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Let $V$ and $W$ be vector spaces over a field $mathbb{F}$. Show that a map $T: V ightarrow W$ is linear if and only if$$Tleft(a v_{1}+v_{2}ight)=a T v_{1}+T v_{2}$$...

Question

Let $V$ and $W$ be vector spaces over a field $mathbb{F}$. Show that a map $T: V ightarrow W$ is linear if and only if$$Tleft(a v_{1}+v_{2}ight)=a T v_{1}+T v_{2}$$for any $oldsymbol{v}_{1}, oldsymbol{v}_{2} in V$ and $a in mathbb{F}$

Let $V$ and $W$ be vector spaces over a field $mathbb{F}$. Show that a map $T: V ightarrow W$ is linear if and only if $$ Tleft(a v_{1}+v_{2} ight)=a T v_{1}+T v_{2} $$ for any $oldsymbol{v}_{1}, oldsymbol{v}_{2} in V$ and $a in mathbb{F}$



Answers

Suppose $\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}$ and $\mathbf{v}_{1}, \ldots, \mathbf{v}_{q}$ are vectors in a vector space $V,$ and let
$H=\operatorname{Span}\left\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\right\}$ and $K=\operatorname{Span}\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{q}\right\}$
Show that $H+K=\operatorname{Span}\left\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}, \mathbf{v}_{1}, \ldots, \mathbf{v}_{q}\right\}$

Here. We're going to be considering an example that has a lot of moving parts. First we'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's how we're going to start. If we'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'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's no harm to take a few moments Drawing pictures are looking at diagrams now, since we'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'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'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's go off to the second part. For the second part, we need to grab to arbitrary elements of t of you. So let's say let t of x and t of. Why be in t of you? It'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'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're still in the subspace you now we know x and why came from you? Because that'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'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's go to Part three in part three, work still going to use the T of X. That's provided, but we also say, Let see be in our. Then again, it'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's see how to show that. First, I'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's X is still in you. So this implies that C Times t of X is in t of you since see Time's X is in you. So we'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.

So this problem we're given transfer in your transformation from V. Too tough for you. Where B and W are finite dimensional. Just say fine. Nine and h subspace off B and age. It's not zero and t is 121 Okay, so we want to show that I want to show you mention team age is equal to dimension off. H all right. So, uh, one thing to notice that if we have have a basis, um, say is itself a JJ. Let's say so. Be one thio the end. So we are considering the mapping off these vectors. Then you won and be, too until p t are also we nearly independent the reason the reason for why this is still lingering. And then it's because, uh, tea is ah, is that 1 to 1? Maybe so. Note that tea is 121 So it turns out this these vectors, the mapping off we want to be p will spend. It's been tee off me, uh, d of H. And since they are p vectors here, so the number of factors is so you mentioned off age. It's equal to and shin of t age, which I won't. Hey, so we're done now, given that addition no additional condition. Say he is on two. So what do we have now? Yeah. Uh, uh. T zone two can directly take, um TV to be w so by our previous, like our previous, uh um, permission. We know that, um, you mention, uh, TV is dimension off W Well, actually, actually, this is a dimension of TV should be equal to dimension. Enough T and dimension of TV is actually dimension enough. W There we go. Curious our our approach to prove that commission of ws dimension is secretive. Dimension off be if if we have tees off.

So we have these linear transformation tea from a vector space v T o r n, and we have, ah interpret well where the finding. I mean a product on V given by the brother of you against V is by definition to you dot tv. We remember that the dot there is in a product that standard in the product inside our end because the U. N TV are factors in our end and so we just need to check that he's in a product against your V defined in the vector space. Big B is actually the product, so we need to check the four actions. So what are we, the 1st 1? Which is the symmetry? So we want to check that V against you is by definition tv dot to you and these by the symmetry off the dot product in rn. This is TV at you. Sorry, not TV and all these by definition you against me. So dysfunction on V is symmetric they only to try that it's leaner. So x plus y against a vector V We want to check the disease X against being plus wagons being so we use the definition of the inner product. Well, these in a product t x plus y dot d y. And now again by linearity off T and by the charity off the dot product. In our end, this becomes tx dot tv bluff the y dot TV And now again, by definition, did his ex against v blast. Why against me Now for the third property we need to check it is or more genius So CX against V. It is, by definition, tee off, see axe dot tv. But now both t and dot product are linear. Well, that I'm a genius. So this becomes C t X dot tv and again, these by definition is c times X against me. And finally, we need to check that the dot product, while the inner product is positive definite off. We need to change X against ex. He's got their equal to zero so exciting. Sex is by definition, tx dot tx. And now because dot is another product on our end, we know that this is definitely great unequalled in zero. And also this can only be zero if t x zero because the season in the product So did your brother have something against itself can be zero only director itself. Zero. But now remember that tea is want one. So the only victory gets within the vector in V that gets maps to the zero vector. Our end zero is zero factor in we and that means that X must be zero. And therefore we have checked for actions showing that isn't a problem is indeed

So we need to verify that this is a valid inner product in riel. Two dimensional space V one, W one minus V one, W two minus V two W one plus four. Interesting. The two double you too. I was thinking ahead and expecting it all. All the coefficients to be one. Okay, so we need to do the inner product of V and V, which is going to be V one squared minus be one V two minus V to V one. Um, plus four V two squared. So I'm going to put this in a in dez, most dot com graphing calculator. Um, I am going to simplify it a little bit. V one squared minus two V one V two plus four V two squared. All right, So I am going to tape in. I'm gonna put in acts instead of ah v one. And why, instead of V two. So this is X squared minus two x y plus four. Why squared? And then I'm gonna tape in is greater than zero, and I see that it fills up the whole space on the rectangular coordinates system fills up everything except the origin, which is Of course, the point V one equals zero V two equals zero. So the inner product is zero. Only when V one equals zero and V two equals zero. Um, which he is allowed. Um, I don't know how this just happened, so I'm gonna scroll back up, okay? I'm gonna put equal zero. Oops. Actually. Gonna put greater than or equal to zero, and it will fill in that that point. No. Yeah. Okay, So that holds to property, too. So we need to start with. Ah, the inner product of V. And you is the ones you one minus V one. You to minus V to U one plus four v to you too. Now, using the community of property, I can just write the use first, and I'm going Teoh rate the's in opposite places. Says u one V two. Whoops. Come on. Um, minus. I just noticed there's minuses here, Um, you to be one plus for you to be to And so that is the same is the inner product of vector you and Vector V. And so that is confirmed. All right. Three k, um, you v All right, well, that's going to be the same. That's going to be okay. You won V one minus K U one V two minus. Okay. You to be one minus four k. You to be to I can factor out the k you won. I don't know why I wrote vector on top of those, be one minus u one V two minus YouTube v one minus for you to be to. And that is the same as okay. Times the inner product of you and V that holds four on this one. We have to have you and the and the inner product of that with W. So that's going to give us you one plus V one. So adding the two together times W one minus you one plus V one times w two minus you two plus V two times w one minus four you too, plus V two times W two Ah ran a space W two. All right, so we can distribute the double use and group all the used together u one w one minus u one w two minus you to w one minus four. You to double you too. So I did distribute the four also, um, and then we have all of the V terms going to run out of room. So I'm gonna start writing it here. Plus be one W one minus the one w two minus V two w one minus four. Is that a minus in the end? Yes. Mystery plus plus plus Plus plus plus Ah V two double. You too. And so that is the same as you. The inner product of you and W plus the inner product of you. No v and W and that whole


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