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Which of the following subsets of R' are subspaces? If the subset is not subspace, explain Why.AlL vectors b that satisfy 6 +6 +6+ 6 = 0. (b) All vectors b w...

Question

Which of the following subsets of R' are subspaces? If the subset is not subspace, explain Why.AlL vectors b that satisfy 6 +6 +6+ 6 = 0. (b) All vectors b with b

Which of the following subsets of R' are subspaces? If the subset is not subspace, explain Why. AlL vectors b that satisfy 6 +6 +6+ 6 = 0. (b) All vectors b with b



Answers

Let $W$ be the set of all vectors of the form $\left[\begin{array}{c}{5 b+2 c} \\ {b} \\ {c}\end{array}\right]$, where $b$ and $c$ are arbitrary. Find vectors $\mathbf{u}$ and $\mathbf{v}$ such that $W=\operatorname{Span}\{\mathbf{u}, \mathbf{v}\} .$ Why does this show that $W$ is a subspace of $\mathbb{R}^{3} ?$

So this question we want to find the be cordoned off this vector X in this, using the spaces so only to find he's buying parts in season one. And to that satisfies the situation by six, 2 11 Now, once again, you could do You could solve this on these through would choose something subversion and finds itself wants us to. Or you could just simply perform more operations on this matrix. Seven seats. Now, this is a pretty meet metric because he sues a two by three countries. Now, you should remember that we only have two confidence here. So all you really need is we're just really two by two. So you take this one and this one or this one or this one or this one. We just really need two of these three. Doesn't matter which one take. So I'm just going to take on st. So then one Fine, you know. Okay, Now I want to get from royal preparations to make this identity matrix, so let's take it. Well, one, I've been the AG three times over too. Thank you to go home. So then he gets 93 blustery, which is there? seven plus three times farmers 15 except possibly give you 22 Now 11 plus three times that you just live in. It has to be. The 2nd 1 remains unchanged. Now we will wounds 1 22 over one. And according to one zone ones, let's turn to which was it, huh? Secretary? Making change. Then you need tripped you Only one. What we're going to minus five because of one that gives you 01 Well, answer it is. It's about to. And so minus y times one. Do I respond? Plus five, huh? And then zero minus five times a heart, which is just 25 to now. Be careful. This is one here. So this is your season. One term and it's just He's so cute. Didn't see someone. Is he into Mama's five over too? Say so too. Is it too?

So in this question, he wants us to find equal hoped Inspector excusing the spaces. So what we needed soul is you need to solve for you need to sell Chris a potion. See you. See you. Bye. You wanna sit in? No. You need to do is you need to multiply. See someone substituting time and in your soul This first set of questions and soft Santini's among another way himself It is buying It's me. Okay, Why the other way can solve it By performing very operations such as this matrix he had become the identity matrix. So to do this Well sorry. Three times one plus Bridget And make that two. So that gives you this pictures. One moment's three three times one plus three is there three times Montree which gives minus line plus five points in one sport 717 times street between one and five minutes. 21 years. Your mind 16. Yeah. Then you go. One two. What was going to write you on one street? Who wants a bull closing form 17 and possible and told me to get out is to die one plus three times right too. They're putting into one that is going to be born. Yeah. Yeah. One. Yeah. And so spying seven support entry, which was a tall told my seven should give you What? So that's your answer. So you said one is easy to find You're too busy for

Okay. Use exercise 27 to show that if a and B are basis for W, then they cannot contain more vectors than be so. I put a question to me seven year and I for my response here. One thing to know about math is that matter isn't always about calculations. There are times where you should be using your words. So in this case, I've just written up my answer here, and you could just read it the first. We will prove my contradiction. Suppose that A and B are basis off and it contains more victims than be. And we will deny the number off Victor's off, eh? As a and B the number of factors in B. So by question 27 it say that since a contains more than being vectors, the victors in a must be really dependent, which could Dixon assumption that a is a basis winning that he must be really independent of each other. So I cannot contain more victims than be now to prove that he cannot contain more because in a way, you repeat the same proof. Just switching a with B

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


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