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(15 pts) 6. Let(1,0,4), I = (1,-1,0) , IlI = (0,1,-1) Frovc that u, Uz, U, form hnsis for R" nud then lind the coorclinate vector of w = (1,2.3) rclative to th...

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(15 pts) 6. Let(1,0,4), I = (1,-1,0) , IlI = (0,1,-1) Frovc that u, Uz, U, form hnsis for R" nud then lind the coorclinate vector of w = (1,2.3) rclative to this basis_

(15 pts) 6. Let (1,0,4), I = (1,-1,0) , IlI = (0,1,-1) Frovc that u, Uz, U, form hnsis for R" nud then lind the coorclinate vector of w = (1,2.3) rclative to this basis_



Answers

13. Exercises $8.2 .30-32$ introduced the standard basis vectors
$$\begin{array}{l}{\mathbf{i}=[1,0,0], \mathbf{j}=[0,1,0], \text { and } \mathbf{k}=[0,0,1]} \\ {\text { (a) Show that } \mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{k} \cdot \mathbf{i}=0} \\ {\text { (b) Show that } \mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=\mathbf{1.}}\end{array}$$

Yeah. Okay. Um so suppose were given a basis of two vectors, which is uh first factors 1 -1. The second vector is 1, 1. And uh were also given a W, which is another factor that is equal to 10 And I want to find the coordinate factor of W. With respect to this basis. Okay, How we do that? Well, let us reformulate this to an easier to handle form. So I'm going to collate the basis factors into this one matrix over here, and I'm going to multiply it by some arbitrary uh you know, like uh column vector. Let's call that X. Y. And then that's going to end up using us back the W which is 10 Okay, so finding the coordinate vector of W. With respect to this basis. Okay. Is the same as finding this column vector X. Y. Under this matrix multiplication. This this problem that I've set up over here. Okay, so how do we solve for X and Y? Well, We just rearrange the terms. Okay, that's equal to 1. -1. 1. The inverse of that Gay Times. 1, 0. Okay. How do we find the inverse of a two by two matrix? Well, again, let us recall that if I have a two by two matrix, A, B C. D. And I want to find its inverse. Well, that's going to be equal to one over the determinant of this matrix, which is a d minus Bc. Okay, and then for the actual matrix I need to exchange the diagonals so it's D. A. And then the off diagonal stays where they are. But I have to put a negative sign in front so negative b negative. See so this is the formula for finding the inverse of a two by two matrix. Okay, so over here I'm going to apply this formula. The determinant of this is one minus minus one. So it's just too, so it's one over to over here and then I have to exchange the diagnose which is just 11 Uh And then over here I have to put a negative side in front of the off dying. And also I have negative 11 Okay. And then I multiplied by the same column vector 10 Well, what is this going to give us, this is going to give us the vector 1/2, 1/2. And therefore we are concluding that the coordinate vector of W with respect to this basis is going to be one half, one half. Okay, so let's clear the screen and work on the second example. So I suppose this time the first basis vector is one minus one, and the second basis factor is 11 And the w. Okay, that we're given is 01 Okay well how would we go about finding the coordinate vector of W. With respect to this basis. While we do the same thing, we're going to collate the basis factors into this one big matrix, multiply it to the column vector X. Y. And set it equal to W. And to finally quote in that vector of W. With respect to the spaces. We just have to solve for X. Y. Well how do we do this? Well we have X. Y. Is equal to 1 -1, 1, 1, Inverse Times 01. And how do we do this? Well again we take the determinant, so 1 -1. So it's 1/2. Okay. And then what is that? Um how do we figure out the matrix part? Well the diagnose have to be exchanged but then they're both ones so they just stay, you know they just stay the same as 11 and they have to put negative signs in front of the off diagnose. So I have minus 11 like this And then we're going to have a zero one. Okay? So uh if you were to multiply this thing out, uh this picks out the second column of this matrix, so this just becomes minus one half in one half. Okay. So therefore we are concluding that minus one half one have this column vector is going to be the coordinate factor of W with respect to this basis uh that I've shown here on the left.

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.

Um Okay, so suppose were given a set of two column factors, which is a basis for our two, And the two common factors are 2 -4 And uh 3, 8 respectively. So and were also given another vector W which is equal to 1, 1. In this case we want to find basically the coordinate factor of W with respect to this basis over here. So let us rephrase this problem into a more easier uh very easier to handle form. So this is actually equivalent to the problem where why were to collate these two basis factors? Okay, and then we were to multiply this matrix by some column factor A B. Then we would yield back the W. Okay, so when we write it in this form, then finding the coordinate vector of W with respect to the basis is basically just trying to find the vector A. B over here. So how do we solve for a B? Well, we just have to rearrange everything, so a B is equal to the inverse uh 234 negative 48 times 11 Right? So I'm just taking the inverse of this matrix and putting it to the other side. And how do we calculate the inverse of A two by two? Well, like just to recall the formula, so I suppose I have a two by two matrix A B C. D. And I want to take its inverse. Well, the formula for that is one over a D minus Bc, which is just the determinant of this matrix. Right? And then times I have to switch the positions of the diagonal elements. So it's D. A. And then for the off diagnosed they stay where they are but I have to put a negative sign in front them, so it's negative be negative C. Okay, so in this case the inverse of this is going to be well, one over the determinant, which is two times eight minus three times decade four, which is 28. And then I have to switch the diagonals, so that's eight and two. And then the off diagonal stay where they are, but I have to put a negative sign in front them in front of them, so it's negative three and four. Okay, and then it's times 1, 1. Okay, so if you work this out, this is going to be Uh 8 -3. So it's 5/28 and 42 which is 6 to 6 divided by 28 but then six divided by 28 you can cancel the factor of two, so you get three over 14. Okay, so uh I'm sorry I wrote this in a row form, but it really should be a column form, that that's the type of. So I'm going to put a transpose over here so that it's a column vector. Okay, So therefore uh the coordinate factor of W. With respect to this basis is going to be five or 28. 3/14. Okay, so let me clear the screen and work on the second example. So I suppose uh we have another basis. One negative one actually, sorry, tipple 11 mm 1 1. And the second vector is 02. Okay, and the W that were given is uh a. And be okay, so again, we want to find the cordon defector of W. With respect to this basis. Well, let's again collate all these basis factors into a matrix, and we're multiplying it to a column vector. A. B. Well, maybe doesn't work this time because W is given in a. B. So let's just say x. Y. Okay, it doesn't matter. Okay, so and then that would give us uh w which is A. B. So now we're in order to find the coordinate vector of W. With respect to this basis, we need to solve for X and Y. Okay, so uh x Y Is equal to uh 1012. In verse A. B. Well, what's the inverse of 1012? Well, that's going to be one times two minus zero. So that would be 1/2 times. Okay? Uh 210 minus one. Right? Just applying the formula in times A. And then time speed. Well, what is this equal to? Uh This is the first the top term is going to be two times a plus zero times B. But divided by two again. So you just get A. And then over here I have minus a plus B. To find it by two, so uh minus A plus B, divided by two. Okay, So this column vector where the first entry is A and the second entry is B minus a. Over to this factor is going to be our coordinate factor of W with respect to uh this basis over here. Okay.

That was. Mhm. Well, given a vector w. and R. four second W has components 1 -2 -1, 3. Just it's no reason to try and in part a were as defined in the cardinal basis for W. That's welcome back to Jersey in our four. We'll say that some parents X, Y, Z. T. If this is in W perp this implies the product of V. Or W. Perp society you Kirk? Which is W. The equal 30 and authorized. Mm. So we have X minus two, Y minus Z. Plus through T equals zero. We have a lot of freedom here. When Children basis vectors 3° of three. New Precise, let's take X. And Y. To be equal to zero and keep it equal to one. Then it follows that Z is equal to three. And so we get detective. Yes, you won. Yes. With coordinates. 00 31 Yeah. Against. Mhm. Just mm. Now we want to find an Ortho dogma basis for W. For we want to find another vector. Mhm. That's that. Um He is not only their parker. V. With W. Zero but also Being a product of v. With you won at the 4-0. So now we get the equation um When we had the four X -2 line, mine is easy. Uh Plus three two equals zero. And we also have the equation three Z plus two equals zero. You can simplify this system. I will Well not only have 2° of freedom, so I'll take uh michigan J issue jewish wow. We get these new systems three x Uh -6. Y mm. Yeah. His muscles Plus 10 t. equals zero from the Seinfeld Kramer. Yeah. Or if you want to keep this a little bit differently, X -2 Y. And they subtract nine, Z -10. Z. equals zero and three Z plus T equals zero. Like sixties. Mhm. So now ah it's like wow let's take TP equals 2, 3. Well then it follows an x equal to zero and it follows that Z. Is equal to negative one. Yeah. And therefore why is you put too Positive five. So we get Mhm. The other vector U. two with coordinates zero five negative 13 Mhm. This is orthogonal who you want. So texting we still have one more degree of freedom. Now a last system is that our vector V. Has to be orthogonal www. But also you won. Um You too. And so we have the above system X minus two, Y -10, z equals zero And three z plus T. But now we are all equal zero. And now we also have that five Y minus Z plus three T equals zero. It's a good take a class seriously, she said especially as a matter of fact once again I'll take two to be three. That's the only degree of freedom they have though. This tells us that Z. Is negative one. And therefore from our 3rd equation, why have you booked through? Mhm mm negative two. Yeah I'm and therefore from our first equations X. Is equal to chords toys. R. Us negative 14. Yes. And so we get a third bacteria. Use three. With the opponents negative 14 negative two. Yeah negative one and three. Well they don't matter. All right. It's right there in that song. That's so and therefore the vectors U. One U. Two and V. Three to form an orthogonal basis for soviet. Perk in the. Yeah. Well I yes in parties were asked to sign an Ortho normal basis for W. Proof. This is similar to part A. But now we want to normalize each of you one through three. Now the normal view one squared is the sum of the squares of the components. Mm So this is a nine plus one which is 10. The norm of U two squared is 25 plus one plus 9 Which is 35. And the normal view three squares, this is 14 squared plus four plus one plus nine, Which is equal to 210. Therefore, a set of Ortho normal basis is given by you once over 10, You two over route 35 And you three over route to 10. This forms an Ortho normal basis for the subspace. W perk. All right.


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