let-f-mathbbrn-rightarrow-mathbbrm-be-a-linear-transformation-let-t-be-an-affine-subset-of-mathbbrm-

Question

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

Gideon Idumah · Accepted 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

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be solving problems over 26 of section 2.3, which is based on linear independence. Ah, transformation or not, linear independence. I&#x27;m sorry. It&#x27;s based on determinants and Cramer&#x27;s rule and, uh, linear transformations and volumes. So here we&#x27;re given that Tia&#x27;s transformation, um from our himto are in and we know that p is a vector and s is a set in r r m s as is a set in, um our, um So let we can let this is a proof so we can let the vector v be any vector any vector from the set s which is on our M. And by definition, we know that people s is a set of all vectors because s a set of all vectors, um, better of the form p plus V that we did find your love. Uh and we know that V isn&#x27;t s so applying t to ah, typical vector and people cess. We have, um t and we apply people. Savi innit would give us tp plus t v. And this is this vector is in the said denoted by t plus s her teepee plus t s because we know that he is then s This proves that t maps the set people says onto the set T p plus t s. So this proves that, um, t as t maps people us us onto t p plus t s was what the initial problems asking which stated that we need to show that the image of people says under T a translated set t put people&#x27;s t s in rn. So there&#x27;s just a simple proof so shit.

Rakvi . · Answer

So, first of all observer that he told the one is a transformation from M in our two women are there is it takes. And anyway, in Matrix and our ports and in very in matrix and you&#x27;re given to you to anyone individually So we just have to compute. What happens for each of them interests is so t two t one off is just even off a and then you applied took even if it so he won off is just a minus a transpose. So this is summon bay in metrics and they do off a matrix is a matrix place It&#x27;s trance fools. So you take this and you add X transpose No transport off some of two matrices. It&#x27;s some off transports off individual metrics so you can open up records like this. A transport a V minus eight transports is just gay transpose minus transport off transports, which is just e. Okay, now, as you can see, this is zero. So no matter what metrics, what matrix You input into t 2 50 when he will get zero. So therefore Kato off even is the zero transformation

Gideon Idumah · 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.

Problem

In Exercises $21-26,$ prove the given statement a…

Problem 20 Hard Difficulty

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

Answer

In conclusion, $f\left((1-t) x_{1}+t x_{2}\right)$ is in $T,$ so $(1-t) x_{1}+t x_{2}$ is in $S$ meaning
that $S$ is affine subset.

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

Linear Algebra and Its Applications

Heather Z.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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In conclusion, $f\left((1-t) x_{1}+t x_{2}\right)$ is in $T,$ so $(1-t) x_{1}+t x_{2}$ is in $S$ meaningthat $S$ is affine subset.

Linear Algebra and Its Applications

Heather Z.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Samuel H.

Michael J.

Joseph L.

In conclusion, $f\left((1-t) x_{1}+t x_{2}\right)$ is in $T,$ so $(1-t) x_{1}+t x_{2}$ is in $S$ meaning
that $S$ is affine subset.

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

Linear Algebra and Its Applications