let-t-mathbbrm-rightarrow-mathbbrn-be-a-linear-transformation-and-let-mathbfp-be-a-vector-and-s-a-se

Name: let t mathbbrm rightarrow mathbbrn be a linear transformation and let mathbfp be a vector and s a se
Uploaded: 2019-08-23
Duration: 2 min 44 s
Channel: Numerade
Description: Let $T : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ be a linear transformation, and let $\mathbf{p}$ be a vector and $S$ a set in $\mathbb{R}^{m} .$ Show that the image of $\mathbf{p}+S$ under $T$ is the translated set $T(\mathbf{p})+T(S)$ in $\mathbb{R}^{n} .$

Question

Let $T : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ be a linear transformation, and let $\mathbf{p}$ be a vector and $S$ a set in $\mathbb{R}^{m} .$ Show that the image of $\mathbf{p}+S$ under $T$ is the translated set $T(\mathbf{p})+T(S)$ in $\mathbb{R}^{n} .$

Viswesh Uppalapati · Accepted 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.

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.

Ashley Boni · Answer

So we&#x27;re given that the vectors v one v p. Span are and tease alone your transformation on We want to show that teest zero transformation. So, um, if X is any vector in on our end t of X is equal to zero. So if X is been any vector on our end and all the vector of these span our end, we can write our vector X as a linear combination of the V&#x27;s. So, in other words, um, just for some Constance C, we can write X equals C one V one plus C to be to all these reserve vectors plus dot, dot dot Ah, C p VP. So now let&#x27;s take t of both sides. T of X is equal to t of C one V one plus up to C P V P Tease Lanier s so we can pull Constance out. So we have C one t of V one close that dot dot c P t of v p, and were given that each individual t of the eye is equal to zero for I goes from one to pee. So from star, which is ah this given Here we have C one time zero plus c two times zero plus all the way up to C P Times zero, which is just equal to zero. So therefore t of any vector X is equal.

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

Michael Jacobsen · Answer

here. We&#x27;re going to be considering an example that has a lot of moving parts. First we&#x27;re doing dealing with the transformation T. Which is mapping V two w. But the most important thing about this transformation is that it is linear. The next big piece of this puzzle is that we have a subspace you of the vector Space V, which is code comes from the domain of tea. So we have a subspace moving around and the second moving piece of this puzzle is this space t of you. This is contained in the CO domain W And here is our goal. We want to prove that this is, in fact, a sub space of w not just a subset. Well, here&#x27;s how we&#x27;re going to start. If we&#x27;re going to prove that this is inside W we should say exactly what t of you is that sets our notation and it helps us move through the solution. So we have by definition, that t of the subspace you is going to be equal to the set of images which are denoted by t of X, such that ex itself is in. You so t of you here It is a set of the images t of X which we&#x27;ve indicated. And the only requirement is that the vector X had to have come from you, which is a subspace of V. This takes a moment to digest, so there&#x27;s no harm to take a few moments Drawing pictures are looking at diagrams now, since we&#x27;re doing a subspace proof are proof comes in three parts And the first part is to show that the zero vector is in t of you. Well, to start this one, we can say Since t is linear, this zero vector we&#x27;re looking at can also be expressed as t evaluate at the zero Vector. Since linearity allows us to take tea mapping is your vector always back to the zero vector but notice what we&#x27;ve shown So far the zero vector has been written as t apply to the zero Vector. So this is your vector here is playing the role of X and it is definitely in you because a subspace you always contains its zero vector. So we can say that this implies that zero vector is in t of you. Since the zero Vector here is in you So let&#x27;s go off to the second part. For the second part, we need to grab to arbitrary elements of t of you. So let&#x27;s say let t of x and t of. Why be in t of you? It&#x27;s sometimes helpful to write out what our goal is. We need to show that t of x plus t of y is also in t of you. So let&#x27;s start off by writing down TF x plus t of y and see what we can do from here. Well, the first thing we could say is because the mapping Trenti is linear. This is the same as T of X plus y. And that means we have tea of a vector X and X plus y is in you because you is a subspace, the linear combinations x and y. If we take X plus, why we&#x27;re still in the subspace you now we know x and why came from you? Because that&#x27;s how we define the set t of you itself. So since we have and X plus y here matching this X where X plus y is a new it follows then or imply it&#x27;s implied that t of X plus t of y is in t of you notice that the only way that this conclusion here is sound is if we use the fact that T is both linear and you is a subspace of V at the same time for part two. Now let&#x27;s go to Part three in part three, work still going to use the T of X. That&#x27;s provided, but we also say, Let see be in our. Then again, it&#x27;s important to stay our goal either mentally or just writing it down. And our goal is to show, or we need to show that C Times T of X is in t of you. Let&#x27;s see how to show that. First, I&#x27;ll start with C times t of X, the vectoring question and now we use linearity again. This is the same as tee times. See Adex And then at this stage, we use the fact that you was a subspace of V. That means since X came from you see, Time&#x27;s X is still in you. So this implies that C Times t of X is in t of you since see Time&#x27;s X is in you. So we&#x27;re able to now verify parts 12 and three for the definition of subspace. And so our conclusion is the following t of you is a vector subspace of w. So what we have shown altogether is if we have a subspace you than the set of all images is also a subspace of the co domain.

Problem

Let $S$ be the parallelogram determined by the ve…

Problem 26 Medium Difficulty

Let $T : \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ be a linear transformation, and let $\mathbf{p}$ be a vector and $S$ a set in $\mathbb{R}^{m} .$ Show that the image of $\mathbf{p}+S$ under $T$ is the translated set $T(\mathbf{p})+T(S)$ in $\mathbb{R}^{n} .$

Answer

$T(\mathbf{p})+T(S)$

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

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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$T(\mathbf{p})+T(S)$

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Alayna H.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications