exercises-1-4-display-sets-in-mathbbr2-assume-the-sets-include-the-bounding-lines-in-each-case-give-

Question

Exercises $1-4$ display sets in $\mathbb{R}^{2}$ . Assume the sets include the bounding lines. In each case, give a specific reason why the set $H$ is not a subspace of $\mathbb{R}^{2}$ . (For instance, find two vectors in $H$ whose sum is not in $H,$ or find a vector in $H$ with a scalar multiple that is not in $H .$ Draw a picture.)
(PICTURE NOT COPY)

Brian Ketelobeter · Accepted Answer

Yeah. Good day, ladies and gentlemen. So today we&#x27;re looking at a problem. Number one from section 2.8 in linear algebra on the problem asked us to consider this region here. H which a zai tried to a zai tried to clarify. Here includes the boundary here and some and the entire interior here. And our goal is to show that this guy is not a linear subspace. Okay, how do we do that? Well, first off, what does it mean to be a linear subspace? Well, um there&#x27;s three things you have to check. You first have to check that the zero vector is in H. So, in other words, in this case, the zero vector, um, is the vector 00 This is an R two, and in this case, we can see that, Yes, because it&#x27;s on the boundary. It&#x27;s clearly in there aan den, the second condition. You have to check that, Um, if you have two vectors, um, the one and the chew, uh, in, um h um Then, um then we have the one plus of the two is also in H, and while I&#x27;m not gonna prove it, um, it&#x27;s pretty it&#x27;s pretty easy to check that. In fact, this would actually hold in this case an H s case that you could add any two vectors. You could note that the easiest way to see it is that if you want to be too are in age, then they have positive coordinates in both X and the Y direction. And when you add them together, you&#x27;re gonna end up with you&#x27;re going to come up with, Um, the coordinates are gonna be positive in both the X and the y of you one plus B two. So in that case, it would also be in there. But the final condition is that if, um if we have the want if the final condition is the, uh in h, um, and a scaler lambda and, um, what we call a scaler here, which, in other words, is just a real number. So lambda in our so remember this is just a really number. Um Then Lambda V is in there. Then Lambda V is also in H. And now it turns out, of course, that this is pretty. This is where it fails. So let&#x27;s just take an example here. Onda take the T equal to let&#x27;s take the t equal to the vector 11 OK, so clearly V is in H. Okay, it&#x27;s positive. Real numbers on debts. Take Lambda T equal to negative one Lambda Thio equal to negative one. Clearly, that is a real number. Okay, um now when you multiply Lambda Times E, what do you get? Well, Lambda Times V is just going to equal to obviously negative one. Um, negative one. And it&#x27;s pretty clear that this is not in h eso. Since it&#x27;s not in age. That means that h is not a linear subspace. Because it fails, it fails. This condition fails the third condition. So since it fails their third condition we have age is not a linear associate age, not a linear cell space, not a linear cell space. Uh huh. Linear subspace off, Um, are too. Okay. And that&#x27;s what we needed to check. That&#x27;s what we needed to show s Oh, that&#x27;s it for this problem. Thank you very much. And have a nice day

Gideon Idumah · Answer

given this dag room say this is the dark room. And then we asked something like these. Um yep, something like these. And then this is field. This is field. So where the show why base is not is subspace. So take, for example, I pick a point here say this is point you and then said, picking points here and in this V. Now let&#x27;s assume that&#x27;s you equal to say zero comma and sue because this is on the white sands is on. Then v equal toa same minus three. Common zero. You see, that&#x27;s you plus V will be called to what&#x27;s minus three. Coming too. So minus three comments will be somewhere here. Rights. So this will be you, close v so we can see that&#x27;s you. Plus V is not in bitch because it is just this shaded region, right? So that implies that&#x27;s it is not closed under audition on ends is not a suspicious off parts. Anand&#x27;s lesson Age is not a subspace

Brian Ketelobeter · Answer

Okay. Good day, ladies and gentlemen, um, this is problem number three from section 2.8 in linear algebra. And the question asked us to show that, um, we have a region given this region H, which is isa drawn a little better in your textbook and the way I did it. But you you get the rough idea in my on here. Basically, it&#x27;s this entire region in here that sort of colored in here. So it&#x27;s this region here. What? Yeah, it should have and do it. So this region here, um, and the boundaries are sort of right here. If you you could take a look at the picture in the book, it&#x27;s a little clearer than what I could do. But basically, there is a boundary line here. Um, and this region in here is H. And we want to show that this is this is not a linear subspace of our too, So Okay, so what does this mean? Well, what we have to check is the three conditions in the definition of the linear subs. So, the first condition, um, to show that linear subspace you have to check first you check that check. Um, the vector, the zero vector here, which is the zero. I&#x27;m sorry. Zero vector, which in this case is just zero zero eyes in H. In this case, it it clearly is. This would be the zero right here. The zero would be, of course, this guy here, which is 00 and that&#x27;s clearly in age. So this is our condition one our condition too. Is that ah condition, too? Is that if we have two vectors V one? Um, and the two in h uh, then, um, the one plus the two is in H also, um, now eso This is our second condition, and I I&#x27;ll be honest with you. I think you you could if you wanted to use this as a counter example, but I don&#x27;t think it&#x27;s the obvious. I don&#x27;t think it&#x27;s the easiest one. Okay, I think there&#x27;s actually an easier one, and it&#x27;s using scaler multiplication, which is the third one. So I&#x27;m I&#x27;m gonna cheat or not really cheap, but I&#x27;m gonna go with the easiest one. Um, and that&#x27;s three, which is scaler Multiplication, which is, um, to check. Oops. Uh, my love. Uh, no. Uh, yeah. uh, scaler Multiplication eso It&#x27;s h is closed under scaler multiplication scaler. Our multiplication on. Basically, what this means is that, um taking take, uh, just a lambda A Really, Um take Lambda in our, um take Lambda in our and V and H um, then Lambda Times the is in h. Okay. And really, this has to be true for every riel number. So every scaler and are so okay. Well, with this being said, then what sort of the obvious way we can apply it? Well, we know that the of course, the edges of this are in the space, so in particular than, um, this point right here, um, which I&#x27;m gonna just label as Maybe because I don&#x27;t know what the the these lines the equations for these lines are. So I&#x27;m just gonna say this point V is equal to we&#x27;ll just call it maybe one zero. Okay. Really? You could call it x zero if you prefer. Maybe X zero is probably better, but it some point acts, right? I mean, some point acts, which is positive number. Um, and the vector the X comma zero is in h. Okay. And the way I&#x27;ve chosen it. I&#x27;ve chosen it on the boundary. Right. So, um, that if I take if I take lambda equal to say to So if I take Lambda to be equal to two, um, then Lambda V is going to equal to, uh, to x comma zero. Okay, now, clearly, um, if this is acts, then two x zero is gonna be out here. So this is gonna be and, uh, the and based on the way I chose it, it&#x27;s clear that this guy is not in age. This guy is not in h. Okay, So since he&#x27;s not in h, that means, um h is not closed under scaler. Multiplication. Hence, h is not a linear. So space, not a a linear subspace. Subspace. Sorry. Subspace. There we go. SS linear subspace off are to Okay, so the the idea is that, um, because the skit, it&#x27;s not closed on your scaler. Multiplication. That means it&#x27;s not a linear subspace. Okay, Andi, that is all we needed to show. Okay, um, thank you very much. And have a nice day

Gideon Idumah · Answer

since you were given this diagram, Um, it&#x27;s goes this way. See, of it&#x27;s tripped line passing through this point and they have something like these. And then this is the shit of the region. I want to show why this is not a subspace. So, for example, I picked this to be used rights. Saddam is This is my you. If you look at the mine issue, you&#x27;re mine. Issue is somewhere here, so this would be like minus one more displayed by you so we can see that&#x27;s minus you is not in bitch, which implies us age is not closed on the scale. A multiplication because we only multiplied by skill are minus one. So it&#x27;s not closed on the scale. A mortification on this Each is not the subspace ce off arms.

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problem number 32 of section 4.1 on, um, basically was to self spaces of the vector space V, H and K um, the intersection of H and catering as h er upside down U K, which basically just means H intersection intersection. Kay is a set of vian in capital of E that belonged to both hnk. So whatever age contains K must contain if it if h and K R h intersection k were to be, uh, was to be ah subspace of v you to show that h intersection Kerry is a subspace of e. So that&#x27;s basically like the beginning, Like the simple pond in the beginning of the section that we don&#x27;t need to prove. Three cases first is that the zero vector exists in h er each intersection k. So, since they&#x27;re both subspace is, we know that zero must be an age and zero must be in K. And since they&#x27;re both in both the sets individually, we know that we know that zero is in the intersection of hnk because the intersection of agent cages include, um, all the elements that are included in both. The both of the subspace is individually next. We need to prove that it is closed under a tradition. So if you take you envy in age ah, Union K R H intersection kay. Then if we do, then, um, we know that you and V is an h individually which, because we know that it is in the intersection of both sets. So you plus V must also be an age since h as a subspace and similarly the same is true for case since u N v e r N k Ah, since we assumed that you would be our in the intersection of Agent K, you know that you plus V is also in K because kay is also a subspace. So we know that, um, So we proved that it is close under, uh, vector addition. And lastly, we need to prove that it is closed under skill or multiplication. So if you take a vector v n h intersection kay and we take see to be a scaler, you know that V is an age so cv any, um, since eight of the subspace, any any value that is scaling the must also be an age because the dimensions of we are not changing because it still has the same amount of entries. So and since agent subspace we know that CV is also an H and similarly the same is true for case. Since he is also in K, you know that CV is also in K So this is also close this H intersection case also close under, um Skilling Multiplication. So we have proved that h intersection kay is a subspace so space of the vector space v Capital B And lastly, we need to show given example and are too where, um, like where The intersection of two subspace is not in the There&#x27;s not an example or not intersection of the union. So it&#x27;s the opposite. The union to subspace is is not always a subspace of, um are too. So in our two, we know that the x axis Sorry, the x axis and the y axis. The values of the via access are individually both subspace is of our two. But when we combine them, they&#x27;re not ourselves space of our two. Because if we take it if you combine them and let&#x27;s take is that the corner. 01 is in our two and 10 isn&#x27;t our two. But 11 is not in our tea because it does not lie on either the two axes. So this is a simple counter example.

Daniel Pezzi · Answer

Hello. For this question, we need to show that this subset of our two is not a subspace of our two in the vector sense. Um, and this is just the set of all vectors such that this algebraic relationship holds which geometrically is just the unit circle now, because the zero vector is in this place, there&#x27;s two things that weaken Dio. We can either find two vectors that when we add them to that air in H that we add them together are not in a church or we could take a vector in h multiplied by a constant and end up outside of age. I&#x27;m going to dio the constant one because it&#x27;s simple essentially, because you have to be closed under multiplication of arbitrary constants. Any geometric set in our to for it to be a subspace needs to include infinity in some sense, What I mean by that is if I take a vector, say one zero, which is this point right here as a vector, and I multiply it by, say, I don&#x27;t know, five, right, Because this if h was a vector space, then this would have to be in. This would have to be in H because F is sorry. Five is an arbitrary constant in our field. That is not the case. As if I look over here, I extend this a little bit five times 10 Is this vector this point here? This would be the point five zero which has escaped our region here. So because of that, this is not in H um, and so we can see that this is not a subspace of our to If you wanted to do the vector addition way. Ah, kind of a cheap way to do it is just to do the toe add this vector to itself. If I take 10 and I add it to the vector 10 that I get the vector to zero. And again I&#x27;m adding to elements of my of my supposed vector space. So I should end up back in that space. But that is not the case. I end up with this this vector to zero, which is not in my in my set. So h is a subset of our two because it contains elements of our two. But because it does not have the closure properties that we want, it is not a subspace

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Exercises $1-4$ display sets in $\mathbb{R}^{2}$ …

Problem 1 Easy Difficulty

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

Linear Algebra and Its Applications

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Alayna H.

Kayleah T.

Michael J.

Absolute Value - Example 1

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Linear Algebra and Its Applications

Grace H.

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Absolute Value - Example 1

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

Grace H.

Alayna H.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications