let-s-be-a-convex-subset-of-mathbbrn-and-suppose-that-f-mathbbrn-rightarrow-mathbbrm-is-a-linear-tra

Name: let s be a convex subset of mathbbrn and suppose that f mathbbrn rightarrow mathbbrm is a linear tra
Uploaded: 2021-01-02
Duration: 2 min 7 s
Channel: Numerade
Description: Let $S$ be a convex subset of $\mathbb{R}^{n}$ and suppose that $f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is a linear transformation. Prove that the set $f(S)=\{f(\mathbf{x}) : \mathbf{x} \in S\}$ is a convex subset of $\mathbb{R}^{m} .$

Question

Let $S$ be a convex subset of $\mathbb{R}^{n}$ and suppose that $f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is a linear transformation. Prove that the set $f(S)=\{f(\mathbf{x}) : \mathbf{x} \in S\}$ is a convex subset of $\mathbb{R}^{m} .$

Lucía Guerrero · Accepted Answer

and this exercise, we are going to prove that the image of a comeback set on their linear transformation is also comebacks. So let&#x27;s take X and Y in FFS so X can be written as if off X hat on Why can be written us f uh, why hot where X had on why had are elements of s then 40 in the close intervals Everyone, we have one minus three times X plus three times Why is the same as one minus t times ffx hat plus three times f of my heart on since official in your transformation, we have that this is the same as if off one minus t time Sex hat plus t times y hot on now sues IHS was a comeback set on x hahd on why hot were elements of s. Then we conclude that one Manistee time sex hard pressed eat them So I had It&#x27;s also in development affairs on since this number isn&#x27;t this if off this if off one minus tedium sex had plus stadium. So I had is an element off FFs on this implies that all minus t times explicitly times y is an f off s So ffs is a comeback set as we wanted to prove

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

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.

Chris Trentman · Answer

so given a conduct set and to positive numbers, C and D were asked to show that CS plus D s going to be equal to C plus D s. So if we in other words, stretch the elements of a convex set by sea and stretch the elements of a complex set ID and then add thes two new sets together we get a set, which is three elements of s stretched by C plus D. So we&#x27;ll let s be convex set. Let&#x27;s see, be greater than zero and d be greater than zero. Now let X be an element of CS plus ts and we confined S one and s two in s such that X is equal to see X one are sorry CS one plus D s to then it follows. The X is also equal to C plus d times. See over C plus d. We can divide by C plus Decency plus D is greater than zero since both c and D or greater than zero s one plus de over C plus D s to now because S is a convex set. It follows that see, oversee plus D s one plus D over C plus D. As to linear combination of elements from s also lies an s. So it follows that X is an element of C plus D s. So we&#x27;ve proven one direction. That is, that C s plus D s is a subset of C plus D s. Now to prove the other direction Similar exercise. So we&#x27;re going toe Let X be in the set C plus D s. This means we confined. A vector s in the set s such that X is equal to C plus D. Yes, and of course, this is equal to CS plus D s. After distributing the scaler, we could say, even distributing the vector across the scale Er&#x27;s And this is clearly an element of C S plus D s. So X is an element of CS plus ts and therefore it follows that C plus D S is a subset of CS plus D s. Combining this statement and the previous statement we get that CS plus D s equals C plus D s, which is what we wanted to prove

Chris Trentman · Answer

rest to prove that a point in a context set is an extreme point of the set if and only if the set minus that point is still convex. So we&#x27;re going to let the be an element of s where s is Convex set. And now let&#x27;s prove the yeah, it direction first. So suppose that V is an extreme point of this And let t b the set of all vectors X in s such that X is not equal to our extreme vector. The now if x and why are in t them it follows that the line segment by X y is an s. Since Esa&#x27;s convicts now the is an extreme point of s so follows the V does not lie on the line segment y z. If it did, it would have to be one of the endpoints x or y, which it&#x27;s not not lying line saying the next why and therefore falls X Y is a proper subset of tea. So it follows that t is convex is what we wanted to prove now to prove the other direction Who&#x27;s that he is not in extreme points were actually going to prove the contra positive, then this means that we confined two vectors X and Wyness such that the lies in line segment X, Y and V is not equal to X and V is not equal to why has to find the interior of the line segment. So it follows that from the previous paragraph x and Y R N t. But the line segment X Y is not a subset of tea because V allies on the line segment India&#x27;s not in T so follows. T is not convicts. We have proven what we wanted to prove.

Problem

Let $f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m…

Problem 13 Easy Difficulty

Let $S$ be a convex subset of $\mathbb{R}^{n}$ and suppose that $f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is a linear transformation. Prove that the set $f(S)=\{f(\mathbf{x}) : \mathbf{x} \in S\}$ is a convex subset of $\mathbb{R}^{m} .$

Answer

See proof.

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Linear Algebra and Its Applications

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

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

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