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Let U and W be subspaces of R*; Uzspan{(1,1,0,-1}, (1,2,2,-2), (1,2,3,0}} and W=span{(2,3,3,1), (2,3,2,-31, (1,3,4,-31}. Find basis for U+W and dim(U+W);...

Question

Let U and W be subspaces of R*; Uzspan{(1,1,0,-1}, (1,2,2,-2), (1,2,3,0}} and W=span{(2,3,3,1), (2,3,2,-31, (1,3,4,-31}. Find basis for U+W and dim(U+W);

Let U and W be subspaces of R*; Uzspan{(1,1,0,-1}, (1,2,2,-2), (1,2,3,0}} and W=span{(2,3,3,1), (2,3,2,-31, (1,3,4,-31}. Find basis for U+W and dim(U+W);



Answers

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

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?

Hello there. So for this exercise we got these three factors B one The two and B. three. And these three vectors is a super space in our four in this case. So we need to find a basis for the orthogonal complement of this based off of you. So how to build this whole? Let's recall the definition of the orthogonal complement is equal to the vectors X. On the space. In this case four such that X. Is an internal to the bacteria V. For all the the space of you. And this condition here is they have anticipated. X. V. Is equal to zero for all the. No. So the point is that we can obtain your final compliment. Just be considering the generators of this space. And what I mean is um considering B. One, B. Two and B. Three. So let's be a generic point X on the compliment. Then this implies that these vectors satisfied that x. v. one X. Movie too and six B. Three. All these are equals to zero. Because if this vector X. And any any vector X. In the compliment will be also known to the generators. Actually there is a generic vector of the form X. One. Uh huh. X. Two X three 64 So this becomes a system of linear equations that have matrix representation that corresponded just built the the vectors as columns of this matrix here. So I mean put in here one for 5 2 days be one 2130. This v. two And the 3 -1 three 21 This vector is multiplying X. One X. Two X. Three X. Mhm equals two 00 Okay so we got this system of linear equations and we can, what we need to find is the a solution for digital space orginal space of these interests here. So for that we're going to reduce that metrics to the Russian form the whole system. So we obtain the following we obtain 101 -2 over seven 011 for over seven And 000 zero times X one x two X three X four. And there is the cost to the 000. Okay. So from this you can observe that here we got a free variable that corresponds to explore. And here we got another free variable that response to X three. So what does mean is that the null space he generated by two vectors. Which is equivalent to say that I love you will be generated will be dispatched and of two vectors. Let's call off for one of two. Okay? Um so let's find those vectors. And those vectors will be the will be obtained by calculating the general solution for this system. So the general solution, If X three and X four are free bibles, that means that X three could be close to t the next four. We're going to give the value of our So the general solution X. Is equal to t -1 -110. Lust har And the vector to -40. So so you can observe here that the and any victor in the any vector X. In the compliment Is generated by two vectors by the linear combination of two vectors. These two vectors corresponds to offer one. Mine is a warning one, 10 and alpha two Which in this case is equal to two minus four, zero and seven. So these two vectors expand the whole space dog and even more they are linearly independent. So two vectors selectors that spanned the space. So we know that the space the arsenals compliment will be this pond here of the one of two and Alpha one Alpha 2 are linearly independent. Then the basis for the compliment will be The vectors all for 1/2.

