let-h-be-an-n-dimensional-subspace-of-an-n-dimensional-vector-space-v-show-that-hv

Name: let h be an n dimensional subspace of an n dimensional vector space v show that hv
Uploaded: 2019-08-08
Duration: 3 min 27 s
Channel: Numerade
Description: Let $H$ be an $n$ -dimensional subspace of an $n$ -dimensional vector space $V .$ Show that $H=V$ .

Question

Let $H$ be an $n$ -dimensional subspace of an $n$ -dimensional vector space $V .$ Show that $H=V$ .

Runpeng Li · Accepted Answer

problem. We&#x27;re given age, which is son and dimensional. Subspace, subspace off. Hey, which is also and dimensional. And where will we want to show ages people to be so first to Christine to notice that that stuff space age as the basics with the in vectors and those in vectors has to be the new independent. So already down here. So that is Asia has a basis. Oh, and electors, because age it&#x27;s in dimensional and more over those vectors has to be leaner. Independent, no. Are we nearly independent? Being dependent now spelled it wrong. Independent Dent. Okay, so Oh, I forgot to say that these factors has to be, you know, the independent in B because H e&#x27;s a subspace off. All right, so now we can apply bassist zero, you know, a textbook, these this serum. So it says by our basic faces, the room. So they, the vectors will constitute well constitute a basis for yes. Well, so factors constituted he two, right? So that means issue. Because why? Because Asian each and b r o r both spend by the same set of vectors. So there&#x27;s no other way to find a dis joint Asian V to make this happen. So this is our conclusion. We just have a vectors, that kind that span, uh, both H and visa. These two spaces has to be his two spaces have to be.

Daniel Pezzi · Answer

hello for this question were given three vector spaces H, K and V where H and K are subspace is of the and we were defining Ah, new set H plus K which is just equal to all elements of the form Little H plus little K, where little ages in big H and Will K is within Big K. Um and we&#x27;re asked Thio show two things first, that H plus K is a subspace of V. And the second thing we need to show that H and K are in fact subspace is off H plus K. Now, both of these facts follow almost immediately from this that HK nvr all vector spaces. If we did not have this thing, these would not be necessarily true. Um, for example, because we know that V is a vector space We know that is closed under addition for any of its elements. And we know that H is with envy and we know that K is within V for any elements Little H and little K in Big H and Big K. Because of this relationship here and so this immediately implies that H plus K is within v again because V is a vector space. And then that is because H plus K is is an arbitrary element of this space. The subspace relationship holds now for this because for sorry for for part B, we just have to notice that because Big H and big K R vector spaces, uh, they each have the zero element in them. It is the case that the zero vector is within H, and the zero vector is within. Okay, so in shot a little bit nicer is within K, which means that in the definition of big H plus Bakay weaken, just pick little H a little K two b zero to get that the other is is a member. So, for example, um so this implies that zero plus okay is within H plus K for all little K thin Big K. And this implies that H plus zero is within H plus K for all little H within Big H on. This is the definition of containment. So it is true that big agent big K R sub spaces of H plus K Um and that&#x27;s all we&#x27;ve been asked to dio That&#x27;s it

Jimmy Yao · Answer

so here since exit seemed the n a y, it seemed be so according to the and linear space we know ex miners Why, it&#x27;s also you be soon in here since the basis is to be one until be in so we can write surly near combination of the one. Me too in all the way to be and into the condition we have is X b one because the I is he going to why the I So this is equivalent to say that x minus Why the I is thinker to there. So therefore, we plug this into this We know a I do you want the I because into the one in the eye First for the wind to a one be, um b I it&#x27;s the culture hero. So in here our i is you go to 12 or the way to end So we can writes this and under questions into the matrix injured this is B B one b one interview one b two Consider this is the one bn in Oh, so we and b I in here is Ian. Ian in the vector is a one a two can be and so you could 200 So? So this is also called the Grand Matrix. In the since the one and you&#x27;ll be in. It&#x27;s a basis of the space. So they securing matrix is it seemed, vegetable. Since its thean vegetables, that&#x27;s means a one and a two in a n have to be zero. So this means X minus. Why is he couldnt zero? That means ecstasy. Could you work?

Ernest Castorena · Answer

okay for this question, they give us the Matrix. A is equal to 11 negative. 112 negative. 35 negative. Six 50 Teoh. Negative three. And they give us two other directors. Negative. 2750 and three Negative. 80 Fine. Okay. What do you do with these? Well, they actually want us to show their basis more than all space. Okay, Okay. Let&#x27;s do some of our technique. Considered him in that. First off, this matrix is three by four. So it&#x27;s a member from there is an r four directors and are what&#x27;s the rank of this matrix? Well, it should be of dimension to its rank. Should be to Why is that? They&#x27;re claiming that this is the basis, so let&#x27;s figure that out. We&#x27;ll have 11 negative. One one, Teoh. Negative. 35 negative time. Zero to negative three. Yeah, these rows. So what can you see at the bottom row is three times the first euro, plus the second round to the end of with three negative. Three on the two young deport, zero on five. Yeah. So this is where done information. We start performing elimination. We&#x27;re gonna end up when my trying to the time in what leads to the zero victor. What&#x27;s then the colonel? We&#x27;re going to end up 11 negative one 10 and then it&#x27;s multiplied by t subtracted from the second room. So end up with negative Fine. We end up with seven and, uh, see multiplying it by negatively. Zero 0000 They&#x27;re up, and we can already see that it&#x27;s a drink, too. So it&#x27;s not busy anymore. With that and get sidetracked. Let us proceed with their goal. It&#x27;s definitely a Brinks you. Let&#x27;s show that these vectors Aaron, are basis. So are there in the kernel. 50 negative three. One negative 15 to negative. Three. A negative six one And let&#x27;s multiply it by the vector. Negative, too. 750 Multiply it out. We actually see that we get 000 on the same for our second rector. We multiplied by the Vector three negative 805 and we end up with the vector 000 the great So they definitely ah are inside the null space, and they&#x27;re definitely literally independent. Why is that? Well, it&#x27;s clear that this isn&#x27;t a scaler multiple of this. If it were, then they come out with the system of equations like, Okay. And three should be equal to a negative to you. Time that scaler negative eight should be equal to seven times that same number. And we can continue for the other injuries. But it&#x27;s already tried to solve for this. There&#x27;s no such T that makes both of these work. Okay, so they are linearly independent there, in the two dimensional vector space, they spend our whole space, okay.

