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SoruPuanWhich of the following sratemenrs true?Let V=Pz and W = {p(t) ePz: Degree of plt) is exactly =7} . WIs subspace of V.Let V= R2 and w= { :xis any realnumber ...

Question

SoruPuanWhich of the following sratemenrs true?Let V=Pz and W = {p(t) ePz: Degree of plt) is exactly =7} . WIs subspace of V.Let V= R2 and w= { :xis any realnumber Wis not subspace of V.Let V=R2 and w={l :xis any realnumber x+6 WIs subspace of V.Let V=Mzz and w-([a a,b,C,d are any realnumbers and W Is not subspace of V,d=o&b- 7c=ok.

Soru Puan Which of the following sratemenrs true? Let V=Pz and W = {p(t) ePz: Degree of plt) is exactly =7} . WIs subspace of V. Let V= R2 and w= { :xis any realnumber Wis not subspace of V. Let V=R2 and w={l :xis any realnumber x+6 WIs subspace of V. Let V=Mzz and w-([a a,b,C,d are any realnumbers and W Is not subspace of V, d=o&b- 7c=ok.



Answers

In Exercises 17 and $18,$ all vectors and subspaces are in $\mathbb{R}^{n} .$ Mark each statement True or False. Justify each answer.
a. If $W=\operatorname{Span}\left\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right\}$ with $\left\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}\right\}$ linearly independent, and if $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ is an orthogonal set in $W,$ then
$\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ is a basis for $W$
b. If $\mathbf{x}$ is not in a subspace $W,$ then $\mathbf{x}-\operatorname{proj}_{W} \mathbf{x}$ is not zero.
c. In a QR factorization, say $A=Q R$ (when $A$ has lincarly independent columns), the columns of $Q$ form an orthonormal basis for the column space of $A .$

