let-w-be-the-set-of-all-vectors-of-the-form-shown-where-a-b-and-c-represent-arbitrary-real-numbers-3

Question

Let $W$ be the set of all vectors of the form shown, where $a, b,$ and $c$ represent arbitrary real numbers. In each case, either find a set $S$ of vectors that spans $W$ or give an example to show that $W$ is not a vector space.
$\left[\begin{array}{c}{a-b} \ {b-c} \ {c-a} \ {b}\end{array}ight]$

Daniel Pezzi · Accepted Answer

Hello. For this question, we need to either show that w can be written as this span of a set of vectors or show that it can. It cannot be so. Our eventual goal is toe have W as the span of some set of vectors. So when they use V, you and W But we know that if this were to be true, there would have to be three vectors here because we have three arbitrary constants. A be and say now, Ah, remembering what the definition of a span is is just all vectors of this form, and I&#x27;m gonna, uh, use the constants given in the problem. So if w can be written as this span of some set of vectors, they must be able to break this form into the sum of three vectors multiplied by the constants A b and safe. So I eventually want this, and now I start eyeballing If I look at this, uh, this vector in the first lot, I have one times in a so that means I have to put a one here. I have no A&#x27;s in the second or fourth slot, but I have a negative a in the third slot. Now, here on the second vector, I have a negative B in the first slot, positive being the second slot. None in the next two. On following the same procedure, I have zero in the first lot, a negative, see in the second and then sees in the 3rd and 4th slots. So you&#x27;ve successfully broken up this form into an arbitrary, constant times a vector which Aiken now label Well, V was was my first one. A na arbitrary, constant times a second vector, which I can now relabel w then arbitrary, constant times. The third vector, which I can Now, uh, this is you. And then I can label this as w So we have shown that this, uh, this set is actually the span of a set of vectors v u and W, which are 10 negative. 10 negative. 1100 and zero. Negative. 111 And we&#x27;re done

Viswesh Uppalapati · Answer

in this video and we selling problem number 16 of section 4.1. And the problem gives us A said W of the given form below. And it says that A, B and C represent our very real numbers. And it also is in each case of whether to find a set of vectors that spans W are given example to show that w is not, um, vector space. So there is an easier way to do this problem. First, we need to find out if they&#x27;ll be really is a vector space. Before we could find a possible example that fits and works for everything and to determine that it has to fit three cases. One is that, um it has to contain the zero vector. So nobody has to contain the zero back there. Ah, which is just 000 on this case. It has to be closed under scaler additions. Clothes under scaling multiplication and tow prove that it&#x27;s not a vector space. We only need to prove one of these wrong. So if one of these is wrong, then we know that the W is not a vector space, so there&#x27;s an easy way to do this. All we have to do is that the choose A equals zero and vehicle zero&#x27;s A equals B equals zero, and we can see if, ah, zero like there really is then this set. So we got 000000 It was 100 here. I just put the this that said definition into its individual components of every variable. So the first column is just a CZ. The second is the B uh, variable, and it&#x27;s coefficients, and the 3rd 1 is technically see. But it has. No, it has no variable so within it. So if you choose a equals B equals zero, we can plug in the values. The first column will be all zeros, the second column movie zero Vector. But the third column has a one that doesn&#x27;t change to a zero, so the resulting vector is 100 So we know that this set will never have the zero vector. So it is not a vector space

Viswesh Uppalapati · Answer

in this problem. In this video, we&#x27;re gonna be selling problem number 18 of section 4.1. Um, here we&#x27;re given Ah, the definition of a set. Uh, W which contains the vectors of the form, all vectors of the form that is given to us with A B and C representing arbitrary real numbers. And it asked us to find a set of vectors s that spans W. Or it also is approved that W&#x27;s not a vector space. So first, we would want to split these in tow every single variable or its components. So here the a component would be eight times before 01 negative too. Plus B times 301 zero plus C times +00 11 So these air these added together would give us the said definition right here. And and we know that, um, since any of accurate over you in in the set w can be written as this set of linear combination. So let&#x27;s call this just V instead of abuse to avoid confusion. But since any set, uh, are any set of vectors, V would be in the form of this linear combination to find right here, which is the same thing as the definition right here. We know that if we choose a equals B equals C equals one, these individual vectors should span the set W so we know that s that Spans. Um said w will be 4012 3010 and 0011 because, ah, a linear combination of these individual vectors are multiplied by any variable A, B and C would just would also be in the set over you. Therefore s is a set of vectors that spends W.

Daniel Pezzi · Answer

Hello. For this question, we need to either demonstrate that the set W which is the set of all vectors of this form, where nbr arbitrary real numbers is a vector space. Or we need to show that it&#x27;s not, um, swell or it&#x27;s going to be that it&#x27;s not. And the the key issue here is the second slot, which is fixed at four, which means that we don&#x27;t have a non arbitrary constant in the second slot. So all vectors in the space need to have four in the second slot. But if vector spaces are closed under multiplication by arbitrary constants. So, for example, if I have the vector zero for zero, which is within W, I just said and be equal to zero. Um, what we call this vector V just for convenience. And I multiply the by two when I get the vector 08 zero, which is not in W, which is which demonstrates the W is not a vector space because it is not closed under multiplication by arbitrary constants, and that&#x27;s it

Abdul Vahid M · Answer

So here. Tell you with Yeah, Comma, be here. Let that be a stripper, Seiko. And that you bigots, Chilies Utama. Exactly like this. Cuban? No, you have to show you. Let&#x27;s that&#x27;s yes. I mean this This that&#x27;s this associate. So because take me, Alexis can say a 10 state of the question Do Plus people. So you rusty, you get like this. Let&#x27;s be Are yet you are conceited. Be Let&#x27;s every he ended. Okay, you can get out of this. Let&#x27;s see Stevens through this the best big is did the director. You know this is going so give me an army. A planet&#x27;s played. It is up a corrective additional. This is no, They&#x27;re still going up a steep second. Thank you. On the 80. From here to the end. No, c c I come back together. Secrecy and the principles of the name submission. Everything&#x27;s do this. It&#x27;s no that is it. Yeah, beans This about you. Let&#x27;s see, Best in. And that&#x27;s the actual guts to beat this. So I really think I used the concept of associative you. So see, if you so But this year so basically you can&#x27;t I just there? Let&#x27;s see, because it his hand

Jason Li · Answer

I want to prove U minus p is equal to U plus minus fee for factors. You and factors be so we have you might this mean is equal to to be minus and now, using our definition of vector subtraction, we have this equal to hey minus d c he dynasty. We also noticed I sense a c and e t r real numbers he had. This is equal to a plus. Negatives, bus negativity. But then this is just addition. So we have This is equal to you. Okay, Plus negatives. But we also noticed that this negative is nothing but a negative one. So we have a baby negative one times negative one time, Steve. And using our definition escape their multiplication, we can bring them negative one out. So we have a baby, plus maybe one times and bringing this up here we have. This is equal to replacing a B C d. Lift you and B, we have new plus negative one times. And this is equal to for us. Thank you. Be so following a chain of inequalities for the qualities we have. Therefore U minus fee is equal. You did

Problem

Let $W$ be the set of all vectors of the form sho…

Problem 17 Easy Difficulty

Answer

$W$ is a span of set $\left\{\left[\begin{array}{c}{1} \\ {0} \\ {-1} \\ {0}\end{array}\right],\left[\begin{array}{c}{-1} \\ {1} \\ {0} \\ {1}\end{array}\right],\left[\begin{array}{c}{0} \\ {-1} \\ {1} \\ {0}\end{array}\right]\right\}$

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

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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$W$ is a span of set $\left\{\left[\begin{array}{c}{1} \\ {0} \\ {-1} \\ {0}\end{array}\right],\left[\begin{array}{c}{-1} \\ {1} \\ {0} \\ {1}\end{array}\right],\left[\begin{array}{c}{0} \\ {-1} \\ {1} \\ {0}\end{array}\right]\right\}$

Linear Algebra and Its Applications

Heather Z.

Caleb E.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Caleb E.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications