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Let W be the set of all vectors of the form shown on the right; where a and represent arbitrary real numbers. Find a set S of vectors that spans W or give an exampl...

Question

Let W be the set of all vectors of the form shown on the right; where a and represent arbitrary real numbers. Find a set S of vectors that spans W or give an example or an explanation to show that W is not vector space_9a 4b8b - 4c9bSelect the correct choice and fill in the answer box as needed to complete your choice0 A A spanning set is S = (Use comma t0 separate answers as needed )There is no spanning set of W because W does not contain the zero vector: There is no spanning set of W because W

Let W be the set of all vectors of the form shown on the right; where a and represent arbitrary real numbers. Find a set S of vectors that spans W or give an example or an explanation to show that W is not vector space_ 9a 4b 8b - 4c 9b Select the correct choice and fill in the answer box as needed to complete your choice 0 A A spanning set is S = (Use comma t0 separate answers as needed ) There is no spanning set of W because W does not contain the zero vector: There is no spanning set of W because W is not closed under scalar multiplication. D. There is no spanning set of W because W is not closed under vector addition.



Answers

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.
$\left[\begin{array}{c}{4 a+3 b} \\ {0} \\ {a+b+c} \\ {c-2 a}\end{array}\right]$

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's not, um, swell or it's going to be that it's not. And the the key issue here is the second slot, which is fixed at four, which means that we don'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's it

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'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'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'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're done

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'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'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's an easy way to do this. All we have to do is that the choose A equals zero and vehicle zero'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'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'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

This time s is the set off all polynomial of the form A plus B X cubed plus C acts to the fourth power. Let me go back here. So I'm going Teoh Call it C X to the fourth power plus be X to the third power. Plus okay, such that A, B and C are really numbers. So I noticed that there's no first order term. There's no second order term. Um, there's also Noah fifth or sixth order turn. So let's think about this a one If we take one solution and we add to it another solution c b a f you de All right, that's going to give me c plus f x to the fourth power plus B plus e x to the third power plus a plus d. And if a, B, C, D. E and F are all real numbers, well, then, um, this Israel and this B plus c. Israel and this Israel. Now it is possible that could be zero. They could be negative. Um, but it doesn't say it has to be exactly. Doesn't say that that the degree has to be four. So, um, the term the coefficient of the X to the fourth power term, maybe zero The coefficient of the exit. Third power term, maybe zero. And, um, the constant may also be zeer. So a real number times X squared was a real number. Times X cubed was a real number. Um so this holds a to if you take a constant times, see excellent fourth power plus P x to the third power plus a that's going to give us K c x to the fourth power plus K b x 1/3 power plus k a. So again, um rial number rial number rial number, which could be zero. But being zero, having this whole thing turned out to be zero is not a problem. Ah, that zero is in the vector set. And so this hold also.


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