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Question

Let $\mathbf{y}=\left[\begin{array}{l}{4} \ {8} \ {1}\end{array}ight], \quad \mathbf{u}_{1}=\left[\begin{array}{c}{2 / 3} \ {1 / 3} \ {2 / 3}\end{array}ight], \quad \mathbf{u}_{2}=\left[\begin{array}{r}{-2 / 3} \ {2 / 3} \ {1 / 3}\end{array}ight], \quad$ and $W=\operatorname{Span}\left\{\mathbf{u}_{1}, \mathbf{u}_{2}ight\}$
a. Let $U=\left[\begin{array}{cc}{\mathbf{u}_{1}} & {\mathbf{u}_{2}}\end{array}ight]$ . Compute $U^{T} U$ and $U U^{T}$
b. Compute projy $\mathbf{y}$ and $\left(U U^{T}ight) \mathbf{y} .$

Doruk Isik · Accepted Answer

in this problem were given six vectors. You want you to know you three at the one beat and beat and were testifying the basis for a TSH system H K and H plus. Now let&#x27;s write estimation. Reduce your for and that ISS 10 000100 to negative 100 it was You can see we have to give a Tele Vince one and one. So we have to pivot columns. It means that the person second probable or genetics will be the basis. So the basis for a church will be you want. And you too first and second hole. Let&#x27;s write the system. Can you reduce tropical? Or that is you come to 1000 0100 three negative to zero zero again. As you can see, he had to give it elements and it means that we had to give its columns. And those are the 1st 2 columns. And on the basis for a system kay, that will be the one and me too. Now I write, uh, h I wrote H plus day using exercise 34 in section 4.1 and in a reduced role. Econ form This system looks like 1000 0100 to negative. 100 1010 to negative 300 Ford, negative. 463 and zero. Now we haven&#x27;t won&#x27;t be ableto element here. First of all, we have a second favorite element here in second problem. And we have another pivot Telemann here in the orto. So our give its columns will be the 1st 1 the 2nd 1 and the portal. So it under basis for this system will be The first column is you one second billings YouTube and Fort Polo is the war.

Michael Jacobsen · Answer

In this example, we have a set of vectors, V one V two V three, which are defined to the left, and we&#x27;re told that these are linearly independent. Then for P one p two p three indicated here we&#x27;re going to show whether or not P one through P three is in the and fine set for s. And in the span of s to do that with a minimal amount of work, we can make the following matrix. Let&#x27;s consider the Matrix that it has calm Zvi one v to V three and then we augment with p one p two and p three. Then this matrix is going to be of the form 1030 for the first column, then to negative 104 corresponds to be to we have next negative 1 to 11 corresponds to be three. Now, when we augment with just p one that be five negative 353 we can roll Reduce this matrix here as indicated, and then determine whether or not P one is in the span of s Then from the linear combination, we could determine if it&#x27;s in the a fine set of s. But if we did it this way, we&#x27;d have to do two more different role reductions. So that&#x27;s why we&#x27;re continuing to augment with P to MP three. So with one reduction, we can make all determinations for all pet vectors p one through P three. So next augment with P two, which is negative. 9 10 9 13 Then the final vector P three is four to eight and five. Now we&#x27;re going to begin role reducing this matrix. And since this is a four by six matrix, it&#x27;s a good idea to use a calculator or computer algebra system for the row reduction here. After using a calculator, I obtained one zeros 00 for the first column. Then it&#x27;s 0100 Then we have 00 10 And next, after augmenting and roll reducing the next columns become to one negative 10 then too negative for 30 And finally, we have 0001 as the last column for the role reduced echelon form. Now for the vector P one, we can write the linear combination. See one V one plus c two v two plus C three v three where the skill er&#x27;s C one c two c three come from this calm which corresponds to P one. And likewise this is for P two and P three And then the coefficients are found to be these provided here or rather, these coefficients here. So we find that P one is equal to two C to be one plus the two minus v three and that when we look at these coefficients which read to than a one from here a negative one. From here the coefficients sums to a positive two altogether. This shows that P one is not in the fine set for us. However, we do have that P is a linear combination of V one v two, v three and so we do know that P one is in the span of s. Now let&#x27;s turn our attention to the second vector p two with the last example in mind. Let&#x27;s consider the coefficients provided here if we some these coefficients together will have five minus four, which is equal that one. And that tells us that p two is in the fine set for s. And it&#x27;s a linear combination using scale er&#x27;s too negative for three. And so we also know that P two is in the span of s as well. Last we turn our attention to P three. Where for p three were considering this column here. Now note. If we row reduced without P one or p two, then we would have had a pivot. And they&#x27;re right. Most column. And that tells us already that p three is not in the span of S. And this is because if we form a vector equation, p three equals C one V one plus C two v two pussy three v three. This last pivot in the far right column tells us the vector equation is inconsistent. Next, it&#x27;s important to note that if P three is not in a span of s that immediately and directly implies that P three is not in the fine set four s either. So this concludes our Anil Alice iss of P 13 p three using the set s

Wen Zheng · Answer

The problem is, find half to Crumb. Where after days, the culture you tee times weighty. You, too, is they want you next to wan you to prom. That&#x27;s they tow three zero or on when she is the Goto he squire his kill first of a computer with peace Prom. This&#x27;s the culture one Duty three squat. So we have weighty with two basic or two chew or eight on the way to prom. That&#x27;s equal to one war. Well, then, after prom, if they caught you, you prompt he I&#x27;m with you. Us. You&#x27;re too not adopt eighteen just so we have to prom. It&#x27;s equal to three four not productive. Two for eight. Plus Want to connect one Not productive one for twelve. This is the conto six plus thirty two. Last one plus eight minus twelve. This place they call Tio thirty five

Rakesh Kumar Sharma · Answer

That&#x27;s one of the questions. So let&#x27;s start with this part that is marketing. And they dio did you find out using my three Gylfie 18 Coastal minus indication where to go GSK taking I didn t Any of this is actually once all this is be this in addition or instead of you g a few Do you know that you drink usually? Yeah, just by changing the valuable sentry. So we content over their you know, De or yet? Yes, we can write on my list 3 to 0 g FX, the FX and they feel there is a studio for part week are seeing the question is Indication minus 3 to 0 images. Do you think this video geode gx into the extra picked on the radio? This We already have this this they are decent organization. My escape to Zito gs you don&#x27;t be. We&#x27;re getting delusion minus 3 to 0 g o. Next is your chance of 86 using kill. So the video this year already have That is were there before. So this is this push

Emma Fink · Answer

to start off with. The first step in this problem is to organize the terms by decreasing exponential power. So you start off with 4.2 t cube post 5.3 t squared, minus 1.8 t squared minus 6.14 minus 9.1 points, 7.3. And you could do this because it&#x27;s just adding to polynomial so the parentheses don&#x27;t really matter. Um, you combine like terms like 5.3 t squared, minus 1.8 e squared becomes 3.5 t squared and negative 9.147 point three becomes negative 1.8. And from there you can supply simplify further. So your final answer is 4.2 t cubed plus 3.5 t squared, minus 6.40 minus 1.8.

Problem

Let $\mathbf{y}=\left[\begin{array}{l}{7} \\ {9}\…

Problem 17 Easy Difficulty

Answer

$\hat{y}=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$
$U U^{T} y=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

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$\hat{y}=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$$U U^{T} y=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

$\hat{y}=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$
$U U^{T} y=\left[\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right]$

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications