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[-n1 Points]DETAILSUWHOLTLINALGZ 4.2.060_If the set {U1" "z U3} is a basis for R3 and 4 = ["1 u3]; what nullity(A)?eBook...

Question

[-n1 Points]DETAILSUWHOLTLINALGZ 4.2.060_If the set {U1" "z U3} is a basis for R3 and 4 = ["1 u3]; what nullity(A)?eBook

[-n1 Points] DETAILS UWHOLTLINALGZ 4.2.060_ If the set {U1" "z U3} is a basis for R3 and 4 = ["1 u3]; what nullity(A)? eBook



Answers

Find a basis for the subspace of $R^{4}$ that is spanned by the given vectors. $$(1,1,-4,-3),(2,0,2,-2),(2,-1,3,2)$$

So to find a basis of a subspace, we just want to see what's the smallest. We can make our set of vectors while keeping the span the same. Now recall from the chapter that the span of two sets of vectors is the same. If you just get rid of a vector, that's a linear combination of the other ones. So what that means is we want to find which vectors are linear combinations of the remaining. So let's see one example, If we look at the vectors v one plus B. Two, We see that we get the vector v. three. So this implies that V three Is a linear combination of U. one and V two. So this means that the span of the set, the one, one, V two, the three envy four Is equal to the span of V one, V 2 and before. So we can ignore this vector when creating our basis. Now we also note the V one minus before Is equal to V two. So that means V two Is a linear combination of V one and before. So I can ignore The Vector B. two. And now we look at our victors be one and before. Well, neither of these vectors are a linear combination of the other. Or in other words, in either of the vectors are a scaler multiple of the other, which means we want to be for must be linearly independent. So then from earlier in the chapter, we know that this span must equal the span of the one envy four. So since we have a span of a set which is linearly independent, that tells us Our basis must be the set V one, V 4.

Now for this question. Suppose V one v. TSU victory is a busies for our three now from problem for exercise That scene we of that's w equals a spon Vito Miners V want v three miners. V one is a blaine in all three. No, we know that this sets w plus v Juan is a blend of you translated because you're a DMV you want It's translated for view on this plane. This plane that is W plus V one would be Parlow will be parallel, so w and at the same time it will contain the one. Now, by definition, we know that ve to manners V one is in w right because the bees goes to spawn of days. So we are vey to minors. V one is in w This would imply that's V two minors view on, which is in W plus v Juan, which is the cost of itsu would be in w plus viewer because this is in W. Then this is V Juan. So this would be this is in w the want similarly so let's go about two dots. So this implies that we can just see Okay, I have it. Yes, so that implies that v two easy w plus view and that's what I wanted. I lights. So we have that. Similarly, similarly by definition, victory Miners v. Juan is in w. So this implies that Victory Miners v. Juan, plus V Juan isn't w and plus view on which is actually 1/4 just victory because V one were constant. This one this is in w plus v want. So this implies that's w close v one. No, Molly, it's contents. Everyone we've established that sits contains the one we've showed its components V two. On that we just showed that it also contains victory. Now this implies that's by the definition off w blows v Juan on dhe. If you also look at the serum one in your textbook, this would imply dots. I find off the Juan V to victory is the plane w blows V Juan dots. Contains v want VD

That was. Mhm. Well, given a vector w. and R. four second W has components 1 -2 -1, 3. Just it's no reason to try and in part a were as defined in the cardinal basis for W. That's welcome back to Jersey in our four. We'll say that some parents X, Y, Z. T. If this is in W perp this implies the product of V. Or W. Perp society you Kirk? Which is W. The equal 30 and authorized. Mm. So we have X minus two, Y minus Z. Plus through T equals zero. We have a lot of freedom here. When Children basis vectors 3° of three. New Precise, let's take X. And Y. To be equal to zero and keep it equal to one. Then it follows that Z is equal to three. And so we get detective. Yes, you won. Yes. With coordinates. 00 31 Yeah. Against. Mhm. Just mm. Now we want to find an Ortho dogma basis for W. For we want to find another vector. Mhm. That's that. Um He is not only their parker. V. With W. Zero but also Being a product of v. With you won at the 4-0. So now we get the equation um When we had the four X -2 line, mine is easy. Uh Plus three two equals zero. And we also have the equation three Z plus two equals zero. You can simplify this system. I will Well not only have 2° of freedom, so I'll take uh michigan J issue jewish wow. We get these new systems three x Uh -6. Y mm. Yeah. His muscles Plus 10 t. equals zero from the Seinfeld Kramer. Yeah. Or if you want to keep this a little bit differently, X -2 Y. And they subtract nine, Z -10. Z. equals zero and three Z plus T equals zero. Like sixties. Mhm. So now ah it's like wow let's take TP equals 2, 3. Well then it follows an x equal to zero and it follows that Z. Is equal to negative one. Yeah. And therefore why is you put too Positive five. So we get Mhm. The other vector U. two with coordinates zero five negative 13 Mhm. This is orthogonal who you want. So texting we still have one more degree of freedom. Now a last system is that our vector V. Has to be orthogonal www. But also you won. Um You too. And so we have the above system X minus two, Y -10, z equals zero And three z plus T. But now we are all equal zero. And now we also have that five Y minus Z plus three T equals zero. It's a good take a class seriously, she said especially as a matter of fact once again I'll take two to be three. That's the only degree of freedom they have though. This tells us that Z. Is negative one. And therefore from our 3rd equation, why have you booked through? Mhm mm negative two. Yeah I'm and therefore from our first equations X. Is equal to chords toys. R. Us negative 14. Yes. And so we get a third bacteria. Use three. With the opponents negative 14 negative two. Yeah negative one and three. Well they don't matter. All right. It's right there in that song. That's so and therefore the vectors U. One U. Two and V. Three to form an orthogonal basis for soviet. Perk in the. Yeah. Well I yes in parties were asked to sign an Ortho normal basis for W. Proof. This is similar to part A. But now we want to normalize each of you one through three. Now the normal view one squared is the sum of the squares of the components. Mm So this is a nine plus one which is 10. The norm of U two squared is 25 plus one plus 9 Which is 35. And the normal view three squares, this is 14 squared plus four plus one plus nine, Which is equal to 210. Therefore, a set of Ortho normal basis is given by you once over 10, You two over route 35 And you three over route to 10. This forms an Ortho normal basis for the subspace. W perk. All right.

Here. In this example, we have a set be that consists of two vectors which forms a basis for our to and what we like to do here is given a particular coordinate vector for X relative to this basis be is coordinates five and three. We would like to determine what the vector X itself ought to be. To do this, let's call the first vector here V one and the second vector V two. Since we already have it The coordinate vector relative to the basis be it then follows that the vector X itself is going to be equal to first take five which comes from the first cornet times, the first basis vector V one, then add to it the second coordinates three coming from the coordinate vector times the second basis vector B two. So this is our vector X and all we're doing now is determining what this is equal to. So we have five times V one, which is three negative five plus three times V two, which is negative +46 So working this out bit by bit. The first vector scaled by five is 15 negative 25 and the scaler multiplication of three times. This vector is negative. 12 and 18. So altogether. After performing the vector addition, we see that X itself is equal to three negative seven, provided that has this as it's coordinate vector.


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