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Let $V$ and $W$ be vector spaces, and let $T : V \rightarrow W$ be a linear transformation. Given a subspace $U$ of $V,$ let $T(U)$ denote the set of all images of the form $T(\mathbf{x}),$ where $\mathbf{x}$ is in $U .$ Show that $T(U)$ is a subspace of $W .$

$T ( U )$ is a subspace of $W$

Calculus 3

Chapter 4

Vector Spaces

Section 2

Null Spaces, Column Spaces, and Linear Transformations

Vectors

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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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.

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