5

0 with 6(C3S5U4Q3) Find the coordinate vector [U]B of u7 2 3 respect to basis B = D1 = 02 03 1 0 1 3 56-1342 :10 0 -13 3 :104 :6...

Question

0 with 6(C3S5U4Q3) Find the coordinate vector [U]B of u7 2 3 respect to basis B = D1 = 02 03 1 0 1 3 56-1342 :10 0 -13 3 :104 :6

0 with 6 (C3S5U4Q3) Find the coordinate vector [U]B of u 7 2 3 respect to basis B = D1 = 02 03 1 0 1 3 5 6 -13 4 2 : 10 0 -13 3 : 10 4 : 6



Answers

In Exercises $3-6,$ the vector $\mathbf{x}$ is in a subspace $H$ with a basis $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\} .$ Find the $\mathcal{B}$ -coordinate vector of $\mathbf{x} .$
$$
\mathbf{b}_{1}=\left[\begin{array}{r}{-3} \\ {1} \\ {-4}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{7} \\ {5} \\ {-6}\end{array}\right], \mathbf{x}=\left[\begin{array}{r}{11} \\ {0} \\ {7}\end{array}\right]
$$

Hello there. For the following exercise. We need to grade the coordinate vector for uh vector B. But we have represented here as two minus 13 in our three. And we need to represent this in uh the discordant in vector relative to the basis as Okay. Two. That means that we need to find a linear combination of the vectors in the basis such that it represents this vector V here. So in equations that represents the following. So here I'm going to put the vector V. And this Should be equals to see one last two plus C. three. And the vectors of the basis. So C1 times the first element in the basis, s. C two times the second vector in the basis. S and plus C. Three times the third venture in the basis. S. This will imply that the vector V written in the basis relative to the basis. S will be the vector C one, C 2 and C three. So technically we need to find a solution for these coefficients C one, C two, and C three. And we have three equations that we have a linear system. So we can represent this as an extended metrics. And then use the well known methods, methods that wind that we know to solve this linear system. Okay, so here we have the matrix and the extended part will be the vector that we want to write in this basis as. So the vector too minus 13. Okay, at this point you can start to do some row operations but at this point it's really clear what's going to be the solution because we have a triangular shape. So we need to start to replace each value and we will obtain the solution. So in this case first we solve 43. That corresponds to this value here for the third column C three. There's a C two and C one Soc three in this case is equal to one. Then that implies that T2 equals 2 -2 and finally C1, it's equals two, 23. Okay. And well that implies that the vector v, the coordinate vector relative to the basis as corresponds to the vector 3 -2, 1. Now that now we need to repeat the procedure but now with the following basis and it's exactly the same. So first we put here the vector v 7 -8, 9. And we need to find a linear combination, see one, C two, C three of the The vectors in the basis, 5 -12 three. 123. And the vector Here -456. Again here we have a linear a linear system. And the solution is that C one, C two, C three. See one in this case as it goes to 162 is the cost two and c three is equal to zero. So after finding the solution for that linear system, we can grade the vector B according to victor. We relative to the basis as us one, two and zero. But this of course, is a factor in the basis as.

So first I write the given Vector 56 So and AIDS of the even or the basis elements and the one on zero one second line He's one three 10 said he's one You one, too. The last one. You, Sirio, if you want 23 and the victor, we're looking for his like C one. C two c three c four four Women's So to solve that, I write the gear maker called to see one bite bite Very silly man and so on. And then you multiply. She want all the entries so we'll have. Maybe if she wants zero c one C to three C two, then we see two here plus C three. Do you see three two c three zero Make if C four to C four cream two by four, Then here you have native C one plus C two plus C three last year. Physical to fight and the same for other entries. Be right system of equations that give us natives you want. Plus C two plus C three equals to five. See one plus three c. Two I see for is a close to six. Maybe see two plus C three plus two c four physical to seven. Finally see one plus two C three US three C four is equal to eight. So we have this system of a question that we need to solve the consulted by. Go see an elimination, for example, what we write these negative form one one and then zero foresee for so as I write. And one one juan zero find the same for the other equations. One, three zero, one native one Jill Sorry, What may be one Juan True one zero 23 Nothing. It's a quote 5678 And then we need to, uh, remove. Eliminate one side, the off the mattress, for example. Here we can ah, multiply the sick and roll by negative once they're off central, then we have removed this element will be zero a Similarly, for this one this one we need to find multiply This roar is sure to this run And then when we have all these zeros, you can easily get or final vector the cheese and a related to be used to call to see one. She's 64 divided by three 12 and this is the victor looking for

You get like before for this problem, they give us a maitresse e and I want to read it as a linear combination of other mattresses. And they also let us know that these matrices for my basis for the rector space of square matrices where to buy team jerseys in the real numbers. Okay, all that's nice. We know it's a vector space, but we need to convert this into the standard rectors were used to. Well, we know what matrix is completely described by its coefficients or entries and there are four of them, so we can translate them. As in our four matrix, we need to translate them consistently so the we can perform our algorithms on them ended up with an interpreted solution that makes sense. Well, my ordering is the first to row is the 1st 2 entries. The second row is the second to injuries. So they correspond like this 10 on two and hopefully you can figure out the last one. That's your Ah, negative. 123 I get. So now this is a question. Hopefully you could answer it right now we have one vector over here. You want to write as a linear combination of others. So we need to find the coefficients where if you add them together, you get this. So that means we have a linear system with four unknowns and we want to solve for what those unknowns could be to get this rector on the right. So let's see it. 13 negative 10 Headway solver. Linear combinations. We is augmented matrices and are Gaussian elimination methods and weaken Deviate from that as long as we interpreted back in terms of equations correctly, which is usually the faster thing to do on something like an exempt 567 Hey, I like it. This is the system of unknowns we have to solve. So let's start during our elimination. Well, it's the best thing I want to do. You? How about I keep row on the way it is for now. Why is that? Well, that leading term is nice. I don't add it to road to. So I end up with zero for one negative 1 11 Let's keep this road the same and let's add it to row for a swell. I think it so I have zero doubt in the first column makes up for the pivot. Let's see now I want to choose a We're here. Let's choose Row three. I like having that negative there. So we're gonna use road 3 to 0 out all of the other entries and columns. I'm just gonna add it to roll one. So I end up with 0 to 2, so I'm going to add it to row to after multiplying it by four. So I end up 057 39. Okay, the and I'm going to add it. Throw for I could just add it like that cause it'll cancel out nicely. I end up with 45 20. Okay, this is reducing. Very well meant seeing. Which is this time. I guess. All you for this time is there out everything else Negative ones. Your 404 I'm going, Teoh. Divided by two you and subtracted from Murrow one. So I end up with zero here on a negative 1/2 and two here. Not see. I'm gonna take road to and add Teoh Negative 5/4 times before so make their columns. Men. So under with 000 3/4 on 14 I'll do the same for the third rule. Oh, divide our robot four and subtracted from around three. I end up with zero negative. 10 3/4 0 that's nice. On to on the bottom row. 00 for five. 20. Okay, we can interpret it back into equations like before. Let's just eliminate on my time, okay? And I'm Euro two is the obvious tracing, but we'll end it with negative. 1000 34 Over. Three. We'll end up with 000 3/4. Poor thing. Zero negative. One 00 Negative. 12 and 00 40 Negative. 220 over three. Yeah, So each of these tells us exactly what our solutions our co efficient for this first maitresse Ian are Linear combination is negative. 34/3. The Coalition for the Second Matrix is, let's see 14 times four divided by three. So 14 times. Where is 56? I believe yeah, de six. Over. Our co efficient for 1/3 1 is the just 12 under coefficient for fourth main. Tracee will be let's see, divided by four. Well 20. The writer by four. That should be in 55 live. Negative. 55/3. Okay, Now, if you take thes co visions, you multiply them by these matrices, you add them together, and you should get this specter on the again because he could verify. That's correct. And this is our major see in ah component form with this basis.

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.


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