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Let $V$ be a vector space over a field $mathrm{F}$ such that $V=V_{1} oplus V_{2}$ for subspaces $V_{1}$ and $V_{2}$ of $V$. If $f_{1}$ and $f_{2}$ are bilinear for...

Question

Let $V$ be a vector space over a field $mathrm{F}$ such that $V=V_{1} oplus V_{2}$ for subspaces $V_{1}$ and $V_{2}$ of $V$. If $f_{1}$ and $f_{2}$ are bilinear forms on $V_{1}$ and $V_{2}$, respectively, prove that there is a unique bilinear form $f$ on $V$ whose restrictions to $V_{1}$ and $V_{2}$ are $f_{1}$ and $f_{2}$, respectively.

Let $V$ be a vector space over a field $mathrm{F}$ such that $V=V_{1} oplus V_{2}$ for subspaces $V_{1}$ and $V_{2}$ of $V$. If $f_{1}$ and $f_{2}$ are bilinear forms on $V_{1}$ and $V_{2}$, respectively, prove that there is a unique bilinear form $f$ on $V$ whose restrictions to $V_{1}$ and $V_{2}$ are $f_{1}$ and $f_{2}$, respectively.



Answers

Suppose $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\}$ is a linearly dependent spanning set for a vector space $V .$ Show that each $\mathbf{w}$ in $V$ can be expressed in more than one way as a linear combination of $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}$ . [Hint: Let $\mathbf{w}=k_{1} \mathbf{v}_{1}+\cdots+k_{4} \mathbf{v}_{4}$ be an arbitrary vector in $V$ Use the linear dependence of $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\right\}$ to produce another representation of $\mathbf{w}$ as a linear combination of $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4} . ]$

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 we're trying to show that this vector is orthogonal to W. And we know that Alfa is given right there without formula. That's the angle between them. Um, so V minus Alfa W died with W, so I can distribute those. So that means V W minus Alfa times W times W. And we're seeing if that equals zero, we're going to show that it does equal zero to be orthogonal. So now it's plug in the angle. Here we have that angle given to us. So v dotted with W minus Alfa. That's this. That's V dotted with W over the magnitude of W squares. And we still have w and w Let's keep simplifying. This means, um, be dotted with W minus v dotted with W and the magnitude square does the same thing as W times W. And we still have w times dub you out here. Keep simplifying. Um, W times w and these cross out. So we're left with v times W minus B times W. And anything minus itself does equal zero. Therefore, they are orthe ago

Problem number 66 from the given statement, the conditions employ that F minus G is a traditional and source free, but this can happen with non constant vector fields. For example, for F equals X squared minus y squared negative two X y and G equals two x y and X squared minus y squared. We have the cross product off Colonel, and death equals the cross product of the Kern and G. And could the project F minus G equals zero or F minus G, his a reputation and source tree? But both letters do not live for by a constant. Clearly hence, the given statement is Schwartz.

So I need to prove that you the inner product of you with V plus double you is the inner product of U and V plus the inner product of you and W. Okay, so this is very much like, um, Property three, which says that, um So I'm gonna write property three. I am going to write V plus W comma. You equals you w plus V w. Okay, well, this is the same as you, V plus W because of property, too. So Property three is here, okay? And one property, too is here. And then we can use just, um this is supposed to say you up here. Oh, my goodness is supposed to say that's re do this. I think that is very wrong. Um, you double you. All right? I wrote this completely wrong. It's supposed to save you and w you all right. So let's let's keep going. Um, Then we can use property to to just switch these two around. So that's going to give us you the plus. You w Okay. So, again, you could write the's just like last time in a different order, probably starting here and then writing that and then writing that and then writing that


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