In this video, we're going to find a basis for the road space of the following Matrix The Matrix, which all call p and it's equal to the farming 123 in the first row, 567 and the second row in 9 10 11 in the third row. We're also gonna be finding ah basis for the column space of this matrix Now to find ah basis for the roast base of this matrix. A nice method that we can use is first Rover do some matrix to row echelon form Ro reduce the matrix to row echelon, former to a row echelon form of this matrix and the second take all the non zero votes Take all of those non zeros of the non zero rose. So what I mean by non zeros is the roads that have at least one non zero value in them. And then these non zero rose will form a basis for the road space of this matrix p. And the reason why this method works is because of falling to fax. The first factors to row equivalent matrices have the same roast based so of P and P prime our row equivalent That implies that their road spaces are the same. So the road space of P is equal to the road space of P prime. Now, the second fact that allows us to use this method is the fact that the non zero rose in a row echelon form matrix are linearly independent from each other. So non zero rose in a row. Echelon matrix in a row, echelon form matrix matrix are linearly independent. So check this out. If we just roll, reduce this to a row echelon form of P and then take the non zero rose than those non zero rose, that is the span of those non zero rose will be the same is the span of P or excuse me of the row factors of P and those non zero votes will be linearly independent. So let's Rover Houthis to a rash one form positive young. Give that a try. So I'm assuming you've given it a try. So this matrix Heathrow equivalent to the falling matrix which all right, right here. 123 in the first row, zero negative, four negative, eight in the second row and zero negative. Eight negative 16 in the Sergo and then we can ro ro reduce this further. Although do sit down here we can reduce this matrix further to the falling matrix. 123 in the first row, zero negative. Four night of eight in the second row and 000 all in the third row. Now remember, this guy has a roast beef. This major stand here is the same roast base as the matrix p and also the roast beef said this guy is the span of its role factors that by definition of a row space. But remember, the span of these row vectors includes as one of as one of the vectors in the spanning, set the vector zero comma zero comma zero and since effective zero comma zero comma zero is a linear combination of these other two vectors. Namely this vector time zero plus Is this nectar time? Zero. It can be removed from the spanning set and we still get a new set without the Spectre Zio that has the same span as these three vectors. So this set right here The vector the set containing the following vectors The vector one comma, two commas, three and the Vector zero comment. Negative four Comma negative. Eight. This set right here is a linearly independent set whose span is the same or who span is the same as the span of the rows of this matrix. That is to say, who's span is the same as the row space of this matrix right here in this. This matrix right here has a rose face that's the same as P. So therefore, this guy has a span as the sit. That's the same as the rose base of P and also synthesis linearly independent. We can say this is a basis for the matrix of for the road space of the Matrix P and moreover, since the basis or rather since the road space for the Matrix P is the span of all three of its row vectors and each of its row vectors has three components, then the span of these three row factors as all the linear combinations of these three row vectors and therefore all since all in your combinations of these three row vectors will also have three components, we can see that the roast face of P, the road base of P, each vector in the road space of P will have three rial number components. So therefore it is a subspace off our three the space of all vectors having three Rio number components. Now to find the calm space for P, we employ a similar method. We first rove reduce the matrix to ah ro echelon forms were overdue p to a low echelon form of P and second, we locate locate the pivot columns. We locate the pivot columns of P or rather off P prime, which in all stages all call p prime the a row echelon form of the Matrix piece. Are they produce P to a row echelon form, say P prime. I just call it p prime Locate pivot columns of P Prime and the pelvic commas API primer. The same is the pivot cons of P. So then the pivot columns of P. The Pitta Cons of P, which we found by finding the pivot columns and P prime, which correspond exactly to the pivot columns in P. There's a pill that Collins and piece of pivot columns in p form a basis. Four. The car in face of p form a basis for the column space calm space of P. So check this out. We actually already row reduced this guy. We actually are rather we already road reduced p and this is to a row echelon form. And this is a role echelon form of the matrix P. So in this row, echelon form this row echelon form of the Matrix P reveals where the pivot cons of P R the pivot columns of P are the ones with the leading in trees, the ones with the leading entry. So this column right here is a pivot column in this column. Right here is a pivot concert. Therefore, this column right here is a pivot column of P and this column. Right here is a pivot column of P and those two columns. Those two columns form ah, basis those two column vectors rather former basis for the confidence of P. So a basis for the columns basic P, we can say are the following two vectors or rather, the set containing the following two vectors. The vectors, the second training the two vectors. One of the vectors is one comma, five common nine and the other vector is two comma six Kama 10 and again since each of these vectors, or rather, since each vector in the column Space of P is a linear combination off vectors that contain three components. Every vector in the column space of P contains three rial number components, so therefore the columns basic piece of subspace of our three.

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.


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