in-exercises-3-6-verify-that-leftmathbfu_1-mathbfu_2right-is-an-orthogonal-set-and-then-find-the-ort

Question

In Exercises $3-6,$ verify that $\left\{\mathbf{u}_{1}, \mathbf{u}_{2}ight\}$ is an orthogonal set, and then find the orthogonal projection of $\mathbf{y}$ onto $\operatorname{Span}\left\{\mathbf{u}_{1}, \mathbf{u}_{2}ight\}$
$$
\mathbf{y}=\left[\begin{array}{r}{-1} \ {4} \ {3}\end{array}ight], \mathbf{u}_{1}=\left[\begin{array}{l}{1} \ {1} \ {0}\end{array}ight], \mathbf{u}_{2}=\left[\begin{array}{r}{-1} \ {1} \ {0}\end{array}ight]
$$

Ashley Boni · Accepted Answer

So first we need to check that you wanted me to or thought Gotham by taking dot product So you won. Uh, not You too. So we get zero minus one plus one and that zero So those are welcoming each other. So now we need to find the doctor. Call it white hat. Um, what is the orthogonal projection of? Why? Onto the span of you wanted me to. So we find this by projecting Why onto your bomb. So you do that by taking dot product of the two over dot product of you one with itself and not apply. Buy you one and add to that the same thing. But projecting onto you too. Eso Let&#x27;s compute all the dot products off to the side. Why dot You won gives us minus 24 minus four. Close one. So negative 28 plus warm negative 27. You won dot product with itself a 16 close one plus one. So we&#x27;ll get 18. Why dot product with you too get zero plus four plus one. So five and finally you to doubt product with itself. Zero plus one plus one. So we get to so our first coefficient is negative. 27 or her 18 which we can simplify two minus three halfs. And 2nd 1 you get five over, Chip. So now we just want to multiply these coefficients through to the vectors s o negative. Three half&#x27;s times you won give us 12 over to just six. And then they gave 3/2 times negative. One three halfs. Negative. 3/2 times, one minus that halves and then multiply it by five laps. And to you, too, you get a zero moon five halfs, five halfs. Adding these together 606 three halves and 5/2 says eight halves before and 5/8 minus three. Half says two halfs more s. So that is our director weren&#x27;t interested in

Ashley Boni · Answer

first, let&#x27;s make sure you want to need to our or thought so we&#x27;re just gonna take without product. So we get lying is 12 plus 12 plus zero which in fact gives the zero. So these are thought of each other. So now we want to find the orthogonal projection off. Why? On to the span of wanting you to s? So we have a formula for this. So on vector, Why hat? Um it&#x27;s equal to the projection do you want? Which is why one you wonder if one time they want and the projection of why onto you too. So we take the dot product with you too. You too dot product with itself times you too. So let&#x27;s go ahead and find these coefficients off to the side. So why I got you. One is equal to 18 plus 12 plus zero. So we get 30. You won dot product with itself. You&#x27;re square all the components and I&#x27;m together. So we have nine put 16 which gives us 25 now for the 2nd 1 we have. Why dot You too negative. Four times six is minus 24. Three times there is mine. Zero so negative. 24 plus nine gives us negative 15. Lastly, YouTube out product with itself, you get 16 plus nine, which is 25. So now I can fill in these coefficients. Ah, so this 1st 1 is a 30 over 25 and the second is negative. 15 over 25 so we can simplify a little bit, get six months and minus 35th. So now let&#x27;s go ahead and play. He&#x27;s out 6/5 times. You won so 6/5 times three is 18 over five, 65 times four is 24 or five and then we get a zero. No, we have negative 3/5 times. You too. So negative. 3 50 times negative for is 12 over five negative, 3/5 times Theories negative. Nine over five and then a zero. So adding is together 18 plus 12 gives us 30 divided by +56 24 minus nine is 15 divided by five is three and 0.0 s. Our final answer is the Vector 630

Ashley Boni · Answer

so wanna rate the vector V as a combination of a vector in the span of you one and one in the span of you&#x27;re 23 and four. Uh, so first we want to find the one in the span of you won. And we can do this by, uh, using the formula for the orthogonal projection of a vector onto you want. So let&#x27;s call this vector be hat and it&#x27;s given by we thought you want over a year warm. Got you one, um, all supplied by defector you want? So do some scratch work be dot You won. So we have four times one plus five times two minus 3 10 1 plus three times one. So we get four plus 10 and minus three and 30. Cancel. So we get 14. And for the denominator, uh, you want a product with itself, We get one times one plus two times two close one. That&#x27;s one close one times one. So we get four plus three, which gives us seven. So this quality on the outside here is 14 over seven More too. So we have two multiplied by one sort is gonna multiply each component by two, Giving us 24 22 Um, and since all the use for another thought, little basis for horror, for we can just take B minus free hat, let&#x27;s see what we get. And we know that, um, they will be in the span of the other directors in you. So v minus. We had ISS four minus two by minus four. Negative three minus 23 months is too. So we get to one, maybe 51 So that&#x27;s the vector we need to add to be had to get back to be so. Therefore we can write be as a sum of 24 to 2 plus 21 negative 51

Foster Wisusik · Answer

Okay, This question wants us to show that this is an orthogonal basis for our three. And then we&#x27;ll use that to decompose a vector into a linear combination of these three. So we&#x27;ll just test each of these dot products. So we want to dot You won in you too. U one and u three. And you too. And you three. And if they&#x27;re all zero, we have a basis. So you one daughter with you two would be negative. One plus zero plus one, which is zero you one daughter with you three b two plus zero minus two. Should be zero. Then you two dotted with you three would be negative for plus four minus two, which would again be zero. So these are a basis. So next we want to write the vector X, which is equal to eight negative for negative three as a some of these unit vectors. So it&#x27;s, um, constant times you want, plus another constant times. You too. Plus another constant times. You three. So writing this out as a system, we would get Alfa Times 101 plus beta times negative one for one plus gamma times 21 negative too is equal to eight negative for negative three or as a system, we get Alfa minus beta plus two. Gamma is equal to eight zero Alfa plus for beta plus gamma is equal to negative four. Or for our third equation, Alfa plus beta minus two. Gamma is equal to negative three. And now we can solve this using a matrix So we could right. This is 101 negative one for one 21 negative, too times Alfa Beta Gamma is equal to eight. Negative four if three. And the solution to this three by three system gives us Alfa beta and gamma equal to five halves. Negative three halves and to respectively. And now that we have these thes air linear combination factors so we can write that X is equal to five halves you one minus three halves you too, plus to you three. And that&#x27;s our answer

Foster Wisusik · Answer

Okay. First part of this question wants us to show that these vectors are an orthogonal set and a basis for our three. So if they&#x27;re in orthogonal set, that means that they&#x27;re dot products between each other will always be zero. So we&#x27;ll test u one u two, u one and u three and you too. And you three So you won dotted with you to that would be six minus six plus zero, giving us zero. You one daughter with you three would be three minus three plus zero, giving us zero. And then you two dotted with you three would be two plus two minus four, giving us again. Zero. So since we have no orthogonal set, we know this does span are three. So now we&#x27;re looking to solve an equation of the Forum. X is equal to a constant times. You won, plus another constant times. You too, plus another constant times. You three. And in our case, filling in for everything. X would be five negative 31 you want would be three negative 30 You too would be 2 to 1, and gamma would be 114 and we can actually rewrite this nicely as a matrix. So weaken right? 321 negative. 321 014 times the unknowns Alfa Beta and Gamma would give us five negative 31 And since we know that these vectors air spanning, set in this matrix we know it has an inverse. So we could just use a calculator to get that. And we know that Alfa Beta Gamma would just be equal Thio this matrix once we take the inverse of it. So you take the inverse of the matrix and multiply that bye five negative 31 And doing this calculation tells us that Alfa Beta and Gamma are equal to four thirds, one third and one third, respectively. So now that we have our alpha, beta and gamma, we can just plug in. So X is four thirds you one plus one third you too. Plus one third you three. And that&#x27;s our answer.

Problem

In Exercises $3-6,$ verify that $\left\{\mathbf{u…

Problem 3 Easy Difficulty

Answer

$\left[\begin{array}{c}{-1} \\ {4} \\ {0}\end{array}\right]$

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

Linear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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$\left[\begin{array}{c}{-1} \\ {4} \\ {0}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Catherine R.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Caleb E.

Samuel H.

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

Linear Algebra and Its Applications