suppose-leftmathbfv_1-ldots-mathbfv_4right-is-a-linearly-dependent-spanning-set-for-a-vector-space-v

Question

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 $\mathbf{v}_{1}, \ldots, \mathbf{v}_{4} . ]$

Doruk Isik · Accepted Answer

I think this problem We&#x27;re given a set off the nearly dependent vectors. We want me to be free and report. We have to solve this. We&#x27;re gonna use the humans. Let&#x27;s try it back to w uh, using an arbitrary number K. So W&#x27;s came one of me One plus get to be too plus k for before. What we know is this We know that this be Zarley nearly depend by so they depend on each other so well, let&#x27;s assume not be one is equal to, uh, some constant A to B two plus a three d three plus a Ford We pour hear Keyser arbitrating And here a czar Arbitrary. So let&#x27;s write to be one in 31 in terms of B to B B m w will be, um, a one a two we two plus a three d three plus a four before plus K to be two plus three v three plus cape or we fall. So then w will be equal to a one a two plus que tu we to plus one a three plus three Well, three plus J won a four plus four me for So since case and a CZ arbitrary. And since we can, right, let&#x27;s say V two in terms of We want me three Emmy for orchid Correct. Be three in terms of other valuables are directors. It means that there&#x27;s more than one way Jax Prospector read, So Dan W. Can be expressed as a linear combination off their vectors, we warn me to briefly and in numerous ways.

Donald Albin · Answer

The question reads, if vector one in vector to our vectors in a vector space V and you want you to and you three are each linear combinations of them. So I&#x27;m going to write Vector you one is a linear combination Vector V one and vector V two. Likewise Vector. You too is a linear combination of the two V vectors and vector you three is a linear combination of the two vectors. Prove that this set is linear ley de pendent. Um, So what I&#x27;m gonna do is I&#x27;m going Teoh solve you the first equation for V two. So it&#x27;s going to say V two equals You won minus a V one, Overbey. And now I&#x27;m gonna substitute that into the second equation. You, too, equals C V one plus de V two, which is you won minus a the one Overbey. All right, so now I have you to related to you one and V one. So, no, I&#x27;m going to take the 3rd 1 and solve it for V two. You three minus E V one, you three minus eveyone over F. But now these two are the same. So I can write. You won minus a V one over B equals you three minus E V one over f. And so I&#x27;m gonna keep going with the red one now. And before I continue, I&#x27;m going to think about Does this make sense? I believe so. So I&#x27;m gonna solve this for V one. So first. Well, I&#x27;m gonna cross multiply, so it&#x27;s gonna be f you won minus F A V one equals B. You three minus b e view one. Okay, so let&#x27;s great this again. I gotta be careful saying that 1st 1 f vector you, um minus B. You three glows F a V one minus b E V one. All right. And now we get f times vector. You being careful there minus bu three over f A minus. B e equals V one. All right. Now I can take this and substitute it into this. So it would say you to equal c V one plus de Overbey. You one minus a V one. But now we know what V one is. It&#x27;s f vector. You won a minus. B, you three over f A minus, bi. Hey. Okay, let&#x27;s keep going. Vector. You too. Uh, constant times V one. I gotta put that in there. Also vector you hips CV. Okay. Plus de Overbey. You minus D a Overbey times f A minus B f vector. You one minus b. You three already getting there? This is supposed to be a one here, making sure that I&#x27;ve got ah, everything written correctly. Okay, Vector. You too equals C over f A minus bi e f vector You one minus see Be over f A minus bi e vector You three plus de Overbey vector You minus de a a half over b times f a minus bi you times vector you one thus d a b o dab over be Oh, the B cancels out So we&#x27;re not gonna dab anymore. Uh oh. Was gonna say we could dab at the end of this, but no, the dad is gone. A que Such is going to be d a over f a minus bi Ah, the a over f A minus bi e you three. So you two is just combining like terms C f over f a minus bi, do you? Plus do you Overbey minus D a f over B f A minus. Be name you want mine s or you know what I&#x27;m gonna do? Plus, um, notice that the U three terms have a common denominator already. So this is just gonna be de a minus C b Cool. So you to is a linear combination of you want and you three. Ah, because this ends up just being an unreal number, and this ends up just being a real number, and so they are linearly dependent. Um, I&#x27;m just interested to see this whole thing here. What does it become? Because, um, we could write it as c f be over f a minus bi me be minus no plus de f a minus b d e over f A minus bi e be minus d a f over b f A minus bi e. Okay, so if we group this together, the denominator is be times f a minus bi e and the numerator cfb plus d f A minus b d e. I would have thought a B would cancel out here minus d efs. Ah, interesting C f b plus de if a minus B d e mine is d A f and so we cant cancel out to be. Can we cancel out an F A minus bi? No. Because that first term, neither has an FAA in it, nor a be in it. So I guess that&#x27;s what the first term would be in its not very neat. All right, It is possible. Maybe even likely, that I made a mistake in here somewhere. But the point remains, even if somehow I did make a mistake. Um, they are a linear combination of each other You to is a linear combination of U one and U three, and so that set is linearly dependent.