To true or false were given. That w is the span of X one x two x tree where x one x two extra linearly independence and then V one V two V three is orthogonal and double Do you okay? Does this mean that V one b two B three is a basis for W? Now always make sure you check your definitions. So let's go to page 340. What is north organist set a set of vectors you want up to U P. They say in Oran is orthogonal means that that each pair of distinct vectors are Thorgan all so you I don't you j is equal to zero for every eye not equal j that's between one and p. Okay, So please notice the very important fact that this doesn't rule out the case where all the vectors in my set of zero okay, so two vectors could be orthogonal to each other simply because at least one of the vectors are zero. So this is where the meaning of war Thorgan ality slightly departs from our intuition. Perhaps I mean is 00 orthogonal 210 the pictures, you know 00 is just the origin right here. And then one zero's this vector. The answer is yes, there are dog. No, Because if you take the dot product, you get zero. So the set of vectors where the set, where every single vector is the zero vector is still a North organo set. So you have to be careful. Okay? Just because you have a North organo set doesn't mean that you can declare it is linearly Independence. Okay. And then the linearly dependent said cannot be a basis. So of course. So this statement is false. Okay, this statement is false. I don't need to come up with a counter example. So, some advice for thick for thinking of counter examples. It's usually best to choose the simplest on the most extreme counter example. We need to come up. Where? With a statement. Where? Um, so we need to come up with a case where this statement right here fails. Okay, this is what it means to have a counter example. Okay, well, let's take our three. Okay. And let's w b So let x one x two extra b x will be the standard basis. It's too. It's a lot extra. Next to extreme be the stunt basis, then w is the entire are three is the entire vector space of our three. And there's no there's nothing wrong with doing that. Okay, so and then also, what will we choose for V one? V two v three. So let's take the one b two, b 32 equal zero vector. Okay, so this is the simplest counter example I can think of. And then the most extreme case where the or Thorgan ality condition fails to guarantee linearly in the linear independence is when you know each of the vectors are the zero vector. Okay, so so x one x two extra or certainly linearly independence. Okay, on dhe V one v two V three. Well, they're definitely orthogonal in w right. There are Fogell, are three all right, because I mean, this condition right here certainly satisfied. Right? V one dot ve to zero. We wound up the 30 and vi tue dot be 300 So that's fine. Then we should check that the conclusion actually fails, and it certainly does. Right? So in this case, even though x one x two extra linearly independence. And even though the one V two V three are Fogell and W V one V two V three cannot be a basis for W. This fails right here simply because, you know, this is a linearly dependent Sets. Okay. Now on to question two are were asked if X is not in the subspace w than X minus the projection of excellent w is not zero. So this is true, Okay? It's pretty obvious. It should be pretty obvious, too. That is true, right? But we can't just, you know, we can't just wave our hands and justify it. We're gonna actually have to come up with the proof. No, this comes from obviously the orthogonal decomposition Terram. Right? So you can take any vector X and split. It decomposes into projections w plus some factors that where that is in, you're Thorgan of compliments. Okay, so we're going to use something to do with this. All right, So let let me show you my proof. Well, it's a little bit awkward to work with. You know, this ex not in a subspace. And then this is not zero. Right? So let me show you a little magic mathematical magic trick, if you like. So saying a implies b is completely equivalent to saying that not be implies, not a okay. If I live in London, then I live in England is equivalent to saying if I don't live in England, then I certainly cannot live in London, right? And so this right here this fact, we use quite a law in my all right, I'm gonna use this here. So instead of showing that X not in the subspace w implies that this is not zero I'm going to say, OK, well, suppose that this guy right here is zero. Then I'm going to prove to you that X must be in the subspace. W shouldn't. This is a little bit confusing. Sorry. I'm talking about the same Max here. Okay, so here's what I'm going to do. I want to show that X minus. This equals zero, implies that X is in W. So let's do that. Well, let's call the skies ET from here, right, It's called the C. So Z is equal zero, right? And when I want to show, the ex is in W, Right. So So, um, this is how I'm gonna do it. X is equal to this plus Zed rights. And I want to show the exes and w So Zed is equal to zero. Right? If you rearrange us, you find that set is equal zero. So is that is certainly in w. Hey, why is that? Well, w is a subspace, right? And any subspace is a vector space in his own right. And vector spaces must contain the zero vector. And so w must contain the zero vector as well. Okay, this guy right here is in W as well. Why is that? That's from the or so Colonel Decomposition theory. Right? We say that you can split X into a vector that's in w and effectiveness in the orthogonal complement of w. So this guy started in W No. Eggs is the sum of two vectors that Aaron W. But I just said that w is a vector space. I take two vectors and w I add them. I must stay in w. And so ex isn't w. And that's it. We're done. Okay, now, right? What? I said out clearly line by line. Okay. In the queue, our factories ation ese click. You times are where a has linearly independent columns were told. Okay. Is it true? Is it true that the columns of Q. Forman Ortho normal basis for the common space of a Well, if you go to page 359 where they discussed the cure factories ation, they say the following If a is an n by N matrix with linearly independent columns. All right, then, lips then they can be factored as que times are where Q is an n by N matrix. Whose columns? Right, listen up. Those columns, Forman. Ortho. Normal basis for the column. Space of A They literally just say that. And so this statement is true. Okay, this statement is true, provided that are you know, it provided that we are actually factoring this in this way, right? But that they tell us that we're in a queue, our factories a shin right when a cure factories ation and so the cure factories ation. When you do that, you get a Lynn, you know, matrix that as linearly independent columns split into a matrix whose columns from the north a normal basis for the column space of a times some, you know, upper triangular in vertical matrix. All right, so this is certainly true. You know, it's basically a direct rip off from what they say in page through 59. All right, so we're done now with the questions, but I want to go back to question one to discuss some interesting subtlety. Okay, so, you know, feel free to and the video here. But if you're interested, then listen up. So let's go back to this question yet. The very start X one x two x tree. And then we said x one x two x three are linearly independence and we want 3 to 3 or four, Colonel. Okay, so we found that this does not imply that B one b two B three must be a basis. All right, now, let me change a statement a little bit. Okay? So instead of making this fourth, Organon and w let me make the condition on this set a bit stronger. I a little bit more stringent. Okay, so let's say that I say that this is a set of non zero orthogonal vector's right. The counter example used for this was when we had, you know, least 10 At least a zero vector in here. What if I completely disallowed that? What would happen with this statement? Still be false, Or would it be true, Paul's And ponder this, and then if you think it's true, come up with a proof. All right? The answer is yes. I think this would be true. So this would imply this. Okay, Now, why is that? Okay, So why is that? Well, you know, you can't just you can't just wave your hands and, you know, kind of argue by your intuition. You have to actually come up with a proof using the theorems and stuff that has been derived in the book. Okay, well, why is this so I'm gonna kind of sketched this out to you. So page 340 they're in four this time. Says that if you have an orthogonal set of non zero vectors in RN, then the set is linearly independence. Okay. No, we know therefore, that this sets must be linearly independence. Now, the question is, does it span? Well, the final ingredient you need is theory 12 on page 229. Where it says that Well, okay, this staring right here. Go to it and check it. Right. And in this case, W is a three dimensional vector space. Okay, the dimensional W is three, and we have a linear, independent set of exactly three elements. And w so by the room 12 on page 229 this set right here must automatically be a basis for W. Okay, so since we've realized that B one B two B three is a linear in independent sets, theorem 12 from Page Turner, 29 again guarantees that this must be a basis for W. So I hope this helps you appreciate why we spend so much time carefully proving these serums. Because without it, it is impossible to convincingly argue anything in the New year. Algebra, even if it feels obvious. Okay, so this is it. That's all for now. Thanks for watching

In this problem there's a linear transformation. Greater acting crawlspace be onto space. Definitely. And in space stuff we have subspace, etc. And we're gonna say let you represent all Xing We anywhere trying to prove that you is a subspace No, We're going to say that you won And you two are two elements in space, you and tur. And to you too we know And this problem settling This is given that these Aaron Z since t easily near a praetor concert t you on wants to see you too is t you want waas You too So this your wants you to is all saying z now You want any too, aren't you? So from this we know that you want waas you jer should be in you So this means that you is blows on there back tradition No, we're gonna use you want to you So we know that do you want isn't easy So if you multiply that with any constant and this should be all saying z right And since I can't easily you to information this will be the tee off. See you are And it should be also Z No, you want wasn't see? Sorry. You always you. It means that see, you won should be in you. And from this we see that here is close under back to for protection. So these are the requirements that we need for subspace subspace. So using this, we can say that you is a subspace. Oh, be

Okay, So for this exercise, we need to prove that they're the theorems. Say that if we got a linear mob f defined from the from the said V to you, then the colonel is a subspace of B, and the image of F is a subspace of you. So let's start by the part A. Let's remember that if we want to show that the um for example, the view is a super space of B, then we need to take two elements on the view one W t on the view and two elements on the field. Let's say Kay, and the only thing that we need to show is that alpha of you want plus beta W two is also an element of the subspace. So this is the way to show that I love you is, for example, our subspace of B. We're going to do the same in this case to prove that the kernel of F is a safe space off beat. So our statements say that the colonel is a subspace of V. I'm going to use this symbol to the note, the subspace. So the first one is that it is well known that we need to map the no vector to the no vector on the on the image. So basically, if we take the zero vector here, we're going to map the zero vector on the image of and and and therefore zero is an element on the kernel of F. So that's one thing. Now let's take the general case. Let's take we involve you on the colonel. Mm. Let's take the and all of you on the colonel of the Mob and let's take our and Vita on the field. So because we got that these two vectors, the and those are part of the colonel that implies that f of B is equal to zero, and f w is also equals to zero. So if we take the map to a V plus, Al Fabby was beated of you. We just, uh what? We want to show this by linearity of the function of the mapping. We got alpha F B plus B to F W, but it is just zero plus zero, which is equal to zero clearly. And therefore that means that eight a V plus vita w are part of the colonel as well so Alpha B plus BTW are also part of the kernel of F and therefore is a space of of U of B. So the kernel of F is a safe space of B. So that's the first part. No, we need to check. But the image of F is a subspace of the set you. So let's remember that F is a fine from B to you and that is linear. So again, we need to first check that. For example, the first is we need to check that the zero vector is on the set, so f of zero should be equal to zero. But, uh, we know that we're here. We're mapping the new vector of the Let me change the notation here. So we're taking the no vector on space. V is equal to the nose to know vector on space You. So from this we got that the new vector? Yes, part of the image of f. Okay, because we can map to the new victory. This condition should be satisfied in order to be a winner map. So that's great. That the first thing Now let's consider you a new prime on the image of the mapping and again, alpha beta on the field. So what we need to show is that if we apply f two of, uh, sorry. So you and we are part of the image of the mapping, so that implies that there exists of HIV and re prime such that on the clearly and be such that f of be, it's going to be equal to you. An f of the prime is equal to you, prime, just by definition. Okay, so now we are going to take the map of a Alpha Al Harbi How far B plus be to be prime. Um, by the narrative of the function we got that this is equal to Alpha F of B plus B to F v prime. We have defined this two maps as you and new prime. So there's equals to offer you plus better you prime. This is what we need to show that this part of the image of F So we take two elements that are part of the image and we reach that alpha. You plus Vita. You are also part of the image of the of the function of this map because if we apply due to. They are part of the image that exists. These two vectors on the on the pro image give us this result here and then we just need to apply this map to this vector. We obtain that these are also part of the image, so that implies that the image of F is a subspace of you.


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