5

() Is2 (a) Do Is V1 , {V1, V2 , V2, V3 V3 } ueds 3 linearly Explain independent why set why not. linearly dependent set? Explain:in Span {V1, V2 7 Whywhy not?...

Question

() Is2 (a) Do Is V1 , {V1, V2 , V2, V3 V3 } ueds 3 linearly Explain independent why set why not. linearly dependent set? Explain:in Span {V1, V2 7 Whywhy not?

() Is 2 (a) Do Is V1 , {V1, V2 , V2, V3 V3 } ueds 3 linearly Explain independent why set why not. linearly dependent set? Explain: in Span {V1, V2 7 Why why not?



Answers

Let $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a linear transformation, and let $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\}$ be a linearly dependent set in $\mathbb{R}^{n} .$ Explain why the set $\left\{T\left(\mathbf{v}_{1}\right), T\left(\mathbf{v}_{2}\right), T\left(\mathbf{v}_{3}\right)\right\}$ is linearly dependent.

Mhm. Okay. Um so suppose we have a set of vectors. Let's call this set uh Little V. And you know, it consists of V one, V two. All the way to some. I don't know if the N where n is the dimension. Okay. And I want to show a statement. Okay. Which is that if you know any of these uh n vectors V one through V N is equal to the zero vector. Then this set of vectors must be linearly dependent. Okay, So how will we show that? Well, we call that uh linear dependence. Well, linear independence, the definition of those is just that if you consider all linear combinations, let's say Z one, B one, C two, B two. If you consider the linear function linear combination in the following way, in order for this linear combination to be zero. Uh if uh this linear combination equaling 20 implies that C one all the way to see. And all these coefficients are all equal to zero. Okay then uh you know like then we know that the one through VN are linearly independent. Okay, so if the only possible linear combination, that zero is the trivial linear combination, then we know that they are linearly independent. If a basis or if a set of vectors are not linearly independent, then they are linearly dependent. Okay, so suppose. Okay, I suppose there's some vector V. K. Okay, over here. Okay, that is equal to zero. Okay then what would happen? Well then that means there's some C k V K over here. Okay, that is equal to zero. Okay, because if VK zero, no matter what coefficient multiplied by that, C k V K must also be equal to zero. Okay, so that means I can pick okay, I can pick C K. To be equal to any arbitrary real number. Okay, I can pick C K. Uh to be I don't know like 10, for example, without loss of generality. Okay, I can fix the equal to 10. And this whole linear combination would still be zero, right? And therefore the trivial linear combination where I have all the coefficients being zero is not the only linear combination. That would yield zero. Okay, I can pick any arbitrary C. K. Okay. And still get this linear combination to B. Go to zero. Okay. And this means that the set of vectors would violate uh the linear independence condition, and therefore this set of vectors must be linearly linearly dependent. Okay, And that's basically it.

In this problem, we need to prove that if anyone V two is linearly independent set of vectors and the three does not lie in the span of you on V two, then we one, V two, V three is a set of linearly independent vectors. Now, for this, first of all, note that since he wants me to is a linear independent set of vectors, so we one cannot be written as a scalar product of V two. So that means that everyone is not equal to K times V two for any real number key. Also, it is said that the three does not belong to the span of We want V two. So this means that we tree cannot be written as a linear combination of V one and V two. Now, next suppose that it is possible to write V one as a linear combination of V two and V three. So we should be able to write V one is equal to K two, V two plus K three, V three. Where K two and K three are real numbers. Is that both of them are not equal to zero? Now, if we assume that the three is equal to zero, then in that case we want will be equal to key to be too. And this will contradict the fact that everyone cannot be written as a scalar multiple off V two. So this means that K three cannot be equals to zero and hence K three must be not equals to zero. Now, since K three is not equal to zero, we can rewrite this equation as key three. V three is equal to be one minus K two, V two. Which implies that we three is equal one divided by key three times we want minus key to divided by key three of V two. We can do this since K three is non zero. But here we can see that the three has been written as a linear combination of V one and V two. And this will contradict the fact that the three does not belong to the span of we want to. So this means that the one cannot be written as a linear combination of V two and V three. This assumption is wrong. So in a similar manner, we can prove that we too will not be able to be written as a linear combination of the one and V three. So what we obtain is that the one cannot be written as a linear combination of V two and V three. We do cannot be written as a linear combination of V one and V three and V three cannot be written as a linear combination of V one and V two. We get this from the first part. Over here. V three cannot be written as a linear combination of V one and V two. Later on, we prove that we won cannot be written as a combination of V two and V three. And we also show that we too, cannot be written as a linear combination of V one and V three. From all of these. We can conclude that we want, we do V three is linearly independent set of records. Yeah.

Here. This state's proved that if the set containing vector V one in Vector V two is linearly independent and if V three is not in the span of the vector of you one in Vector me too. Then the set of vectors, v one, V two and V three is linearly independent. So if it's not in the span, then Vector V three does not equal any constant times. Vector V one plus any constant times Vector V two. All right, so Vector the three is not a linear combination of vectors V one and vectors v two. And so we could right that if we want the set of vector V one, Vector V two and Vector V three to be linearly independent. That means that some constant has vector view one plus some constant groups times vectored v two plus some constant times. Vector V three cannot equal zero. All right, so I'm gonna try to prove by contradiction. So I'm going to let d one v one plus d two v two And you know what? I better write this over again. That t one v one Plus it d two v two plus de three V three equal zero for some really numbers, do you one de to Andy three. If that were true, then by subtract d three v three from both sides of the equation I get and then I divide by negative d three on both sides. I would get the three equals negative D one over de theory V one, minus D two over d three v two. And if I, uh, say that d one over D three is some other constant C one and I mean negative, do you? It went over to three. And if I let negative de through to over 83 b some other constant C two then V three is C one V one plus C two v two. But we already said that V three does not equal. See one V one plus C two V two for some C one C two that are really numbers, which is a contradiction. And so that means therefore d one v one plus de two v two plus de three v three do not equal zero the opposite of our initial statement

Yes, So here. So we want to argue. That's so B one b two B three before it's Kenya independence. So first we consider the linear combination Nike. So let's do a one B one plus a two b two purse. His dream. This we just a four before it's the country barrel and we consider. So if they're leaning into pensions, the wisher conclude that all the, uh or the A's should be zero. So the first thing is, if a four, it's Nico Todaro. So if If so, our for Miller reduced to released director right into according to the condition we already know B one B two b sweetie is leaning independents, so we can deduce a one. We go to a to be continued Ansari. It's Kiko Trujillo. So so here, every since zero for the second. So if a four is not Yuko Todaro, so then where happened? So then what happened is if we can write a four before it's minus a one B one us into B two is the three is because it's not there. We can do by the the right hand side by a four. So before its miners, he went over a four. Be one minus into over a four meets U minus a three over a four b. C. So here we can see before kids that spin by, we want to research. That's the condition. Tear us before it's nothing. The span. So this means this condition is not true. So a four had to be zero. So if a four had to be there Oh, all the coefficient had to be there. That this means if we have the linear commission and also coefficient had to be Barrow, then we know B one b two b three before I leave me needing chips.


Similar Solved Questions

5 answers
What is the sum of the following series2 n2 _1" n=2A)372B ) divergesC) 1/203E)1
What is the sum of the following series 2 n2 _1" n=2 A)372 B ) diverges C) 1/2 03 E)1...
5 answers
[ 0 7 Uun 2 8 2 2 2 I [ 1 1 M [ 1 L L L H 2 2 8 2 3 3 8 1 1 MH 8 [
[ 0 7 Uun 2 8 2 2 2 I [ 1 1 M [ 1 L L L H 2 2 8 2 3 3 8 1 1 MH 8 [...
5 answers
Determine an "optimal" 6-month order schedule (i.e-, Z1~Z6) using any available solvers (e.g , Excel solver; AMPL; etc )Month (t) Demand Unit cost Setup cost (Kt) Holding cost (ht _1000 50 20001200 50 2000500 50 2000200 50 2000800 50 20001000 50 2000 10
Determine an "optimal" 6-month order schedule (i.e-, Z1~Z6) using any available solvers (e.g , Excel solver; AMPL; etc ) Month (t) Demand Unit cost Setup cost (Kt) Holding cost (ht _ 1000 50 2000 1200 50 2000 500 50 2000 200 50 2000 800 50 2000 1000 50 2000 10...
5 answers
Caiciccdtcopeptidoglycan cell walls[Choosc ]mitochondria[Choosc ]mcmbranc-bound nuclcus[Choosc ]chloroplasts[Choosc ]ccllulosc ccll walls[Chooscstarch[Choosc ]glycogcn[Choosc ]chitin cell walls'[Choosc ]Ckxtcthal dicestion[ Choosa |
Caiciccdtco peptidoglycan cell walls [Choosc ] mitochondria [Choosc ] mcmbranc-bound nuclcus [Choosc ] chloroplasts [Choosc ] ccllulosc ccll walls [Choosc starch [Choosc ] glycogcn [Choosc ] chitin cell walls '[Choosc ] Ckxtcthal dicestion [ Choosa |...
5 answers
The sum of the elements in the sixth row of pascal triangle is(1) 32(2) 63(3) 128(4) 64
The sum of the elements in the sixth row of pascal triangle is (1) 32 (2) 63 (3) 128 (4) 64...
5 answers
Find the pH (to the nearest tenth) of the substance with the given hydronium ion concentration.Grapes, $5.0 imes 10^{-5}$
Find the pH (to the nearest tenth) of the substance with the given hydronium ion concentration. Grapes, $5.0 \times 10^{-5}$...
1 answers
Generalize (4.5) as follows: given 13 points $P_{1}, \ldots, P_{13}$ in the plane, there are three additional determined points $P_{14}, P_{15}, P_{16},$ such that all quartic curves through $P_{1}, \ldots, P_{13}$ also pass through $P_{14}, P_{15}, P_{16} .$ What hypotheses are necessary on $P_{1}, \ldots, P_{13}$ for this to be true?
Generalize (4.5) as follows: given 13 points $P_{1}, \ldots, P_{13}$ in the plane, there are three additional determined points $P_{14}, P_{15}, P_{16},$ such that all quartic curves through $P_{1}, \ldots, P_{13}$ also pass through $P_{14}, P_{15}, P_{16} .$ What hypotheses are necessary on $P_{1},...
5 answers
And the vectors & = 37 _ 2j and & = 2i 3jpoints) Consider the point d3.What is the distance of P to the lines parallel to & and &, respectively; and going through the origin? b What is the distance of P to the plane spanned by and w through the origin?
and the vectors & = 37 _ 2j and & = 2i 3j points) Consider the point d 3. What is the distance of P to the lines parallel to & and &, respectively; and going through the origin? b What is the distance of P to the plane spanned by and w through the origin?...
5 answers
12. Provide reagents necessary to carry out the following transformations. Several steps may be required and there may be more than one correct answer: (9 pts)13.Provide reagents necessary to carry out the following transformations. Several steps may be required and there may be more than one correct answer: (9 pts)OHHsCo
12. Provide reagents necessary to carry out the following transformations. Several steps may be required and there may be more than one correct answer: (9 pts) 13.Provide reagents necessary to carry out the following transformations. Several steps may be required and there may be more than one corre...
5 answers
(sxueu 01) 'X Oool Je !*0 Sl MquaisSIla pue zW Oool eaje aeuns asoym Kpoq xoejq WoJ} JaMod Jueipej [eJO} a4} aleinoie: osiv 'Mei uueuzliog-uejas 94} ssnosia (q
(sxueu 01) 'X Oool Je !*0 Sl MquaisSIla pue zW Oool eaje aeuns asoym Kpoq xoejq WoJ} JaMod Jueipej [eJO} a4} aleinoie: osiv 'Mei uueuzliog-uejas 94} ssnosia (q...
4 answers
Consider the following vector fieldFlx;y2) Zyz Inxi (9x 10yz) j xz kFind the curl ofF evaluated at the point (9. 1,3).(6) Find the divergence of F evaluated at the point (9. 1,3).
Consider the following vector field Flx;y2) Zyz Inxi (9x 10yz) j xz k Find the curl ofF evaluated at the point (9. 1,3). (6) Find the divergence of F evaluated at the point (9. 1,3)....
5 answers
One of the following is the set of linearly independent eigenvectors for 32 A = -2 2a Noneb20(8) (%_(2} (12)(2} (-15)
One of the following is the set of linearly independent eigenvectors for 32 A = -2 2 a None b 20 (8) (%_ (2} (12) (2} (-15)...
5 answers
Use induction to show that 1 + 2+22 + 232n-1 2n _ 1
Use induction to show that 1 + 2+22 + 23 2n-1 2n _ 1...

-- 0.021206--