in-exercises-15-and-16-list-five-vectors-in-operatornamespanleftmathbfv_1-mathbfv_2right-for-each-ve

Question

In Exercises 15 and $16,$ list five vectors in $\operatorname{Span}\left\{\mathbf{v}_{1}, \mathbf{v}_{2}ight\} .$ For each vector, show the weights on $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ used to generate the vector and list the three entries of the vector. Do not make a sketch.
$$
\mathbf{v}_{1}=\left[\begin{array}{r}{7} \ {1} \ {-6}\end{array}ight], \mathbf{v}_{2}=\left[\begin{array}{r}{-5} \ {3} \ {0}\end{array}ight]
$$

Sean Thrasher · Accepted Answer

Okay, so we&#x27;re asked to find five vectors that sit in the span of V one and V two and were asked to give the weights of each factor. So Vector X that sits inside the span will be a linear combination of U and V two. So the simplest ones. So the simplest choices of a one and a two is first, we&#x27;ll take a one in 82. Equal 00 is always in the span of any vectors, right? Because the because the span is a subspace right in the subspace contains the zero vector, since it is a vector space. And so 000 is one possible answer, and the weights are zero for both of them. The next one is to take this one to be one, and this one to be zero. So 71 minus six sits also inside the span of V one and V two and minus 530 Also spit sentence in the span of the wannabe to the weight on V 10 The weight on V two is one in this case. Then you could take you could just add them. We want to be too So take a one and a two to be both won. And when you do that, you get this. I didn&#x27;t component by component, or you could do the one minus V two. So the way on V one would be one the way envy to would be minus one. And so you get 12 minus two minus six. And these are five vectors there. Sit in the span. Um, of course, you could choose any anything for a one and a two you like, but this is the simplest one that I could think of.

Runpeng Li · Answer

Okay, so for problem 16. 16 we are given two vectors, and we and were given this ban off this two vectors. So that&#x27;s the space, um, space spend by these two vectors. First is three. There are two, and Victor, too, will be connected to their three. So we need to write how? Yeah, We need to write down five victors in this space. So let&#x27;s first to use two up one. Say, this is our X one. To affects one mine minus T o B will be one minus B two. So this will be six minus negative, too. So that&#x27;s hate. And zero and two times 24 minus three. So that&#x27;s one that&#x27;s right. Down another one. Say, um, three off B one. Tow us Be, too. So, in this case, three times 39 minus two. So that&#x27;s seven. So zero on the second entry and three times too will be six and six plus three will deny. So this is our second vector. Okay, 3rd 1 let&#x27;s try elective V one us say plus V two. So this will be negative three plus next to for centuries. So this will be negative five again zero for a second entry and negative two plus three. So that&#x27;s one. Okay, export. So this time we try connected for the one us. Me, too. So this will be negative. 12 minus two. So he&#x27;s selective. 14 and zero on a second entry and connective aid plus three. So it&#x27;s next five. Okay, so that&#x27;s right down the last vector. Say it will be four of the one or a V one, because to off me too. So this will be 12 minus four. So that&#x27;s hate on the second on the first entry there on the second entry and aid 6 14 on this on 1/3 entry. So here&#x27;s our bye vectors that I write down. You can write down any other factors by taking different ways for or be want to be, too. But this is the This is my answer. In this case,

Sean Thrasher · Answer

Okay, so we got this vector equation right here and were asked to write it in on equivalent system of equations which have done here. So this is the answer How I get to that while the left inside this is six x one minus x 15 X one, Right for this vector right here. I factored this X one inside, and then this is similarly this Okay. And then the left hand side you, When you add these two vectors together, you get this. Okay? And this is equal to this on the right hand side and for two vectors to be equal, they have to be equal component by component. So this right here has to equal this. I&#x27;ve done that there. This has to equal this. That&#x27;s that. This has to equal this, and that&#x27;s that. And we&#x27;re done.

Michael Jacobsen · Answer

in this example, we have five vectors provided and what we&#x27;d like to do is define a vector space h to be the span of vectors, v one, V two, V three before and V five. So now the H is the span of this set of vectors. One thing to notice is that the set here does not form a basis. The reason being is that we have five vectors, but only four entries per vector. And whenever we have more vectors than entries per vector, the set here would be linearly dependent. So we know it spans. And if we want a basis, what we need to do is to decide which vectors to delete so that we still span h and form a basis. And as you might have suspected, the trick to determining all of those things at once is to form a maze tricks and ro reduce. So let&#x27;s start off by saying, Let a matrix a be given as the one be to the three before and be five as its columns. Then this matrix is going to be 10 negative. 32 012 Negative three Negative three negative 416 and one negative. Three Negative 87 last. It&#x27;s 21 negative, six and nine. So this is going to be possibly an interesting role reduction. Since we&#x27;re dealing with a four by five matrix, let&#x27;s see what we get into. We&#x27;re only going for echelon form, let&#x27;s remind ourselves. And so from this pivot, there&#x27;s just two inch, two issues We need to eliminate this and this entry. So here I&#x27;ve copied the first rows of this matrix and now let&#x27;s multiply row one by positive three At the results to growth Row three, what will obtain is a zero to negative eight negative five and zero for the next step, multiply row one By this time a negative two and then add the results to the last row. This operation will give us all together a zero, then negative three Looks like a positive 12. Then a five and a five. So with analyzed, this new matrix we have a pivot. Here was euros below. We have a pivot here as a zero above, but keep in mind, we&#x27;re just going for echelon. So that&#x27;s nice but not needed. But we need to eliminate the two and the negative three. So let&#x27;s start by re copying this matrix Rose ones and two. Mm. So the matrix is copied and we know have a zero here and here. So let&#x27;s look at what the pivots are. This time we have a pivot and column one a pivot and calm to as well as a pivot in Column four. And that&#x27;s all we have. But I also noticed that column one correspondent to V one. Then we have the two and V four. So what we need are V one V two, Envy four If we collect these particular vectors where you will still span, but they will also be linearly independent, since they corresponded to the pivot columns and a reduced echelon form. So we&#x27;re ready to state our conclusion. The set consisting of V one V two and before we already know it spans because that was given. But we also know it is linearly independent now, by definition, a set of vectors that spans and is linearly independent forms a basis. So we can say, therefore, this set forms a basis. We should say a little bit more, and we have to be careful. It doesn&#x27;t form the basis for our four. It forms a basis for what its fans, which is H. It&#x27;s just a subset of our four, so be sure to say it forms a basis for H itself.

Doruk Isik · Answer

and this problem, we&#x27;re given a sending electors knew where to find the basis for this that so let&#x27;s form our system for big from the given lectures. Endless side of this matrix site is our system a CZ you can see hey has four rows and by colors. So it means that a mite hair enough given elements for Pete Rose, since the number of problems is greater than the number of rows. So we know that this system will spend the space. Um, next thing to do is to find the reduced role for or by this system and that Is he going to 1000 0100 000 Sorry. 0010 Negative one negative. 300 Negative to link two by two and zero. Since we know that this settled vectors, I&#x27;ll spend the space Don&#x27;t be basis. Dundee column Give its columns will be the basis for the system. Hey, what are you giving calls? The 1st 1 the 2nd 1 and the 3rd 1 will be our give. It holds so many faces where we set a will be the first told him off a 1001 The second column Obey negative to one negative one. And her column? Um A, which is six negative one, too.

Taylor Shimono · Answer

so for a problem 16 We are going to want to calculate Vector Z and we know that he has given to us by this equation using our vectors u V and w so we could go ahead and start by finding Thies killer multiple of each of our vectors here. So we know that we want Thio will supply vector you by our skill er seven. And so to do that we can go ahead and essentially just, uh, multiply each component of vector you by seven. So this will give us a negative seven. Three times seven is 21 2 times seven is 14. Next, we know that Vector V is just going to be a scaler multiple of one. So it&#x27;s going We&#x27;re going to use the original vector V that were given and finally four vector w We are going to be multiplying by the skill er negative 1/5 So negative 1/5 w is equal to five times negative. 1/5 is negative one zero times Anything will still be zero and negative five times negative. 1/5 will give us positive one so we can go ahead and add each of our vectors together. You know that Z is going to be equal to seven. You, which we said was Vegas seven 21 14. Um and this is going to be plus V, which is negative One negative. Too negative too. And finally a plus. Negative one face still be so that will be plus negative +101 And we know that when we want to add vectors together, we are going to add their corresponding components. So that tells us that Vector Z is equal to we can go in and start by adding seven years to V. So we will add Negative seven plus one will give us a negative six. 21 plus negative too will give us 19 and finally 14 plus negative to with it was 12 and we know that we still have to add our last vector negative one Finn still view to this, so that&#x27;ll be plus negative +101 And finally we find that Vector Z is equal to negative six minus one. So that&#x27;ll be negative. Seven 19 plus zero, which is still 19. And finally our Z component is 12 plus one, which is 13

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In Exercises 15 and $16,$ list five vectors in $\…

Problem 15 Easy Difficulty

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

Linear Algebra and Its Applications

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Absolute Value - Example 1

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

Grace H.

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

Linear Algebra and Its Applications