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Let $u=(2,3,1), v=(1,3,0)$, and $w=(2,-3,3)$. Since $(1 / 2) u-$ $(2 / 3) v-(1 / 6) w=(0,0,0)$, can we conclude that the set ${u, v, w}$ is linearly dependent over ...

Question

Let $u=(2,3,1), v=(1,3,0)$, and $w=(2,-3,3)$. Since $(1 / 2) u-$ $(2 / 3) v-(1 / 6) w=(0,0,0)$, can we conclude that the set ${u, v, w}$ is linearly dependent over $Z_{7}$ ?

Let $u=(2,3,1), v=(1,3,0)$, and $w=(2,-3,3)$. Since $(1 / 2) u-$ $(2 / 3) v-(1 / 6) w=(0,0,0)$, can we conclude that the set ${u, v, w}$ is linearly dependent over $Z_{7}$ ?



Answers

Linear combination Let $\mathbf{u}=\langle 1,2,1\rangle, \mathbf{v}=\langle 1,-1,-1\rangle$
$\mathbf{w}=\langle 1,1,-1\rangle,$ and $\mathbf{z}=\langle 2,-3,-4\rangle .$ Find scalars $ a, b,$ and $c$ such that $\mathbf{z}=a \mathbf{u}+b \mathbf{v}+c \mathbf{w} .$

In this exercise were given a set of linear dependent vectors in R N Let's start out by writing just what it means for this set of vectors to be linearly dependent, it means is that what this means is there exists scale. Er's x one. Let's go with C one rather see to and C three, which are not all zero such that the following linear combination is satisfied. Where we take C one times V one plus C two times V two plus C three times V three and this will be equal to the zero Vector where we said not all of these vectors air Zero. And that means that this vector equation has a nontrivial solution. Let's inject a new element into this situation. Let's define T, which will map from the vector space are in into RM. So that t is linear here. The most important statement about T is that this transformation is linear. So now we wonder about that are set of vectors that could be the images of tea that would be specifically TV one TV too, and TV three we know that the set of vectors v one v two v three is already linearly dependent. So the question becomes, What about this set? Should it be linearly, independent, linearly dependent? Or do we have not enough information to tell? Well, one way to determine the outcome here, using our transformation T is to start back up at this equation here. Let's apply the transformation t on the left hand side. So that's tea of C one times V one plus C two times V two plus C three times V three. Then on the right hand side will use the fact that since T is linear t of the zero Vector is still the zero vector so we can write in the zero back there here. Well, this equation so far, it's giving us closer to determining if this set is linear, independent or dependent. But we haven't gone further enough. Let's begin using more properties of linearity. Linearity allows us to write this left hand side here in the form of C one times T a. V one plus C two times TV, too, plus C three times t at V three and this is set equal to zero vector. Sylvie in your itty allows us to break apart the vector addition as well as the scaler multiplication so that we get to this new equation. Now let's recall what we said about C one, C two and C three. We said that they are not all zero, and here they are in this equation. So that means that this vector equation has a non trivial solution. And it's also a homogeneous vector equation because the right hand side is equal to the zero vector. Now see what it means for a set to be linearly dependent. This set is linearly dependent now because we have a nontrivial solution to this equation, and so we can now state our conclusion TV one TV, too, and TV three as a set is linearly dependent. So the punchline to this example is that linearly dependent sets are mapped into linearly. Independent sets provided our transformation t is linear now, sadly, we might want to extend this. What if the set was linearly independent? Well, tragically linearly. Independent sets could still be mapped to a linearly dependent sets. So the statement here that this set is linearly dependent on Lee works for dependence

I think this problem We're given a set off the nearly dependent vectors. We want me to be free and report. We have to solve this. We're gonna use the humans. Let's try it back to w uh, using an arbitrary number K. So W'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'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'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'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'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.

Before we get started into this problem. I do want to say that this is a proof. Um, so if you're not comfortable with proof writing yet, or maybe you haven't taken improves class P. Pardon me, please feel free to leave comments below in this video. I'm happy to explain any of the steps. Um, if you're not comfortable with these kinds of concepts yet, So we have a very interesting question here, we're told, Can we find 22 dimensional vectors which we call you and V, And we know that their subspace is of our three such that the intersection of these two vectors is a singleton zero. So, essentially, can we prove that we have these two vectors, and when we take the intersection, do we get zero? Meaning are they literally independent? So let's go through this proof, and we're also going to be utilizing a theorem from your textbook as well. So to start this proof, let's first assume that these vectors do have the dimension to like we're told. So let's assume that that that the dimension of you is two. And the dimension of V is to where we know that these vectors, U and V are sub spaces of our three. Okay. Additionally, we can right sets of vectors that are the basis of both of these vectors U and V. So let the set containing V one and V two and the set containing U one u two the basis of V and you respectively. So what do we want to show? Well, right now, we would like to see that the intersection of U and V a zero And now because of that intersection being zero and the fact that we have a basis, we know that B one and B two in this set and this set containing u one u two are linearly independent. However, when we combine these vectors or these basis vectors into one set, we know that they are not literally independent from a theorem in your textbook. So we have that the set containing U one u two, V one and V two is linearly dependent, and that is the part in this proof that we need to prove so. First, let's pick some scholars. Let's let Alpha pardon me. Alpha beta and gamma being are that just means that they're scholars and we want to show that we some way somehow don't have a trivial solution that are that these scholars are not zero that's going to show that this is not linearly independent. So we're going to say you want equals Alpha, you two plus beta V one plus gamma V two, where we know that all of these scholars are non zero. So we know that there is at least one scaler that isn't zero. That's what this means right here. So now we can just simplify. What do we get from this? You one minus alpha. You two equals beta V one plus gamma V two. So further we get that you one minus alpha. You two equals beta B one plus gamma V two, which we now know is in the intersection of U and V. But remember, we have these scholars who are at least one of them is non zero is nontrivial, So alpha Pardon me, you one minus. Alpha two is non zero, which is not equal to beta V two, plus a gamma V two and you one minus. Alpha U two is in this intersection of you and be So what does this mean, what did this tell us? That means our scholars are not all zero. That means that our intersection does not equal the singleton zero because these are not literally independent. We would need every scaler to be zero if they were literally independent. So what we just showed is that the set containing the basis vectors you want you to be wanted me to is literally dependent. And that is the end of the proof. I hope that this made sense to you. And I hope that you can apply this to your current and future studies in linear algebra and possibly even proofs again. If you have questions, please feel free to put them in the comments. And I'm happy to answer any of them.

So we're given the four vectors U B W NZ, and we want to find scaler, Xabi and see such that Z is equal to au plus B B plus C w. So let's go ahead and first plug in u V and W so v you will start up here. This is going to be so It's a times one to one and then we're gonna have plus b times be, which is one negative one negative one and then we're going tohave w so 11 negative one. So that's going to be the right hand side of the question. So now let's go ahead and simplify this down to remember scale er's When we multiply a vector by one, we just multiply each of our components by our scaler. So we're gonna have a to a a and then we're going to multiply be by each of these components here, so it's gonna be be negative, be negative b and then we're gonna go ahead and do the same for W silicate w o picture. This should have been see right here, not w So this should also be see like it. See, see, negative c all right. So now let's go ahead and add each of these components. Remember, we're gonna do a plus B plus C, and that's gonna be our first component here because we add vectors component wise. Then we're going to do the second components. All added together. So it's gonna be to a minus bi plus C, and then we'll do the last component. So a minus B minus c. So now this is supposed to be equal to Ze and XE is supposed to be the vector tu minus three minus or, well, no. So we can set each of these components equal to each other. So first we can have to is equal to a plus B plus sees all right that off on the side here a plus B plus C and then will have negative three. Is he go to the second component. So that's gonna give us to a minus bi plus C, and then lastly will have negative four is equal to a minus. B minus C a minus B minus E. So now we have a linear equation with three variables and three equations, so we can just go ahead and add and subtract thes in some way to solve for A and B, it's actually, I'll call this equation one equation to an equation. Three so actually knows. If we add Equation 12 together, this is going to give us. So it looks like negative, too. Uh, maybe I shouldn't actually put equal here, but negative too. And this is going to be equal to to a cause notice. Here we have the negatives. So when we add those scare those just cancel out. And so then that's going to imply that a is equal to negative one where we divide inside by one. All right, so we have that now and then Let's just go ahead and actually replace a with negative one. And doing that would give us all do it in actually the 1st 2 equations, because doing that would then allow us to just solve for the others. Flu. If we replace these 1st 2 equations, so this is gonna be two is equal to negative one plus B plus C, and then the second equation is gonna be negative. Three is equal to negative two minus B plus c. So now we can go ahead. Add one and I had to over. So that's going to give us three is to be plus C and then down here that's going to give us negative one is equal to negative B plus C. And so the reason why I want to add those two equations together, our plug it in for that because now, if we had these together, the bees cancel out. So we'll get to is equal to to see. And that's going to imply that C is equal to what? Let me go ahead and start boxing these just so we don't lose them. So right now we have a is negative one c is one, and then, lastly, we can just go ahead and plug and see. And I'll just do this into this equation right here and green. So that's going to give us three is equal to B plus one, so we subtract one over and that's going to give us. Two is equal to be. So are three scale er's that we're going to want to use are negative one in front of you to in front of B and negative one infirm see. So if you were to multiply each of those vectors by the scale er's Adam up, that should give us Z


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