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In Excrcises $1-6,$ the given set is a basis for a subspace $W .$ Use the Gram-Schmidt process to produce an orthogonal basis for $W$ . $$\left[\begin{array}{r}{3} \\ {-1} \\ {2} \\ {-1}\end{array}\right],\left[\begin{array}{r}{-5} \\ {9} \\ {-9} \\ {3}\end{array}\right]$$

$\left\{ \left[ \begin{array} { c } { 3 } \\ { - 1 } \\ { 2 } \\ { - 1 } \end{array} \right] , \left[ \begin{array} { c } { 4 } \\ { 6 } \\ { - 3 } \\ { 0 } \end{array} \right] \right\}$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 4

The Gram–Schmidt Process

Vectors

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bust these two vectors I'm calling ex one. Next to they form a basis for W. We're told using Gram Schmidt. Find the North O'Connell basis for W. So I'm just gonna go through and apply the process. So we want to go to a new basis V one V two, where V one V two is a basis for the same subspace w, right. We don't leave that subspace. We don't leave the plane spanned by these two vectors on dhe, we will envy to our orthogonal to each other. Okay, So Gram Schmidt tells us what? So v one is the same as X one. So I put this factor into my new basis as it is, V two is what I'm gonna take x two, and I'm going to shift it around in a deformity, right at another vector to it so that it becomes perpendicular to be one. So I'm gonna adds some multiple of the one, and that's given by X two. Started with the one divided by the magnitude square. I'm gonna drop these arrows because it gets really cumbersome. Okay, so, as always, I really highly recommend figuring out what these guys are first with these quantities are first, do not get too muddled up. So x two daughter would be one while the one was the same as X two x one, of course. So that gives you minus 15 minus nine minus 18 minus three. And that's minus 45. And then the magnitude of the one squared that's going to be three squared, plus one plus two squared, plus one, which is 15. Let's go over to the next page. So V two, which is given by this is X two minus minus 45. Divided by 15 times. Is he okay? So now we want to figure out what the coordinates of the two are, so we just simply substitute. So that's gonna be minus 59 minus 93 plus three lots off. Three minus one to you. That's one. Okay. And so that's gonna give you minus five plus nine. So that's going to be four. Okay, nine minus three is gonna be six minus nine plus six s very minus 33 minus three is gonna be zero. So the one we already know what that was B two. We know what that is. So, um three minus 12 minus one On this new vector that I've produced out of the ground shrimp procedure is a North agonal basis for W. How can we check that? Well, first of all, these two guys are still going to be in W. They couldn't have left W because all I've done to produce this vector here is take linear combinations of Vector that factors that were in W and w the vector space, right, The vector space. You know, if you have a bunch of actors in the vector space and you take a linear combinations of them, you stay in the vector space right on DDE. Never mind that. Let's check that they're orthogonal. So these two guys are sucking also three times for those 12 26 minus six, which is zero that's there dot product zero. So that's fine. It should be correct. And again, I don't have to normalize things because all I asked is to find a North organo basis for W

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