Viswesh Uppalapati · Answer

and this problem, we&#x27;re gonna be solving problem number 10 of section 4.4. Um, so the problem assets, if the given set are all vectors of this given definition, Where to? Where? We have a column vector of two teas there on negative to t. Here&#x27;s some space of our three, which basically just means that, um, it spans are three. So for a given set, to be a subspace of another set, we need to fulfil three conditions. One it has to contain the zero vector to it has to be close under addition and three, it has be closed on her multiplication. So, um, if we just take a vector V at its base form to zero negative to t zero and negative t and we take t to Tia&#x27;s any scaler values. So if we take t to be zero, we got two times 00 negative zero equal, which is 000 which is the zero vector. So the zero vector exists in this set so we can go on to the second condition. Clothes under a scaler are closing tradition. So if you take two vectors in this set where they multiply by a different scale our values So we can have V be to t zero negative t and you be to see zero negative c and we add these two together we get V plus you times are sorry. It would just be replace you. Oh, sorry. We Plus you would equal to to t plus to see you&#x27;re a plus zero and negative t minus Siebel That is just the same thing as adding adding the scaler multiples together t plus c times this vector 201 which are just 20 negative one which is just the coefficients of the initial vectors. So this means that the vector remains the same but the scale of multiple just changes. It just adds and compounds. That means that we&#x27;re still on the same line with varying scaler values and we&#x27;re still within the same set. So it has closed under Skeletor, Earth is closed under addition and lastly, it has to be closed in her scaler multiplication. So if you just do, um, if you just do a scaler scene times the vector V, we got see Time&#x27;s see Time&#x27;s to t zero negative t That is just C Times, T times two 20 negative one Which again, is still this the same coefficients of the set? Um, So we still stay on the same line again? Because I just just the scale of all you were multiplying this by changes. See, times to you just get bigger. But we&#x27;re still scaling the same line or vector. So we stay on the same line. Um, therefore, this is also closed on her skin. Her multiplication therefore, um h is a so space is a subsidies o r three

Michael Jacobsen · Answer

for this example. We start out with age. It&#x27;s going to be a vector subspace of some vector space V, and then we&#x27;re going to make an assumption that we have two vectors U and V, which are also in H. So what we&#x27;re going to do in this situation is verified the following claims. We claim that the span of those two vectors U and V, is contained in H. I&#x27;ll use a symbol subset to say that the span of UV is a subset of age, which is a cinnamon synonym of saying it&#x27;s contained. Let&#x27;s see how to prove this. First, we begin by taking an element. We could call it X if we like that comes from the span. So let&#x27;s say let X be in the span of the vectors you envy. Our goal here is to show that if X came from the span, then it must be in the set. H if we can show that that our work is done and we would have shown containment. So let&#x27;s go on to the next step. Then, as soon as we know that the Vector X came from this man, that means next that there exists scale er&#x27;s I&#x27;ll call them C and D such that X is equal to see Time&#x27;s You plus d times V. The reason we could write that such an equation like this is FX came from the span of U and V. That just means it&#x27;s a linear combination of you envy. And this is one such linear combination now notice. So far we have nowhere used the fact that H is a vector subspace. So let&#x27;s begin doing that next. We know that since H is a subspace of the and we know that U and V came from H, which is also critically important, it follows that See, you and Devi are in H now why is that true? Well, the reason is, if we have a subspace, a subspace is closed under scaler multiplication. And so this is a scaler multiplication of a vector u a vector v which came from H so they must now both b NH as well. Whatever the closure, do we have? Well, another closure property is if we use vector addition now of two vectors that are known to our day b NH then the some of those vectors must also be in age. I will say likewise X which is equal to see you plus DV is in a church. So this was our goal. This shows that the spin vectors U and V are contained in the subspace H, as required.

Problem

Explain why the space $\mathbb{P}$ of all polynom…

Problem 26 Hard Difficulty

Let $H$ be an $n$ -dimensional subspace of an $n$ -dimensional vector space $V .$ Show that $H=V$ .

Answer

In this case $H=V$

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

Linear Algebra and Its Applications

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Samuel H.

Vectors Intro

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In this case $H=V$

Linear Algebra and Its Applications

Lily A.

Catherine R.

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Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Heather Z.

Samuel H.

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

Linear Algebra and Its Applications