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(ek) is an orthonormal sequencc in an inncr product spacc X, and x € X, show that with y CR @kCk; where ak (T,Ck) is orthogonal to the subspace Yn span{â‚...

Question

(ek) is an orthonormal sequencc in an inncr product spacc X, and x € X, show that with y CR @kCk; where ak (T,Ck) is orthogonal to the subspace Yn span{€1, €2,en}

(ek) is an orthonormal sequencc in an inncr product spacc X, and x € X, show that with y CR @kCk; where ak (T,Ck) is orthogonal to the subspace Yn span{€1, €2, en}



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Let $V$ be an inner product space. Show that if w is orthogonal to both $\mathbf{u}_{1}$ and $\mathbf{u}_{2},$ then it is orthogonal to $k_{1} \mathbf{u}_{1}+k_{2} \mathbf{u}_{2}$ for all scalars $k_{1}$ and $k_{2}$. Interpret this result geometrically in the case where $V$ is $R^{3}$ with the Euclidean inner product.

In this problem. We have an Ortho agonal set of vectors, and we're told that these vectors are all orthogonal. And that's how we know that there are orthogonal, um, to make them Ortho normal. We have to also make each of the vectors normal. And so, um, so if we take any of the vectors v i and we divide it by the length of the I, well, that will have the same direction as V I, but it will be, um, normal. And so that's four. Oh, I from one tu que So now our normal set would just be Ortho normal set because they're already orthogonal. And to make them normal, we just have to do that divide by their length. Um, then it's just you won. You, too. Dot dot dot. You okay?

Okay, so let's suppose that be is orthogonal. So v is orthogonal to both, um, to both w one and w two. Okay, then what? We know. Well, since we're a thug and all, then the dot product the dotted with W one is equal to zero, and v dotted with W two is equal to zero. Okay, so I want to prove that v is orthogonal to um too. Well, K one w one plus k two w two for all scale er's we could say for all um scale er's k one and K two. Okay, so let's let k one and K two b any scale, er's. So let K one and K two be any Steelers. Okay. Well, then, um, we have that v Times K one w one plus k two w two is equal to the times K one w one plus v times. Katie W two. Okay, so we're using the result here that while u Times v plus w right is equal to u V plus you w right. We're basically using the distributive property here. Okay, So, um then well, from here, we get that this left hand side is equal So this v Times K one W one plus k two w two is equal to K one times the times w one plus que two times v times w two, which is equal to well, that part is zero. So this is equal to K one, um, time zero plus k two times zero, which well, is equal zero. Right, This is equal to zero. Right? So therefore, v is orthogonal. If we could say therefore, V is or diagonal, I'm gonna abbreviate orthogonal, um, to k one w one plus k two w two for all Steelers. So for all scale er's um k one and K two k one and K two and we are done. So what? All right, take care.

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

Yeah. Were given two vectors you wanted you to from our four. Um as a vector with components 1, 1- two. And you too is a vector with components 01? My dick two negative one. I think. My. Mhm. And we're told that W. is the subspace of R. four. Orthogonal to. You wanted you to in part they were asked to find an orthogonal basis for W. When you're eating. Find an orthogonal basis. We want to find somebody in our four. This has opponents X. Y. Z. T. So such that ah while the inner product of V. With you, one and the other product of V. With you to equal zero. This gives us the two equations. X plus Y. Plus two, Z plus two T. Equals zero. And why? Plus to the minus T. Equals zero. Broken joshi. We have 2° of freedom in this equation. So I'll let TV called a zero. And let's say that uh X is also equal to zero. Well it's not pretty helpful But said that Z is equal to negative one for convenience. And we have that Y. Is equal to positive too. And therefore X is equal to I was korean zero as well. And so we obtained Director You one with you Tony zero two negative 10 His restaurant manager. It's a series. Oh, frank. Find another basis. Specter. Well we know that this basis factor you too has to satisfy the inner product of you too. Yeah. You won. I'm sorry. Can he won here? That was a mistake. You must I'll call this w. one one of the inner product of W. two with you one, the entire product of W. Two with you two. And the inner product of W. Two with W. One all to the zero. This is the north optional basis. And so we have the same system as above. X plus Y. Yes plus two. Z Plus two. T equals zero. As well as Y plus two, Z minus T Equals zero. And finally we have the new equation two y -Z people zero. And so we only have one degree of freedom. Just yeah. Uh Let's say that's Mhm. We take Z equals two for our degree of freedom. Then it follows that Y is equal to one From our 2nd equation T. Is their quarry quote too. Remember that one minus well positive five. And therefore from the first equation X is equal to negative 15 posing. And so we get the second and last vector in the orthogonal set W. Two which has components negative 15 1 25 And we have that vectors W. one n. W. two form a orthogonal basis. Thanks for our subspace. W I want to have to swim because my son is we need to then in part B four for us to find an Ortho normal basis. It said for W. Well this is similar Part A. We just want to normalize all the doctors from Part A. So the norm of W. one, you swear this is the sum of the squares of the components. So four plus one, which is five In the norm of W two squared. This is 15 squared plus one Plus four plus 25, Which is equal to 255 mm hmm. And therefore W one is equal to one over root five times, vector zero to negative 10 Caroline, and W. Two is the vector one over the square root of 255 times. I want -15 125. She seems like my name is. And so this set of these two vectors, W one and W. Two forms in Ortho normal set or in or so normal basis for our subspace W. Chantelle buffet


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