exercises-29-and-30-show-that-every-basis-for-rn-must-contain-exactly-n-vectors-let-sleftmathbfv_1-2

Question

Exercises 29 and 30 show that every basis for Rn must contain exactly n vectors.
Let $S=\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\}$ be a set of $k$ vectors in $\mathbb{R}^{n},$ with $k>n$ . Use a theorem from Chapter 1 to explain why $S$ cannot be a basis for $\mathbb{R}^{n}$ .

Doruk Isik · Accepted Answer

in this problem, we&#x27;re given a setup vectors as and we&#x27;re told that this set of K vectors ah isn&#x27;t an intermission space. And and we&#x27;re also told that Kay is greater than ain&#x27;t so this system point by this vectors would be like a and a system metrics, would we would have a number of roles and K number of homes. So since chase greater than a, then it means that Rosa A would be less than cold. So this implies that, uh, there wouldn&#x27;t be enough. Give it elements for each home. So they cannot the revenge or each hole. And using your abate being section 1.7 or your textbook. This system for the columns can not be linearly independent, so that cannot forma basis.

Doruk Isik · Answer

in this problem, we&#x27;re given a set of vectors. It was represent those this B and we&#x27;re set up. These factors are the news independence and be half in number of factors and barely needle about independent in end dimensional spice. So it be work you former system out of these wreckers, Uh, you would have a number of factors intimations by sword on the cooling system will be an n by n square metres so that we would have given element for vitro since the number of columns is equal number rolls. And since b vectors are the nearly abundant and since you know that number of columns is think of a number of roles so that this given, ah, system given set up back to roll spend space. We can combine this and say that this given system is actually a faces or, uh, the n dimensional space

James Chok · Answer

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&#x27;t always about calculations. There are times where you should be using your words. So in this case, I&#x27;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&#x27;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

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

in this film, we were given a basis that be with directors B one B two up to it and we&#x27;re giving a lecture. Why? And we&#x27;re working in a director space be. And where all the vectors and bases are defied our and so less fun venture, You using the basis wreckers so we can represent us. Why won&#x27;t be one plus where to be too was upto end the end. So these bees are our basis vectors as the fine here. Certain wise would be being, um coordinated vector. So this would be why one up to why and but this one went up The wind is nothing about why itself. So in this case, why would be our, um, Corden vector? And this, actually is the definition off coordinate mapping onto our

Griffin Johnston · Answer

in this video, we&#x27;re going to find a basis for the span of the following set of vectors. Now, whenever I&#x27;ve been finding a basis for a span of vectors, I always like to take a first initial glance and see if there&#x27;s any obvious vectors in the set of vectors that air linear combination That is a linear combination of the other vectors in this set to see if there&#x27;s any obvious linear combination relationships here. Linear dependency, relationships so positivity. And see if you can see any obvious linear dependency relationships in the set of vectors from assuming you&#x27;ve taken a look at it. So the one that stood out to me was that this factor right here, inspector, right here was two times this vector right here. So check this out. The span of this set of vectors the span of the set of vectors is the same as the span off the set of vectors. With this vector removed because this vector is a linear combination off the other vectors in the set, its namely, it&#x27;s the linear combination of two Times Inspector plus zero times this factor plus zero times this specter so we can remove from the spanning set and get a new set of actors who span is the same as the original is who span is the same as the span of the regional set of actors. So this makes our computations a little bit easier. So we now have easier spanning set to deal with, which are right now. So now we only have three vectors to deal with. We need to find a basis for the span of this set of factors right here. And a nice way to do that as we&#x27;ve talked about in previous bit videos, is to make these vectors the rows of the Matrix. So I&#x27;ll show you what that means. So this is what I meant. And now check this out. If we find a basis for the road space of this matrix, that&#x27;s the same. It&#x27;s finding a basis for the span of the row vectors of this matrix. But the road vectors of this made between made this matrix to have row vectors. That is the same vectors that they&#x27;re in this set right here. So if we find a basis for the road space of this matrix, that&#x27;s the same things finding a basis for this. The span of the centre vectors right here, the span of the set effective. So remember, if you want to find a road space based on our previous videos, if you don&#x27;t remeber, that&#x27;s totally fine. But let&#x27;s just review if you want to find a basis for a row space of a matrix. We want to find a basis for the roast base of a matrix. We first we&#x27;ll reduce matrix to a Roche lawn, form to a row, Rashwan form, and then we take the non zero rose. Take non zero rose. And those non zeros form a basis for the road space off this matrix. Now, now, if you want to know why this method works the details, that&#x27;s the widest networks. I talk a little bit about how this works in previous videos on previous problems, so check this out. If you have any questions now, positive video and take take a few tries at Rove reducing this matrix. So I&#x27;m assuming you&#x27;ve had a go at it. So I&#x27;m gonna show the series of Roe Equivalency is that leads us to a row echelon form of this matrix. So this is the series of Roe Equivalency is that leaves us toe this low echelon form matrix and check this out. These are our non zero growth in these two non zero growth. These two non zero low vectors of this row echelon matrix form a basis for the road space of this matrix, which happens to also be based on how we construct this matrix. Ah, basis for the span of this set of vectors which is the same as the spaces for this for the span of this inter vectors. So here is a basis, this set of vectors, the vectors containing the set of actors containing the vector one comma for comma, one comma, three and zero comma, zero comma, one comma, negative one. This set of vectors is a basis for the span of this center vectors right here. So we could do the same thing using the columns using a column space method instead, where we make this set of vectors, we make each of these vectors in the set. The columns in a matrix will show you what I mean by that. So this is what I meant by that. But the column vectors and this majors correspond exactly to the vectors in this set right here. So all we have to do is if we find a basis with a column space of this matrix that&#x27;s the same is finding a basis for the span off the column vectors in this matrix and the span of the column vectors of this matrix is the same as the span of these, the set of actors right here. So we need to find a basis for the column space of this matrix. But we know how to do that, or it&#x27;s okay if you don&#x27;t all review it right here. But check out my previous videos if you want a little bit more of a review as to why this method works, and I also have I will have a lecture video discussing the details of how this meant of how this method of finding a basis of confidence of the Matrix works. But for now, so we first row reduce produce matrix to row echelon for him to locate in the goal of this road reduction series of road reductions to Rochon for Mr Locate, the Pivot columns pivot columns off our original matrix. I&#x27;ll call this matrix like as I usually do. P Looking to locate pivot columns of P in the second step is to take those pivot com&#x27;s take Pivot Collins and those pivot Collins take the pivot calms as basis vectors. Take Piva columns in those Piva Collins of P that we located by Rhodesian P to a row echelon form. There are all call that revolution former produce matrix to ocean form P prime P prime. So the Pitta columns of P will form a basis for the column space of P. So the pivot Collins of P will form a basis for the con space of P. So take a minute, positivity on road use this matrix to a row, Ashlan for So I&#x27;m assuming you&#x27;ve had a go at it. So all reveal now the Siris of Roe equivalent sees that leads us to a row echelon form of this matrix. So this right here is the series of race series of Roe equivalent sees that leads us to a rational in form of the matrix penal. Call this relation on foreign p prime and this reveals where the pivot columns of PR located the difficulties of P. exactly correspond to the pivot calms api prime. So in this case, since the first and second columns of P Prime are the pivot columns or are the pivot columns of P. Prime because those are the columns that contain the leading entries, then that means that the first and second columns of P are the pecans of Pete. Through these two column vectors form a basis for the column space of the Matrix P. And so we have our basis. Or we have a basis for the conference API, which is the same, which is the same by by how we constructed the Matrix p as, ah, basis for the span of the set of actors in that and again, that&#x27;s the same as a basis. The span of this set of actors since the span of these two sets of actors are the same. So the set of excuse one comma contain the CEP containing the two vectors, one comma, four kind of one, comma three and to common eight comma, three comma five. This set of actors right here is a basis for the common base of P, which is the same as span of this set of vectors, which is the same as a span of this set of vector

Problem

Exercises 31 and 32 reveal an important connectio…

Problem 30 Hard Difficulty

Exercises 29 and 30 show that every basis for Rn must contain exactly n vectors.
Let $S=\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\}$ be a set of $k$ vectors in $\mathbb{R}^{n},$ with $k>n$ . Use a theorem from Chapter 1 to explain why $S$ cannot be a basis for $\mathbb{R}^{n}$ .

Answer

Given vectors are in $\mathbb{R}^{n}$ . That means they have $n$ entries. If given number $k$ is greater than $n$ , then using Theorem 8 from Chapter 1 we can say that $\left\{v_{1}, \ldots, v_{k}\right\}$ is linearly dependent set because it has more vectors than they have entries.

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

Linear Algebra and Its Applications

Anna Marie V.

Caleb E.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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Given vectors are in $\mathbb{R}^{n}$ . That means they have $n$ entries. If given number $k$ is greater than $n$ , then using Theorem 8 from Chapter 1 we can say that $\left\{v_{1}, \ldots, v_{k}\right\}$ is linearly dependent set because it has more vectors than they have entries.

Linear Algebra and Its Applications

Anna Marie V.

Caleb E.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Anna Marie V.

Caleb E.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications