let-f-mathbbrn-rightarrow-mathbbrm-be-a-linear-transformation-and-let-t-be-a-convex-subset-of-mathbb

Name: let f mathbbrn rightarrow mathbbrm be a linear transformation and let t be a convex subset of mathbb
Uploaded: 2019-08-25
Duration: 5 min 58 s
Channel: Numerade
Description: Let $f : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a linear transformation and let $T$ be a convex subset of $\mathbb{R}^{m}$ . Prove that the set $S=\left\{\mathbf{x} \in \mathbb{R}^{n} : f(\mathbf{x}) \in T\right\}$ is a convex subset of $\mathbb{R}^{n}$ .

Question

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

Gideon Idumah · Accepted 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

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

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

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.

Chris Trentman · Answer

were asked to prove that the sum of conduct sets is again a convex set. So let A and B he convict&#x27;s now let the vectors x and y B in the set some a plus B. This means that we confined in a in C in A and A B and a D in the set B Such that X is equal to a plus B and why is equal to C plus D. Now let t lie between zero and one Them we have that the point on the line segment x y It&#x27;s going to have the form W, which is tee times X plus one minus t times why you could do the other way around. This is the same as tee times a plus B plus one minus t times C plus d. Of course, this is equal to grouping together vectors from the same set. So we have t a plus one minus t see plus t B plus one minus t the now because a and beer convex it follows that ta plus one minus t See, this is an element of the line segment between A and C. This lies in a and we have that T B plus one minus TD. Since B is convicts and this is an element of the line segment between B and D. This also lies in B, so it follows that w lies in a plus. B. Therefore, it follows that a plus B is convicts, since it contains any points online segment between and between X and Y.

Problem

Let $\mathbf{v}_{1}=\left[\begin{array}{l}{1} \\ …

Problem 14 Hard Difficulty

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

Answer

since $T$ is convex, point $( 1 - t ) y _ { 1 } + t y _ { 2 }$ must be in $T ,$ lence $f ^ { - 1 } ( 11 -$ $\left. t ) y _ { 1 } + t y _ { 2 } \right) = ( 1 - t ) x _ { 1 } + t x _ { 2 }$ in $S$

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since $T$ is convex, point $( 1 - t ) y _ { 1 } + t y _ { 2 }$ must be in $T ,$ lence $f ^ { - 1 } ( 11 -$ $\left. t ) y _ { 1 } + t y _ { 2 } \right) = ( 1 - t ) x _ { 1 } + t x _ { 2 }$ in $S$

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Kayleah T.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications