let-mathbfv-in-mathbbrn-and-let-k-in-mathbbr-prove-that-sleftmathbfx-in-mathbbrn-mathbfx-cdot-mathbf

Name: let mathbfv in mathbbrn and let k in mathbbr prove that sleftmathbfx in mathbbrn mathbfx cdot mathbf
Uploaded: 2019-08-24
Duration: 3 min 13 s
Channel: Numerade
Description: Let $\mathbf{v} \in \mathbb{R}^{n}$ and let $k \in \mathbb{R} .$ Prove that $S=\left\{\mathbf{x} \in \mathbb{R}^{n} : \mathbf{x} \cdot \mathbf{v}=k\right\}$ is an affine subset of $\mathbb{R}^{n}$ .

Question

Let $\mathbf{v} \in \mathbb{R}^{n}$ and let $k \in \mathbb{R} .$ Prove that $S=\left\{\mathbf{x} \in \mathbb{R}^{n} : \mathbf{x} \cdot \mathbf{v}=k\right\}$ is an affine subset of $\mathbb{R}^{n}$ .

Gideon Idumah · Accepted Answer

okay for this question we have Let&#x27;s V b, r N and Kay in our we want to show that one toe shoe. That&#x27;s s which is, of course, tow X In our end. Given that ex dot v equals two K, these are fine subsets. Both are in. So do this by definition on page four nights warn we have to sure you have to show that one minus tse times p lost. See, Time&#x27;s Cube is in s full p com acute in s. So this is in dilemma rights. It&#x27;s much more elaborate for peek A McCue in S on DDE. Seeing is any real number. So now let&#x27;s become a Q B s then, because they&#x27;re in s on by definition of s, that implies that p dot v equals two K and similarly q dot v equals two K. Now, let&#x27;s see be any real number, be any real number. Then we have that want minus c more supply by P. Lost See most like you dot v. We need to show that this is the coast. Okay, this is the way we can show that this thing is an s. So the show these we need to show that dese. Is it war too? Two. No. One minus C Times People lost cq dots V will be equal to one minus C p dots V plus C Q Dots Ving Now beat us. V is K So I have one minus tse multiply by cabe los que times That visa&#x27;s okay, That&#x27;s plus C k. When I saw this, this is called you K minus C k lost CK on. This is a course to K. So this implies that because we have Kay, this implies that s is are fine subsets off our end.

Gideon Idumah · Answer

here one is should&#x27;ve sits s equaled Sue X In our end, Given that f off ex isn&#x27;t see, he&#x27;s on fine subsides off our end. So let&#x27;s p com a Q B in s to show that this is an offense off subsets we need So sure that&#x27;s one minus tse more to abide by p lost See more to ply My q isn&#x27;t s four and you really number now to show us something is an s by different by definition of s we need to show sure dots f off eggs because if anything is an s, then I miss f off one minus c p plus c Q Most be in C. So now let&#x27;s show that this isn&#x27;t see so off f off one minus c be plastic You will be equal to because we know that f is a linear transformation. So this we can break this down to f off one minus c p clothes F off c Q. This can also be broken down. Sue f off I can ride is us p minus C p less. I can pull out it see because it&#x27;s Lena transmissions after c f off cue this. I can write us F p minus a comm plotted C F P close C F off Cuba and then we can ride these oz one minus c Pull off the F off P lost C f off Cue knows things become a Q Isn&#x27;t s dune by definition of s it&#x27;s implies that&#x27;s f off p comma f off. Few isn&#x27;t see because that is our s is defined. No scenes, See is our fine subsets off our M? This would imply by definition, off Alfonso sits. This implies that they are fine. Combination off f p on f off cue use who seem on this is an offering combination off f o p N f off cubes. If you combine these, this would imply dots. Bizweek implied dots one minus c fo being lost C f Q using C, which is what we wanted to show. If you go back to the force Page 12 should this which we have shown. So this implies that&#x27;s this is, um I&#x27;m fine. Some sense

Michael Jacobsen · Answer

in this example, we have a linear transformation f and our goal is to show that if s is in a fine set coming from the domain r n that the images ffs is also a fine in the co domain RM. For proof like this, because we&#x27;re trying to show FFS is a fine, we&#x27;re going to start by saying Let let&#x27;s choose vectors a and be be in the set ffs Well, f of s is a set of images. So this implies. Then there exists two other vectors. Let&#x27;s call them X and y, and these come from the domain or excuse me, the set s from the domain are in such that if we evaluate f at se X, we get the first image which is a and likewise evaluate f at. Why? Then we&#x27;ll get the second image which is being now we have enough important facts that we must use here s f fine. Let&#x27;s go down to the definition of what it means for us to be a fine. It says If we have any vectors p and Q, that&#x27;s X and Y In our case, then this particular combination here is always in the set s for any real number T we pick. So it&#x27;s right down that statement. They&#x27;re by definition, we can say since s is fine. And we just picked X and y to come from s we have that one minus t times X plus tee times. Why is in S and this is for all t in our So what we&#x27;ve used is the definition that s is a fine we&#x27;re trying to show FFS is a fine And so we need the equation much like this, but using not an X and y we want an a an a b in these locations. Then we can go back to the definition and conclude FFS is a fine well, to get to a conclusion like that Noticed. The only assumption we have not used to date is that f is linear. Let&#x27;s try that out next. Well, since f is linear, if we evaluate f at the input one minus t times X plus t times, why then, by linearity we get first one minus t times, eh? FedEx Then add to that tea times f y and recall next that f a Texas A and Ethel why is be Let&#x27;s make those two substitution. This will be equal to one my s T times and a plus tee times a B And this works for all t in our once again. So notice what we&#x27;ve shown. This is an image which means that lies in the set F of s and we have the same kind of equation We were concerned over here, and this worked for a generic element A and B which came from F of s. We&#x27;re ready then for a conclusion. Our conclusion is that this shows meaning the last equation shows that A and B are in the a fine hole off the set s. But then this also shows weaken, say, and so f of s is a fine, and that completes our proof.

Gideon Idumah · Answer

so two shoulders the sides s which is a gold toe x In our end such that f of X is in C. To show that this set is a convict sense convicts subsets off our aim would peak sue points See, let&#x27;s call them X one x two in this under. We sure that&#x27;s the line segments the line segments between Damn, Isn&#x27;t this no leads X one be in this By definition, it&#x27;s implies deaths f off X one isn&#x27;t seen Now this implies dunce f x one de quarto. Why one for why one in c all rights? Because we say f excellent isn&#x27;t seen. So I miss fo x one is altering Why one for which why one has got to see this would imply that X one accord to f involve us off what I want. You just take the inverse of what side? Similarly, let x to be element of Bess. This implies that f of x two by definition, beaten See, which implies that f of x two. Because why too for a white soon in C. This implies that X Toole accord to f involve us off. Why? I want so to show that the line segments between them it&#x27;s between s we must show We must show That&#x27;s one minus C x one lost The ex too must be in s for every zero lesson opposed to see less than a course one. So let&#x27;s show these now off one minus c x one lost Seen X to disappear one Miners seem our x one is f ive us of y one loss seem f universe off white too. This is nothing Bad&#x27;s when you multiply it. Woman listed by these are FBI of us off Why Juan minus C f f s of y one lost seen f inverse off. Why, too, which of course toe f any of us or why one miners? Because f is Nina f impossible genius So I can pull this inside. We&#x27;ll see why one loss as one of us off. See why, Sue? So this is just equal to f ive us because this sort of putting things f is linear. Ft of us is also Minya. So that&#x27;s why I can do all these operations. I can pull out every bus I have what I want minus tse. Why one lost see Why soon? So this is recalled to f Reverse off one minute. Just see why Want plus c? Why soon Now, just pull these down. What obvious? That things is a board to bees? No, Because of these, you call that See his converts. So let&#x27;s go back. That&#x27;s so since C is convicts the points one miners see why one lost See white soon Most be unseen all rights Because this is the convicts combination. Because why one on White? You isn&#x27;t seen now from this three questions, Tom and Nicole is a question star. No, from star. What would our duds one minus c? Why? Want lost? See White? So I&#x27;m moving this f in verse to this side. So this is equal to f off this. So that&#x27;s is this will be equal soon. F off one minus c. That&#x27;s one lost X zoom. Now we said this isn&#x27;t seen, so this most b c so we&#x27;ll show that f off one minus c x one Must be in C by definition. Oh s If this isn&#x27;t see by definition of s, this would imply that one minus X one lost. See X two Must be s on. This implies that this is a conference

Julian Wong · Answer

this problem were asked to prove that and choose K is the same as and choose and minus K for all in case and the natural numbers and end has to be greater than or equal to K. So it is approved. It will start on the left side. Um, n choose K. Ah, this will make it Ah using the the combinatorics formula n factorial over que factorial and n minus k factorial and ah, we would like to arrive at and choose ed minus case. So I gotta change this form so that it will resemble that one of the first moves that I&#x27;m gonna make here swap the denominator. Just the order of it. That&#x27;ll be n minus k factorial times k factorial Um so now you can kind of see uh, n minus. K has taken up the position of the K that&#x27;s on the bottom. Um, I&#x27;m gonna rewrite the K by itself as something else. This will be written as n factorial over and minus K factorial times n minus and minus K factorial. So you can see that second bracket while looks a little bit more complicated. That just cancels up to Cape so there is no difference here. Now you can see some similarities In the first item I had n and then Kay and ah, the cake or responded to those two positions. Now I have ah different one. I have n minus K and n minus k. Ah, in that same in the same positions as three highlighted Kay&#x27;s above. So what we&#x27;re really saying here is that this is the same as n choose in minus K. That was the end of the

E R · Answer

s and sanitation. We have that This all pulling a mules of the form A X squared plus one with a consisting A riel elements. This is the sanitation. No need to see if the subset s is a subspace of e Do this winning 21 should that s is closed. Under addition to that s is closed. Under scaling multiplication, we could do a quick check first for the zero Vector zero Vector does not exist in the subset dress then asked is not a subspace. So starting with the zero after check in this case are zero vector is the zero bowling a mule generate a Z of X is equal to zero and from her form we can let a equals zero. If we do that, we get that are zero vector would be zero times x squared plus one which is not equal zero. So are zero Vector is not in our subset, Which means that this is not a subspace of the So the answer is new. This is not a subspace

Problem

Choose a set $S$ of three points such that aff $S…

Problem 16 Medium Difficulty

Let $\mathbf{v} \in \mathbb{R}^{n}$ and let $k \in \mathbb{R} .$ Prove that $S=\left\{\mathbf{x} \in \mathbb{R}^{n} : \mathbf{x} \cdot \mathbf{v}=k\right\}$ is an affine subset of $\mathbb{R}^{n}$ .

Answer

Therefore, is a subspace of the vector space.
If satisfies, then is a translate of the subspace, so the set $S$ is a flat in.
So, the set is an affine subset of.

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

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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Therefore, is a subspace of the vector space.If satisfies, then is a translate of the subspace, so the set $S$ is a flat in.So, the set is an affine subset of.

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Caleb E.

Kristen K.

Samuel H.

Therefore, is a subspace of the vector space.
If satisfies, then is a translate of the subspace, so the set $S$ is a flat in.
So, the set is an affine subset of.

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

Linear Algebra and Its Applications