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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)

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Algebra

Chapter 2

Matrix Algebra

Section 8

Subspaces of Rn

Introduction to Matrices

McMaster University

Harvey Mudd College

Idaho State University

Lectures

01:32

In mathematics, the absolu…

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Exercises $1-4$ display se…

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Let $H$ and $K$ be subspac…

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Let $H$ be the set of poin…

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Let $S=\left\{\mathbf{x} \…

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Let $H$ be the set of all …

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Given subspaces $H$ and $K…

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Consider the vectors$$…

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Yeah. Good day, ladies and gentlemen. So today we'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'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's on the boundary. It'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'm not gonna prove it, um, it's pretty it'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're gonna end up with you'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's just take an example here. Onda take the T equal to let's take the t equal to the vector 11 OK, so clearly V is in H. Okay, it'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's pretty clear that this is not in h eso. Since it'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's what we needed to check. That's what we needed to show s Oh, that's it for this problem. Thank you very much. And have a nice day

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