let-s-be-a-finite-set-in-a-vector-space-v-with-the-property-that-every-x-in-v-has-a-unique-represent

Name: let s be a finite set in a vector space v with the property that every x in v has a unique represent
Uploaded: 2020-12-10
Duration: 4 min 15 s
Channel: Numerade
Description: Let $S$ be a finite set in a vector space $V$ with the property that every $x$ in $V$ has a unique representation as a linear combination of elements of $S .$ Show that $S$ is a basis of $V .$

Question

Let $S$ be a finite set in a vector space $V$ with the property that every $x$ in $V$ has a unique representation as a linear combination of elements of $S .$ Show that $S$ is a basis of $V .$

Priyanka Sadarangani · Accepted Answer

Okay, So our question is, let s be a finance set. Okay? Finance it in a victory space victory space. We with the property that every X in V has a unique representation. Okay. Unique representation as a linear combination off elements off. Yes. So we have to show that, as is a basis off we Okay, so we know that we know that as is finite dimensional. Okay, is finite dimensional. So we can say that this is it. And dimensional ticket and dimensional. So on this elements are as I and for their values, less than it was toe end and greater than it was 21 Okay. And for that on bet Visa victory space. So this will be fine. So each each eggs and we has a unique representation. Okay? Unique representation as a linear combination off as a linear combination off combination off s. And as to s three. And so on s and Okay, so where definition as an as his spending set for V okay. And in particular, that is also means that zero has a unique representation. Okay, I tried to understand that Nick represent representation, okay? And currently we cancer. That this killer C one C two C three and so on. C and the victory questions. Okay. In the world of the factory question, it will be C one s one plus he two as two plus C three s three and so on. C n s m equals to zero. You know that from this. So this will have a unique solution. And the travel unique solution. And the solution is Jiro on zero. It means thistles, the travel solution. And we know travel solution is always the solution. Okay? And it is the only solution. So it means, by definition, it means as one as two s and so on s and is linearly dependent. Okay, because it has travel solution and only solution. So it is linearly independent. Okay, So any as Iselin independent spending set that does, by definition, definition, as is a basis off movie. Okay? And we can say that as it was to s equals two. We want we do. We three we and you can save as one as two s three s and is a spending set off. We okay? And these are linearly dependent. So the definition as he is a basis off we it is prude. Thank you.

Ernest Castorena · Answer

okay for this question, they give us the Matrix. A is equal to 11 negative. 112 negative. 35 negative. Six 50 Teoh. Negative three. And they give us two other directors. Negative. 2750 and three Negative. 80 Fine. Okay. What do you do with these? Well, they actually want us to show their basis more than all space. Okay, Okay. Let&#x27;s do some of our technique. Considered him in that. First off, this matrix is three by four. So it&#x27;s a member from there is an r four directors and are what&#x27;s the rank of this matrix? Well, it should be of dimension to its rank. Should be to Why is that? They&#x27;re claiming that this is the basis, so let&#x27;s figure that out. We&#x27;ll have 11 negative. One one, Teoh. Negative. 35 negative time. Zero to negative three. Yeah, these rows. So what can you see at the bottom row is three times the first euro, plus the second round to the end of with three negative. Three on the two young deport, zero on five. Yeah. So this is where done information. We start performing elimination. We&#x27;re gonna end up when my trying to the time in what leads to the zero victor. What&#x27;s then the colonel? We&#x27;re going to end up 11 negative one 10 and then it&#x27;s multiplied by t subtracted from the second room. So end up with negative Fine. We end up with seven and, uh, see multiplying it by negatively. Zero 0000 They&#x27;re up, and we can already see that it&#x27;s a drink, too. So it&#x27;s not busy anymore. With that and get sidetracked. Let us proceed with their goal. It&#x27;s definitely a Brinks you. Let&#x27;s show that these vectors Aaron, are basis. So are there in the kernel. 50 negative three. One negative 15 to negative. Three. A negative six one And let&#x27;s multiply it by the vector. Negative, too. 750 Multiply it out. We actually see that we get 000 on the same for our second rector. We multiplied by the Vector three negative 805 and we end up with the vector 000 the great So they definitely ah are inside the null space, and they&#x27;re definitely literally independent. Why is that? Well, it&#x27;s clear that this isn&#x27;t a scaler multiple of this. If it were, then they come out with the system of equations like, Okay. And three should be equal to a negative to you. Time that scaler negative eight should be equal to seven times that same number. And we can continue for the other injuries. But it&#x27;s already tried to solve for this. There&#x27;s no such T that makes both of these work. Okay, so they are linearly independent there, in the two dimensional vector space, they spend our whole space, okay.

Doruk Isik · Answer

I think this problem We&#x27;re given a set off the nearly dependent vectors. We want me to be free and report. We have to solve this. We&#x27;re gonna use the humans. Let&#x27;s try it back to w uh, using an arbitrary number K. So W&#x27;s came one of me One plus get to be too plus k for before. What we know is this We know that this be Zarley nearly depend by so they depend on each other so well, let&#x27;s assume not be one is equal to, uh, some constant A to B two plus a three d three plus a Ford We pour hear Keyser arbitrating And here a czar Arbitrary. So let&#x27;s write to be one in 31 in terms of B to B B m w will be, um, a one a two we two plus a three d three plus a four before plus K to be two plus three v three plus cape or we fall. So then w will be equal to a one a two plus que tu we to plus one a three plus three Well, three plus J won a four plus four me for So since case and a CZ arbitrary. And since we can, right, let&#x27;s say V two in terms of We want me three Emmy for orchid Correct. Be three in terms of other valuables are directors. It means that there&#x27;s more than one way Jax Prospector read, So Dan W. Can be expressed as a linear combination off their vectors, we warn me to briefly and in numerous ways.

Rakvi . · Answer

So suppose the A one feel 31 plus your toe Feel free to all the way a pool. Okay, Be all freaky Is zero That is some linear combination off these vectors. Steel free I zero Okay, inside W Then you want to show that all the eyes are in fact zero. That is all the weights are zero. That is with what you want to show. Okay, so let&#x27;s show this. So using the property of linear transformation, you know that a one off P off Leven is equal to p off even everyone. Okay, Similarly, second term is equality off. It will be too similarly all the way up to give it. Okay, this is 1/4 here. I use the property that he respects tradition 60 off. Even we won, plus de off. It&#x27;ll be it all. The way is just equal to tee off. Even we even close it. We do close, you carry. This is 1/4 0 now you know, used effect that de is 1 to 1. So 121 will tell you that if he off a vector is equal to zero, the vector itself has to be zero. So this implies that even we even plus a 22. Thus all the were toe a Cherokee is actually called a zero. Okay. No used effect that the set of even week A is a linearly independent certain side we so linearly independent. So if some combination off revised zero, that means all the weeds have to be equal to zero. So you have shown that the 80

Doruk Isik · Answer

in this problem there&#x27;s a linear transformation. Greater acting crawlspace be onto space. Definitely. And in space stuff we have subspace, etc. And we&#x27;re gonna say let you represent all Xing We anywhere trying to prove that you is a subspace No, We&#x27;re going to say that you won And you two are two elements in space, you and tur. And to you too we know And this problem settling This is given that these Aaron Z since t easily near a praetor concert t you on wants to see you too is t you want waas You too So this your wants you to is all saying z now You want any too, aren&#x27;t you? So from this we know that you want waas you jer should be in you So this means that you is blows on there back tradition No, we&#x27;re gonna use you want to you So we know that do you want isn&#x27;t easy So if you multiply that with any constant and this should be all saying z right And since I can&#x27;t easily you to information this will be the tee off. See you are And it should be also Z No, you want wasn&#x27;t see? Sorry. You always you. It means that see, you won should be in you. And from this we see that here is close under back to for protection. So these are the requirements that we need for subspace subspace. So using this, we can say that you is a subspace. Oh, be

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

Problem

Suppose $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v…

Problem 19 Easy Difficulty

Let $S$ be a finite set in a vector space $V$ with the property that every $x$ in $V$ has a unique representation as a linear combination of elements of $S .$ Show that $S$ is a basis of $V .$

Answer

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Kayleah T.

Michael J.

Joseph L.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

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

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Kayleah T.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications