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Question

Suppose $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}ight\}$ is a basis for $\mathbb{R}^{3} .$ Show that $\operatorname{Span}\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}ight\}$ is a plane in $\mathbb{R}^{3} .[	ext { Hint: What can }$
you say about $\mathbf{u}$ and $\mathbf{v}$ when $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ is a plane? $]$

Lucía Guerrero · Accepted Answer

from the hypotheses that the set containing V one v two V three services it can be proven that the set off factors 15 minutes free, one on 53 minutes 31 is a linearly independent set. This implies that the space generated by 15 minutes We won on 53 minutes. We won. Has I mentioned to on this is enough to conclude that this space is a plane. Now to prove that the said continue fee to minus V one and federal ministry one is linearly independent. Let&#x27;s do it by contradiction So suppose there linearly independent that iss feta minus V one is a multiple off p three minutes We won so we can write the above equation as K minus one times fy one plus feet too minus K times fee three equal to zero on this is a linear combination. We&#x27;re not all the coefficients or zero on this would imply that the V 1 15 23 are linearly independent. But since the set containing the missile bases in particular linearly independent and this is a contradiction and we prove what we wanted

Gideon Idumah · Answer

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&#x27;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&#x27;re a DMV you want It&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s w close v one. No, Molly, it&#x27;s contents. Everyone we&#x27;ve established that sits contains the one we&#x27;ve showed its components V two. On that we just showed that it also contains victory. Now this implies that&#x27;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

Donald Albin · Answer

prove that the span friend right here Ban of V one V two V three equals the span of V one V two if and only if the three can be written as linear combination of V one and V two. So I&#x27;m gonna write out what these spans are. The span of V one v two V three is V and elements of the such that V is C one B one plus C two v two plus C theory V three with Ah c one, C two and C three being real numbers. But if we write out the spin of V one and V two, it would be all V all vectors in the vector space. Such that V is some D one rial number times view one plus some real number D two times V two. No, I don&#x27;t have space. But ah, I did wonder, right? That um c one c two, c three d one and d to are all rial numbers. So if we equate the two sides, then we see that every vector could be written as C one v one plus C to v Teoh plus C three v three. But that same vector could also be written as D one v one plus de to the to. If we now solve this, we could write it as see one minus D one times Vector one plus C three minus not three Tu minus D to times Vector to equals Negative C three Vector three Now if we divide by negative C three we see that vector The re is D one minus C one over C three times Vector one plus C to minus D to whoops de tu minus C to oversee three times vector to, um, this is a real number and this is a real number. So, um, Vector three must be able to be written as a linear combination of vectors one and vector to

Donald Albin · Answer

here. This state&#x27;s proved that if the set containing vector V one in Vector V two is linearly independent and if V three is not in the span of the vector of you one in Vector me too. Then the set of vectors, v one, V two and V three is linearly independent. So if it&#x27;s not in the span, then Vector V three does not equal any constant times. Vector V one plus any constant times Vector V two. All right, so Vector the three is not a linear combination of vectors V one and vectors v two. And so we could right that if we want the set of vector V one, Vector V two and Vector V three to be linearly independent. That means that some constant has vector view one plus some constant groups times vectored v two plus some constant times. Vector V three cannot equal zero. All right, so I&#x27;m gonna try to prove by contradiction. So I&#x27;m going to let d one v one plus d two v two And you know what? I better write this over again. That t one v one Plus it d two v two plus de three V three equal zero for some really numbers, do you one de to Andy three. If that were true, then by subtract d three v three from both sides of the equation I get and then I divide by negative d three on both sides. I would get the three equals negative D one over de theory V one, minus D two over d three v two. And if I, uh, say that d one over D three is some other constant C one and I mean negative, do you? It went over to three. And if I let negative de through to over 83 b some other constant C two then V three is C one V one plus C two v two. But we already said that V three does not equal. See one V one plus C two V two for some C one C two that are really numbers, which is a contradiction. And so that means therefore d one v one plus de two v two plus de three v three do not equal zero the opposite of our initial statement

Donald Albin · Answer

So we have three vectors. V one, which is 123 V two, which is 456 and V three, which is 79. Determine whether those three vectors the set is linearly independent. So that&#x27;s the first thing that we need to dio. So I&#x27;m just going to go ahead and write them as column vectors and take the determinant. One times five times nine is 45 four times eight times three is 96. Seven times, two times six is, um, seven times 12. It is 84 minus three times, five times seven. That&#x27;s, um let me just use a calculator. 105. Six times eight is 48 and, um, nine times. Two times four, huh? Yeah. I put pluses in my calculator nine times. Two times for two times four is eight times nine is 72. So 45 plus 96 plus 84 minus 105 minus 48 minus 72 is zero. That would mean that these are linear. Lee depended. One of the vectors is a, um you call that, uh, I&#x27;m they try to think of the words vector some oh, media combination one of the vectors is linear combination of the others. Um, so if we just look at the 1st 2 vectors, I do notice, since we&#x27;re now now down to only two vectors in the set, Um, why isn&#x27;t this writing this vector and this vector? They are not proportional to each other. And so, um, there in different directions and will therefore form ah plane in the three dimensional space so they will form a plane. It is possible that with something like this, they would just form a line in three dimensional space. But the two vectors are not proportional to each other.

Donald Albin · Answer

we have three vectors in riel, three dimensional space. A show that the the set of these three vectors is linearly dependent. Well, um, V three is actually just from observation. Negative. Three times V two. Um, negative. Three times one is negative. Three negative. Three times three is negative. Nine and negative. Three times four is 12. So we can remove V three from the vector set. Now, I also noticed that V one and V two are not proportional to each other, so they will form a plane. Ah, that goes through the origin. So be draw a picture or their span will form a plane that goes through the origin drawing picture. So X, Why the Okay, so just going to use red for this one Over two down, one out, five. I&#x27;m gonna say it might be around there, um, over one up three. That way, four might be there. So this point. So it&#x27;s like that and that, um, it&#x27;s going to form a plane in three dimensional space. By the way, those two vectors are not in a line. They kind of look like they are, but they are not in a line

Problem

Show that if $\left\{\mathbf{v}_{1}, \mathbf{v}_{…

Problem 13 Easy Difficulty

Answer

Therefore, the set $\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is linearly independent.
So, by the above result $\operatorname{Span}\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is a plane in $\mathbb{R}^{3}$ .

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

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Caleb E.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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Therefore, the set $\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is linearly independent.So, by the above result $\operatorname{Span}\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is a plane in $\mathbb{R}^{3}$ .

Linear Algebra and Its Applications

Anna Marie V.

Heather Z.

Caleb E.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Heather Z.

Caleb E.

Kristen K.

Therefore, the set $\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is linearly independent.
So, by the above result $\operatorname{Span}\left\{\mathbf{v}_{2}-\mathbf{v}_{1}, \mathbf{v}_{3}-\mathbf{v}_{1}\right\}$ is a plane in $\mathbb{R}^{3}$ .

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

Linear Algebra and Its Applications