one, B two and B. Three. What we need to do is let's remind if these vectors are Yeah. And to do that we need to check if one of them is reading as a combination of the other one. And you can observe that actually The Rector B three Is equal to be one. Yes. Beach. So you can observe that this just by inspection if you want a procedure to check this. So process or agree them to do that. What what I recommend to do is put in these factors in a matrix form. That means putting the vectors as Rolls of our matrix. That 012 -101 -113. Okay. And here then you try to reduce this matrix to the action form by applying the girls procedure and here you will observe that when you want to eliminate this. Mhm Put a zero here or here you will obtain at the end you only two pilots one and hear something like the A. B. Here. C. 00 Yeah that means that one of your rose will become full of zeros. And that means that you is enough to pick you. Only two of the vectors that you're considering this in this sense, If you obtain three pilots after applying the girls elimination Using three pilots, then your three factors are linear. But in this case what happened is you obtain something of this form or you can check based by Inspection that the effect to be three is a linear combination of the. Okay, so first you that your mind that if the set is linear independent and we observed that it is not. So it's enough to eliminate one of the factors that you have here This case I'm going to be three and we're going to use we one and two. So for this victory people for this space is enough. two. Great. That is the span of the factor we want and beat and they said expand the subspace up and they are linear independence. So now we can apply the garnishment procedure to obtain a basis. That is all for now. Okay, so let's remember that for the grant smith procedure we are going to obtain, We're going to pick this had to be one B two. And after applying the Greenwich you're gift in a set of factors Of the one Alpha 2. Both unitary vectors. To do that. We need to Fix one of the vectors here, say alpha one here. I'm used the Yeah. For unitary factor. So this vector here is not unitary. So we fixed one of the batteries in our set Either be one of you to to follow the same order. I'm going to be one. So alpha one. It's going to be B one. That means 012 Is the 1st step. So in or set. In our final set, we have already director of what Then the next step is the second vector offer to Such that this Alpha two is a phony one. And to do that we need to take the vector V two and transform Director of Alpha two Such that this vector is also an out for one. To do that we need to right offer to us the vector V two minus the projection of YouTube on all. For one When we do this we are obstructing the projection. We're eliminating the component of YouTube that is aligned with alpha one and the result will be and Better. That is Arthur 12 for one. So in this case This becomes the Alpha two is equal, two minus one, one minus. And here the projection just to remind you that the formula for the projection of a vector into another one is equal to you. The inner product of these two vectors times that picture to the one that we're projecting, two divided the norm of this square, enormous. So this projection particular case is equal to -250 times 012 And this part you can observe that corresponds to awful. So after subtracting these two pictures, We obtained that Alpha two is equal two minus one minus 2/5 And 1/15. Okay, the second vector orthogonal set. So what we have so far is a set of orthogonal factors Okay over one Equal to 0 1, two. Alfa two equal to -1 -2 and one over deep. But what we need is an Ortho normal set. So so far this is or a phone but we need say that should be normal. That means that the factors are unitary. And to do that we need just to normalize the vectors. That means that alpha one, you need to hurry Will be all for one divided the normal album. This is just One over the square root of 501 two. And for alpha tube we applied the same formula. So we normalize the vector by divider but it's not and we obtain one over The square root of 30 times minus five minus 21 These two vectors for or phone normals. Ah these vectors, so all for one, All for two, the span of these vectors, he is a subspace of you even more. We know that by theorem 6.3.1 Ortho normal set is linear. And then so that means that these two vectors are linearly independent. Therefore we call director. We will be Formed by Alpha one oh two. Is a basis is an is an orphan, success is an old on our mountain basis for done

To find the basis of the subspace of R4 spanned by these four vectors. We just put the four vectors into a matrix where each vectors a column of the matrix. And we find the column space of said Matrix. So to find the columns face, we take our matrix and we do Gaussian elimination. So our first row stays the same. And then we're gonna subtract the first row from the second, giving us a 00 two, negative three. And then I'm going to take our second row or a third road to be the same. And then for our fourth row, I'm going to take the fourth row minus the third row, which is going to give us a zero zero zero negative three. Well, at this point, I see that I can scale this negative three in the last row just to be a one. All right. And then I'm gonna continue my Gaussian elimination. I'm first gonna switch the second and the third row. And then I'm going to use the third road to zero out the third column. So I'm gonna just add the first and the third column together. I'm going to take the second row and subtract the third column. Just keep the 3rd column the same. And at this point, I can see in row echelon form that each of my variables are literally independent, which means all four columns are literally independent. So that means the column space is just the collection of my four original vectors or the set right here.

Okay. Now, in order to find a basis for the subspace of R. For that spanned by these four vectors, we want to get rid of any of vectors. Any of these vectors that are repetitive because they're dependent. So for example, we see that V4 is actually V one plus we can have before B three, V one plus a third of the three. So we can't have the four B three, M. V one all be in the basis because they're dependent. Now, let's see for V two, is there a way That we can write V2 as a combination of the other vectors? Let's see. We can have two times 3, one minus 2/3 B three. So that means that we can't have V 1, 2 and three all in the basis. Now can V to be expressed With just the 1? No. So that means that one example of a basis we can use is V one V two. We can also have V one, V 3 B. A basis. We just can't have Me Too NV three being the basis with V one Or V one, V 3 and V four together. So these are two possible bases.


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