Jimmy Yao · Answer

yes, So here. So we want to argue. That&#x27;s so B one b two B three before it&#x27;s Kenya independence. So first we consider the linear combination Nike. So let&#x27;s do a one B one plus a two b two purse. His dream. This we just a four before it&#x27;s the country barrel and we consider. So if they&#x27;re leaning into pensions, the wisher conclude that all the, uh or the A&#x27;s should be zero. So the first thing is, if a four, it&#x27;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&#x27;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&#x27;s minus a one B one us into B two is the three is because it&#x27;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&#x27;s the condition. Tear us before it&#x27;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.

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.

Donald Albin · Answer

So the question reads that vector V one in Vector V two. The set is linearly independent, And, uh, I wrote that down in a vector space V and let Vector V equal, Alfa V one plus V two and Vector W equal V one plus Alva V two, where Alfa is constant Use definition 4.5 point four to determine all values of Alva, for which the set of Vector V in Vector W is linearly independent. So I took notes on that. I want to look at definition 4.5 point four. So let&#x27;s find that, um, moving through the book definition, 4.5 point four says in its on page to 86 to 86. Ah, finite, non empty set of factors like we have right here. In this case, it&#x27;s ah V and W in a vector. Space V is said to be linearly independent if and only if. Oh, never mind. I got ahead of myself. If the only values of the skill er&#x27;s c one c two done a Nazi K for which you add them up. So it&#x27;s saying see one vector one plus c two vector to does not equal the zero factor. The only values of those scale er&#x27;s for which that is true are see one equals C two, which equals zero. Okay, well, let&#x27;s think about this. So what I&#x27;m going to do is I&#x27;m going to use these statements and, um, I&#x27;m going to, ah, rearrange them and, ah, solve them for V one as a function of V and W. So, first of all, I&#x27;m going to solve each of them for V two and then use substitution. And so the one on the left is going to say that V two equals the minus off Ah, v one. And now that we know what V two is, I can just substitute that into the equation for W w equals V one plus Alfa V two. But V two is V minus Alva V one. Now I want to solve this for view one. So I get double you minus alv. Ah, the equals V one minus noticed that the Alfa distributes in to the parentheses Alfa squared V one and notice that W and V are to be linearly independent. So that means ah, constant like one times w plus another constant like negative Alfa Times V has to be not equal to zero because they are linearly independent. And so I can write that this is not zero. And so this is not zero So v one minus Alfa squared V one is not zero one minus Alfa squared Just factoring out the V one, not zero. We know that V one is not always zero and so one minus Alfa squared is not zero. And when we solve that we get, Alfa does not equal one or negative one.

Rakvi . · Answer

So suppose the A one feel 31 plus your toe Feel free to all the way a pool. Okay, Be all freaky Is zero That is some linear combination off these vectors. Steel free I zero Okay, inside W Then you want to show that all the eyes are in fact zero. That is all the weights are zero. That is with what you want to show. Okay, so let&#x27;s show this. So using the property of linear transformation, you know that a one off P off Leven is equal to p off even everyone. Okay, Similarly, second term is equality off. It will be too similarly all the way up to give it. Okay, this is 1/4 here. I use the property that he respects tradition 60 off. Even we won, plus de off. It&#x27;ll be it all. The way is just equal to tee off. Even we even close it. We do close, you carry. This is 1/4 0 now you know, used effect that de is 1 to 1. So 121 will tell you that if he off a vector is equal to zero, the vector itself has to be zero. So this implies that even we even plus a 22. Thus all the were toe a Cherokee is actually called a zero. Okay. No used effect that the set of even week A is a linearly independent certain side we so linearly independent. So if some combination off revised zero, that means all the weeds have to be equal to zero. So you have shown that the 80

Problem

Let $\mathcal{B}=\left\{\left[\begin{array}{r}{1}…

Problem 20 Hard Difficulty

Answer

So $\mathbf{w}$ is expressed in more than one way as a linear combination of $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ and $\mathbf{v}_{4}$

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

Linear Algebra and Its Applications

Lily A.

Kristen K.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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So $\mathbf{w}$ is expressed in more than one way as a linear combination of $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ and $\mathbf{v}_{4}$

Linear Algebra and Its Applications

Lily A.

Kristen K.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Kristen K.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications