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Given the set WcR', W={ (a,b, C,d) €R4| a-2b+c-4d=0}. Determine whether W is subspace of the vector space R4 or not: ii) Find basis for W and find dim(W)...

Question

Given the set WcR', W={ (a,b, C,d) €R4| a-2b+c-4d=0}. Determine whether W is subspace of the vector space R4 or not: ii) Find basis for W and find dim(W). iii) Construct basis for R4 containing the basis that is obtained in part (ii) .Note: Write all your calculations and steps ofyour solution in details to the below empty field_

Given the set WcR', W={ (a,b, C,d) €R4| a-2b+c-4d=0}. Determine whether W is subspace of the vector space R4 or not: ii) Find basis for W and find dim(W). iii) Construct basis for R4 containing the basis that is obtained in part (ii) . Note: Write all your calculations and steps ofyour solution in details to the below empty field_



Answers

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

In this video, we're going to find a basis for the span of the following set of vectors. Now, whenever I've been finding a basis for a span of vectors, I always like to take a first initial glance and see if there'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'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'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'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've talked about in previous bit videos, is to make these vectors the rows of the Matrix. So I'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's the same. It'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're in this set right here. So if we find a basis for the road space of this matrix, that'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't remeber, that's totally fine. But let'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'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'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'm assuming you've had a go at it. So I'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'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's okay if you don'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'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'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'm assuming you'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'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

Yeah. Okay. Suppose. Were given three vectors 1, -4 negative three, And then 20 to -2 in two negative 132 Okay. And given these three vectors, uh I want to find a basis. Okay, for the subspace that is spanned by them. Okay, subspace of R. For So we consider a subspace of R4. That is spanned by these three factors. And I want to find a basis for that subspace. Well, recall what is the basis? Well, it requires that the basis vectors are linearly independent. Okay. And that they span the space. Okay. But we already given that they span the subspace. Okay. That that's the premise of the question. Okay, So we don't have to worry about the spanning part of the basis requirements. We all we have to do is we need to make sure that, you know, these three vectors are independent. If these three vectors are linearly independent. Okay. Then we're done. Right, Because by definition they span the space that they span. So as long as these vectors are linearly independent, then they form a basis. Okay, So how do we check whether these vectors are linearly independent? Well, the easiest way over here is to simply see that um you know, like there are no ways to there are no nontrivial linear combinations that you zero. Okay. So for example, over here, right? You can let's look at vector to vector three. Okay, so you can see that if I were to subtract, the only way for for us to get to zero in the first entry on the top here is to subtract them from each other, right? Like like uh I get zero at the top but then none of the other entries would be zero, Right? Because if you subtract them, even though the first entry cancels out, none of the other entries who cancel out, so therefore there does not exist. Okay, any nontrivial linear combination of this? Uh these two vectors that would use zero, so therefore these two vectors are linearly independent. Okay, now let's look at the first two vectors. Okay? Um if I the only way okay for us to scale uh for us to get a zero in the first entry is to, you know, like multiply the first one by a factor of two. And then a subtract. Okay, so, well, only up to a scalar constant, but that doesn't really matter. Um So over here I have taught I have a zero at the top. Okay. But then over here you would see that even though I get zero from the top by multiplying this by two and then subtracting uh none of the other entries, Okay, would be zero. Okay, So therefore there are no uh nontrivial linear combinations of these two vectors that would use zero, either and last, but not least. Let's look at the one and 3. Let's look at vectors more than three. And again, you see that if I scaled the first factor by two, and I and I and I subtract the third factor from it, that would get the first entry to zero, but none of the other entries would be zero. So therefore, we conclude that uh none. Uh you can you can basically you get the idea of this exercise right? Which is to just try to figure out is to try to to uh figure out a justification for the fact that there are no nontrivial linear combinations of these three factors that would yield the zero vector, and therefore they are linearly independent, and therefore they span the space that they span and they form the basis. So the answer okay. For the basis, the simplest answer that you can find okay without doing all that that, you know, like reduction, you know, like us or like those matrix manipulations things, is to just say that aha. The basis is the fault is given to us already. The basis is just 11 negative four, negative 3 to 0 to negative two and two negative 132 Okay.

Okay. Now, in order to find a basis for the subspace of R. For that spanned by these four vectors, we want to get rid of any of vectors. Any of these vectors that are repetitive because they're dependent. So for example, we see that V4 is actually V one plus we can have before B three, V one plus a third of the three. So we can't have the four B three, M. V one all be in the basis because they're dependent. Now, let's see for V two, is there a way That we can write V2 as a combination of the other vectors? Let's see. We can have two times 3, one minus 2/3 B three. So that means that we can't have V 1, 2 and three all in the basis. Now can V to be expressed With just the 1? No. So that means that one example of a basis we can use is V one V two. We can also have V one, V 3 B. A basis. We just can't have Me Too NV three being the basis with V one Or V one, V 3 and V four together. So these are two possible bases.

Where the fuck I want this. So We're giving two vectors in R. five. Yeah. You won with components 11341. And you too with components 1 to 1 to one time two. We'll have to find a basis for the subspace W. Bar five orthogonal. To that review one of you two. If the vector V. Lies in the subspace, envy has formed X. Y. V. Uh S. E. T. That's a year. And thanks. Then it follows that the inner product of the that you won in the inner product of these if you two is zero, so we get X plus Y plus three, G. Plus four, S. Plus T. Equals zero and X plus two, Y plus Z. Plus four, S plus T equals zero. I want to solve this system to find basis to do this. Also track the church equation from the 2nd 19. New system, X plus Y plus three, Z plus four S plus T equals zero and X minus 60 to I minus wise. Why Z- Cruzi is -2 Z. And four s. Money for us. A hero. Two months to year zero. All equal zero. Uh Belushi's So looking at this system, we see that there are 3° of freedom. So for example take S. And T. The equal zero. And then you take Z to be equal to one. So he's three days of freedom. And from our second equation it follows that Y equals two. Therefore X is people to -5. Carly. He was just like a pipe dream concert. So we have the 0.-52100. To find other vectors basis vectors. Uh Consider taking, Oh his hair grew long immediately. Yeah, Z. And S equal to zero, but T equals one. Create, create steve. Yeah, everyone's fucking well, the worst thing that could happen, Then it follows from our second equation that Y. is also reported zero. And from the first equation straight X. Is equal to negative one. So we get the vector negative 10001 Finally consider to take uh G. Equal to one and then S. And T. Both be towards zero. Right then. I'm sorry. We already did this instead. Let's take s equal to one and Z. And T. The quarter zero. Then again from the second equation get one equals 0. And from our first equation we get X equals negative six. I would actually see more of it or some. Okay candy did you sloughs? Sorry X should be negative for and maybe six. And therefore he gets instead of three bacterias. Were you good job -10001. Uh negative four 0010 And director negative 52 100 You say that there's one a basis for our subspace W. In our five. Hey?


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