in-excrcises-1-6-the-given-set-is-a-basis-for-a-subspace-w-use-the-gram-schmidt-process-to-produce-a

Question

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 $W$ . 
$$
\left[\begin{array}{r}{3} \ {0} \ {-1}\end{array}ight],\left[\begin{array}{r}{8} \ {5} \ {-6}\end{array}ight]
$$

Ahmad Reda · Accepted Answer

in problem one. We have these two victors as a basis for US office space. W We want to use the Grammy. Schmid poses to produce on our stagnant basis for W. The first system is to consider one of these two victors as V one. The only equals 30 minus one. The second step is to get the two. Based on the projection off the second victor, which is here on the first victim, it equals the second victor, minus the projection off the second victim on the first of it way will project on the first victor. Then we have here the one 30 minus one in the nominator. Here the multiply here by a factor. This factor in the nominator we have the dot product between x two and V one. X two is the second victim. It five minus six and we won 30 minus warm. And in the denominator, we have the dot product off the one and the one. Then in the second term, here we have all victories is Vermont except off X two here, which is the second victor. We want to project on on the first victim. Let&#x27;s calculate it equals 85 minus six minus. We have the dog product is eight multiplied by three plus five, multiplied by zero minus six, multiplied by minus one gives 30. And in the dominator, we have three multiplied by three, which is mine plus zero plus one equals 10 multiplied boy 30 minus one. Then it equals it minus three, which is minus one five minus zero, which is five. And finally we have minus six plus three, which is minus city. And this is V two, then the two victors. First Victor is 30 minus one. And the second victor is minus 15 minus three. This set off these two victors is an orthogonal basis, or signal basis for W, which is the final answer off our property.

Sean Thrasher · Answer

bust these two vectors I&#x27;m calling ex one. Next to they form a basis for W. We&#x27;re told using Gram Schmidt. Find the North O&#x27;Connell basis for W. So I&#x27;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&#x27;t leave that subspace. We don&#x27;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&#x27;m gonna take x two, and I&#x27;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&#x27;m gonna adds some multiple of the one, and that&#x27;s given by X two. Started with the one divided by the magnitude square. I&#x27;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&#x27;s minus 45. And then the magnitude of the one squared that&#x27;s going to be three squared, plus one plus two squared, plus one, which is 15. Let&#x27;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&#x27;s gonna be minus 59 minus 93 plus three lots off. Three minus one to you. That&#x27;s one. Okay. And so that&#x27;s gonna give you minus five plus nine. So that&#x27;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&#x27;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&#x27;t have left W because all I&#x27;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&#x27;s check that they&#x27;re orthogonal. So these two guys are sucking also three times for those 12 26 minus six, which is zero that&#x27;s there dot product zero. So that&#x27;s fine. It should be correct. And again, I don&#x27;t have to normalize things because all I asked is to find a North organo basis for W

Sean Thrasher · Answer

Okay, So when this problem were given two vectors and I&#x27;m gonna call them extra next to we&#x27;re told that they form a basis for a vector Subspace w So they&#x27;re linearly independent, and they span w and were asked to find the North organo basis for W. And we&#x27;re gonna do that using the Gram Schmidt procedure. So how does this work? This is the way I like to think of it. So here&#x27;s X one x to let&#x27;s say rights and this plane of the white Board is the vector space W Okay, so the whole point of doing this is to find a new set of basis V one V two of W such that V one and V two are perpendicular to each other, right? That&#x27;s what it means for them to be orthogonal. So first off, I can call X one the same as V one. Right that can. That can survive in the new basis. Rights unchanged. That&#x27;s fine. What the problem is is that this x two might not be perpendicular to the X one, so I&#x27;m gonna make it perpendicular. So what I want is for my V two toe look like this. This is what I&#x27;m aiming for my V two to be. Hey, so this is everyone. And this is me too. Okay, The question is, how do I get to be too? It&#x27;s clear that I can get to it by taking X two and adding to it a suitable multiple of V one. Okay, I should be able to do that anyway because V two I wanted to sit in the span of X one, the next to so I can&#x27;t do that. Well, imagine I take the sex too, and I project it right. I projected onto V one and I get this vector here if I subtract this vector from X one. Sorry if I subtract this factor right here. The projected vector from X to so x two minus this projected vector. I&#x27;m gonna get me too. Hey, so let me call this projected vector approach. Right. Okay, so v two is going to be x two minus, So some minus this vector. And what is this vector right here? Well, first was gonna be in the direction of the one, so I put the unit vector for V one, right? And what is the What is the length of this going to be? Let the angle between this Beefeater. So it&#x27;s gonna be definitely whatever the magnitude, the length of x two is times the coast sign of Peter. Right X to coast Scientific gives you this. Okay, Now I want to bring in V one original like the original V one, as opposed to the univ. Ector. So clearly this is the same things. V one divided by the magnitude. Okay, Now, finally, I&#x27;m gonna multiply top and bottom by the magnitude of the one, and that&#x27;s gonna give me this and this, my friends, you can recognize this simply being the dot product of you wanna next to, which is very simple to calculate giving the components of the vectors. So v two will just be given by x two minus x two dot v one divided by square. The money to give you one times Be one. So that&#x27;s the intuition that I have at least behind the Gram Schmidt procedure, right? It gets more sophisticated than this, right? But I just wanted to show that to you because it might be helpful. Just a sanity check. How can we be sure that V two will sit in w right, Because I won&#x27;t be one of the two to be, ah, basis for W, right? So at the very least, V two should sit in this vector space. W The answer is because a subspace, like I said, is a vector space right now ve one is in w right, because the same things x one and it forms the basis for W and X two is in w as well, because it&#x27;s a basis for w on all I&#x27;m doing all the Gram Schmidt procedure does is you take linear combinations of the vector that you of the vectors that you already have and taking linear combinations of a vector in some vector space will make you stay inside that back your space. Hey, if I have a bunch of actors in some vector space and I take linear combinations of them, I&#x27;m gonna stay in the back to space. That&#x27;s the crucial property of a vector space. Okay, so now let&#x27;s solve the problem. So this is going to be pretty straightforward, right? So I know what my V one already is. Okay, My V one is zero for two. All I&#x27;ve got to do is figure out V two and I&#x27;ve written the formula more neatly here. So I would do it like this. I&#x27;d figure out what this guy is. First on. What? This guy&#x27;s first? Okay, so separately. X to doubt, it would be one. What is that? Well, that&#x27;s 24 minus 14. That gives you 10. And then the square of the magnitude of the one that&#x27;s four squared. Plus two squared. That&#x27;s 20. Okay. And so this quantity, right, there&#x27;s gonna be 10 over 20. So x two minus 10 divided by 20 comes to be one. So now let&#x27;s actually plug in for the actual vectors. It was 1/2 and then view one was zero for two, and that should give us It&#x27;s a 56 minus two for and then minus eight. Okay, so my claim is that on orthogonal basis for W is given by 042 comma 54 minus states. Okay, now, how do you know that there are so Colonel just is a secondary check. Let&#x27;s stop these two. What do you get? You get 16 minus 16 in that zero. So they&#x27;re also Colonel on dhe. We&#x27;re done

Sean Thrasher · Answer

Okay, So we&#x27;re told that these two vectors which I&#x27;m labeling x one the next two to be consistent with the book forma basis for the vector Subspace W, which sits in our three and what were asked to do is find the North orginal basis for W. And when I solved problem number two, I went through in detail of how at least I like to think about the Grand Shrimp procedure, but now I&#x27;m just going to directly apply it. So we want to go to a new basis V one V two that is a North organo basis and is the basis for the same vector space. W we don&#x27;t leave the vector space that you started out with the vector subspace that we started out with. Okay, so V one is the same as X one. It&#x27;s finding V two that&#x27;s going to take the computation so the formula is take x two and subtract X to Doc V one, divided by the magnitude square. Everyone times the vector V one. Okay, so to make sure we don&#x27;t get into too much of a mess with the computation, let&#x27;s figure out what this is and what this is separately. So x two don&#x27;t be warned. What is that? So that&#x27;s three times minus three X minus nine. Plus some minus four times 14. That&#x27;s, um, minus 56 and then minus 35 rights. Okay. No. What&#x27;s this going to give us? We compete this. So that&#x27;s 35 plus nine. That&#x27;s 44. Okay, that&#x27;s what gives us minus 100. All right, now, finally, what is the magnitude squared of the one? Well, the one is the same as X, remember? So it&#x27;s nine plus 16 plus 25. So that gives us 50. All righty. So this quantity right here that I&#x27;m multiplying be won by that&#x27;s just minus minus 100. Divide by 50. So it&#x27;s too. Okay, So let&#x27;s figure out what V two is. It&#x27;s a V two is equal to let me actually plug in the vectors themselves. I apologize. This might look like a nine. I should be really careful. Okay. 14 in the 97 minus was plus two lakhs of three minus four five. Okay, so that gives us three 14 minus eight. That is six minus two plus 10 minus seven plus 10. That&#x27;s three Hey. So my are Fogell basis is three minus 45 That&#x27;s the one 363 Check that there are signals or three. So take their dot product nine minus 24 plus 15. That gives you zero. Indeed. Okay. And just as a side note, if we were asked to find an ortho normal basis of W instead, then we would have to normalize these vectors, right? So divide by their magnitude. But please note that you want to do that. Really? At the very end. Otherwise the computation will get incredibly messing. So that&#x27;s that problem done.

Jimmy Yao · Answer

So here we&#x27;re going to use Crispian Teoh, get the or second basis for the subspace. So there&#x27;s a we&#x27;d be one here will be one too sweet and b to here will be six months ST zero. So you wanted to do their grins? Me first the vector say if we tap into one, it&#x27;s Nico to be one. It is one too sweet and w two will be V to minors. The one you know, productive crispy too in the norm square of the one Be one here. So here, we&#x27;re going to compute. Do you keeping your protects V one than to be to? So this is one time six class two time minus three. So it is there. Oh, so so here U S c. This will be good to meet you and Nico to six minus 30 So here the orthogonal basis is at she&#x27;s They send this because you know we&#x27;re starting. The one is particular to be to so conclusion the also gonna basis will be one too sweet in there six minutes to eat their own

Jimmy Yao · Answer

Yeah. Here we can see the X one is equal to 201 and x 20 to minus 31 And once ex tree is equal to one minus 38 So I would be one. It&#x27;s ICO two. X one is equal to 201 and I love you too. Can speak 02 x two miners. The units projects between the X two and to be one into the super between B one and B one and B one. So if we compute the X two and the one is to go to two time minus three classes they wrote one house one time one that this will give you minus five into the norm square of the one here It&#x27;s the to casts one squared. It&#x27;s wealthy flight. So you&#x27;re already to this. Nico two x two minus 31 in one in minus minus by five. That would be one is 201 So here we have two minus one into one ginger to then we proceed going to compute with street. So not safe is the formula for history. It&#x27;s extreme miners, miners, the universe extreme gonna be the one the one. Everyone could be one miners the union party two x three and two The productive be teary envy too. Gonna beat you here. So first week. So we&#x27;re going to compute the mystery and the X one the the nuclear VC and X one will be once hung too. Cost zero because eight time one. So this will give you 10 and we already knows that the normal will be won by though into the begin Become your unit project between the extreme there be too. This will give you one time. Minus one plus minus three times one plus eight ton to here is four and here is 16 is well, give you child and we compute the norm to enemy too. I&#x27;ll give you one square minus one square, one square, two years square. They&#x27;ll give your six the pretty good So we have a history is because your ex tree it&#x27;s one minus three aids then miners 10 over by V one V one is 201 In my news travel over six would be to be too, though the two iss minus 11 into to be to here. So here we have one minus. We AIDS in minus 2201 minus two minus 112 So here we have one minus four plus two. We&#x27;ll be one minus two minus one in the ministry. Zero and minus to remind us. Fine in AIDS minus 26 Minus four to Texas. All right, so let&#x27;s some right. So order or someone bases first beat If one there will be 201 and then will be to the to the miners. 112 Then we&#x27;ll be sweet. It&#x27;s minus one minus buy into.

Problem

In Excrcises $1-6,$ the given set is a basis for …

Problem 1 Easy Difficulty

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 $W$ .
$$
\left[\begin{array}{r}{3} \\ {0} \\ {-1}\end{array}\right],\left[\begin{array}{r}{8} \\ {5} \\ {-6}\end{array}\right]
$$

Answer

$\left\{ \left[ \begin{array} { c } { 3 } \\ { 0 } \\ { - 1 } \end{array} \right] + \left[ \begin{array} { c } { - 1 } \\ { 5 } \\ { - 3 } \end{array} \right] \right\}$

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

Linear Algebra and Its Applications

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$\left\{ \left[ \begin{array} { c } { 3 } \\ { 0 } \\ { - 1 } \end{array} \right] + \left[ \begin{array} { c } { - 1 } \\ { 5 } \\ { - 3 } \end{array} \right] \right\}$

Linear Algebra and Its Applications

Lily A.

Caleb E.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Caleb E.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications