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Let $H$ and $K$ be subspaces of a vector space $V .$ The intersection of $H$ and $K,$ written as $H \cap K,$ is the set of $\mathbf{v}$ in $V$ that belong to both $H$ and $K .$ Show that $H \cap K$ is a subspace of $V .$ (See the figure.) Give an example in $\mathbb{R}^{2}$ to show that the union of two subspaces is not, in general, a subspace.

Thus, the intersection of subspaces is a subspace of vector space but union of subspaces neednot be subspace of vector space.

Calculus 3

Chapter 4

Vector Spaces

Section 1

Vector Spaces and Subspaces

Vectors

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Boston College

Lectures

02:56

In mathematics, a vector (…

06:36

02:43

Given subspaces $H$ and $K…

05:08

Let $S=\left\{\mathbf{x} \…

11:42

Let $S_{1}$ and $S_{2}$ be…

03:22

Let $\mathbf{u}$ and $\mat…

02:35

Given $T : V \rightarrow W…

05:10

Exercises $1-4$ display se…

01:40

07:48

01:16

02:24

Let $H$ be the set of poin…

in this video, we'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's basically like the beginning, Like the simple pond in the beginning of the section that we don't need to prove. Three cases first is that the zero vector exists in h er each intersection k. So, since they're both subspace is, we know that zero must be an age and zero must be in K. And since they'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's not an example or not intersection of the union. So it'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're not ourselves space of our two. Because if we take it if you combine them and let's take is that the corner. 01 is in our two and 10 isn'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.

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