let-mathbfv_1leftbeginarrayl1-0endarrayright-mathbfv_2leftbeginarrayl1-2endarrayright-mathbfv_3leftb

Question

Let $\mathbf{v}_{1}=\left[\begin{array}{l}{1} \ {0}\end{array}ight], \mathbf{v}_{2}=\left[\begin{array}{l}{1} \ {2}\end{array}ight], \mathbf{v}_{3}=\left[\begin{array}{l}{4} \ {2}\end{array}ight], \mathbf{v}_{4}=\left[\begin{array}{l}{4} \ {0}\end{array}ight],$ and $\mathbf{p}=\left[\begin{array}{l}{2} \ {1}\end{array}ight] .$ Confirm that
$\mathbf{p}=\frac{1}{3} \mathbf{v}_{1}+\frac{1}{3} \mathbf{v}_{2}+\frac{1}{6} \mathbf{v}_{3}+\frac{1}{6} \mathbf{v}_{4}$ and $\mathbf{v}_{1}-\mathbf{v}_{2}+\mathbf{v}_{3}-\mathbf{v}_{4}=\mathbf{0}$
Use the procedure in the proof of Caratheodory's Theorem to express $\mathbf{p}$ as a convex combination of three of the $\mathbf{v}_{i}$ 's. Do this in two ways.

Lucía Guerrero · Accepted Answer

we can check that. This is expression is correct. Just by doing the corresponding products on the some eso Here we have a p written as a linear combination off the B one b two B three, before on the coefficients of these linear combinations are going to be called. She wanted to see three and C four on. What we want to do is to write p as a linear combination, but involving less factors. So for that, we&#x27;re going to follow Kathy other East procedure so you can go check the details in the chapter on were given this expression to Sophie one minus V. Two plus feet, three minus before sirrah. However, we&#x27;re going to multiply by minus one. So we have minus 131 plus B two minus 53 plus before equals to Zira. I wrote it this way because I want to. Well, I will make reference to this coefficients on. I&#x27;m going to call the thing one thing to think. Three on 34 Okay. So e multiplied by minus one. Because I wanted to have the four have positive coefficient for the more I wanted see four over the four, which is in this case 1/6 to be less than or equal tendency to over the two, which is one third. I was interested in C two over the two because the two is also positive. So now I&#x27;m going to define this numbers they want or B I equals C I minus C 4/84 times three I. So we&#x27;re going to compute them. The one is gonna be a C one, which is one third minus c for a birthday four, which was 1/6 time. Stay one which is minus one B two city, which waas one third, minus 16 times the two which Waas one be tree is going to be C three in this case 1/6 minus 16 times in three, which is minus one on the same for before before is going to be 1/6 minus 16 10 54 which is one on these numbers are one half 16 one third and Syria On. With this numbers, we can write p as a near linear combination. The in this case, it&#x27;s gonna be one half off. Be one pause 1/6 off feet, too, plus one third off feet three. No, I&#x27;m going to write the fees. Explicitly. This is one half of the victors. She 10 +16 off Vector 12 on plus one. Third time&#x27;s a factor for two on board. Yes. So we have done half of the problem. We need to find another way to express this vector p. Again, We&#x27;re going to do it by using this method with it before. So for our second expression, we&#x27;re going to let the big one and feet too, as they were before. But we&#x27;re going to interchange feed three on before, So V three is going to be 40 on before is gonna be the vector 12 Now you can verify that. Be one minus feet too. Minus 33 plus before equals to zero eso this expression is gonna give us the new the eyes on notice that since the coefficient C three and C four, we&#x27;re both 1/6 then the sea ice. They&#x27;re going to be the same as before. Um, so we like this 31 minus 50 minus 53 plus fee for equal zero Because this way, see forever. The four is again. 1/6 on this is less than or equal than see one over the one which is one third we care about you whenever they want. Because they want was Thea other positive coefficient in this expression. So now we have everything to compute the B eyes again. B one is one third, minus 16 times the B one which in this case is one B two is gonna be one third, which is it to minus 1/6 times did, too. Now the two is minus one Be tree is going to pay one sexed minus 16 times the three we choose minus one on before is gonna be 1/6 minus 16 times before, which is one now. These numbers are before is going to be zero B three is one third B one is when sixth on B two is one half. So our second expression is P equals to 1/6 off 31 plus one half off feet too. Plus when third off feed three. But now here you gotta be careful, because remember, we interchange the roles of the three and before so writing the vectors explicitly, we have p equals 16 of the victor 10 plus one half of the victor 12 plus one third of the victor Forests. Hero on this is another way to write P s a linear combination off they for or, you know, victories.

Chris Trentman · Answer

were asked to prove that the sum of conduct sets is again a convex set. So let A and B he convict&#x27;s now let the vectors x and y B in the set some a plus B. This means that we confined in a in C in A and A B and a D in the set B Such that X is equal to a plus B and why is equal to C plus D. Now let t lie between zero and one Them we have that the point on the line segment x y It&#x27;s going to have the form W, which is tee times X plus one minus t times why you could do the other way around. This is the same as tee times a plus B plus one minus t times C plus d. Of course, this is equal to grouping together vectors from the same set. So we have t a plus one minus t see plus t B plus one minus t the now because a and beer convex it follows that ta plus one minus t See, this is an element of the line segment between A and C. This lies in a and we have that T B plus one minus TD. Since B is convicts and this is an element of the line segment between B and D. This also lies in B, so it follows that w lies in a plus. B. Therefore, it follows that a plus B is convicts, since it contains any points online segment between and between X and Y.

Chris Trentman · Answer

so given a conduct set and to positive numbers, C and D were asked to show that CS plus D s going to be equal to C plus D s. So if we in other words, stretch the elements of a convex set by sea and stretch the elements of a complex set ID and then add thes two new sets together we get a set, which is three elements of s stretched by C plus D. So we&#x27;ll let s be convex set. Let&#x27;s see, be greater than zero and d be greater than zero. Now let X be an element of CS plus ts and we confined S one and s two in s such that X is equal to see X one are sorry CS one plus D s to then it follows. The X is also equal to C plus d times. See over C plus d. We can divide by C plus Decency plus D is greater than zero since both c and D or greater than zero s one plus de over C plus D s to now because S is a convex set. It follows that see, oversee plus D s one plus D over C plus D. As to linear combination of elements from s also lies an s. So it follows that X is an element of C plus D s. So we&#x27;ve proven one direction. That is, that C s plus D s is a subset of C plus D s. Now to prove the other direction Similar exercise. So we&#x27;re going toe Let X be in the set C plus D s. This means we confined. A vector s in the set s such that X is equal to C plus D. Yes, and of course, this is equal to CS plus D s. After distributing the scaler, we could say, even distributing the vector across the scale Er&#x27;s And this is clearly an element of C S plus D s. So X is an element of CS plus ts and therefore it follows that C plus D S is a subset of CS plus D s. Combining this statement and the previous statement we get that CS plus D s equals C plus D s, which is what we wanted to prove

Chris Trentman · Answer

we&#x27;re asked you strong induction to prove that when a simple polygon p with consecutive Verdecia be one through the end is triangulated into end minus two triangles. The end minus two triangles could be numbered 12 up their end minus two. So that V I is going to get Vertex of triangle I for I from one end minus two. So let P N b the statement that the end minus two triangles and the number 12 up Teoh and minus two such that the eyes of Vertex of triangle I for i e going one up two and minus two. So clearly we can&#x27;t triangulate simple polygon with anything less than three courtesies. So for of this statement for the basis statement first So in this case, n equals three So a convex polygon p the three birdies scenes is a triangle and thus is the triangulation of P name the entire triangle of the one. Then we have that triangles have been labeled such that the one is a vertex of triangle one. Sorry is trying one so have shown that p three is true and to use strong induction Let&#x27;s suppose that p one starting at one p three up through PKR. True for some K greater than or equal three. We want to prove that P K plus one is true. So let P K plus one a convex Holly Gagne with Vergis sees a one B to weigh up to V K plus one. These are actually consecutive Verdecia section C. They&#x27;re connected by edges each the truth and let Devia diagonal its present a triangulation of peed and minus one triangles. And we had this. Dagnall has to exist. Or else there will be more than an minus two triangles, I mean and minus one triangles. And let&#x27;s assume that the Dagnall D is between Vergis ease V one and V I where we have that I is gonna be anywhere from three decay. It&#x27;s any non adjacent vertex. And of course, if not in this relay Bolivar theses of the polygon. So we have that convicts Holly Gagne keep prime with courtesies. B one B two up to he I You&#x27;re consecutive courtesies waas a triangulation with I minus two triangles and the I minus two triangles can be labeled 12 up through I minus two, such that VJs a vertex of Triangle J. This is because the statement t J p J p i is true since I z going to be less than K plus one. And likewise we have the convicts Polygon p double prime with the consecutive Vergis is V I plus one up to the K plus u K plus one. The one has a triangulation with. So the number of burgesses in this case is going to be K plus one minus I and then from that will subtract two to get the number of triangles in these triangles. And he labelled well as I I plus one up to In this case we get k plus one minus I minus two. So it&#x27;s k minus one minus. I such that we have the J is going to be a vertex of Triangle J. This is because the statement p k plus warn minus I it&#x27;s true again. I is going to be less than faithless one that are equal to one. And so this form the triangulation of P such that the end minus two triangles. I&#x27;m sorry, the K minus two triangles, You should really be k minus one. Sorry came on this one. Triangles we have can be numbered 12 all the way up to K minus one. And so that her text VJ z vertex of Triangle J. And so this shows that P K plus one is true in the US by strong induction. It follows that PN is true girl and greater than or equal to.

Kira Harland · Answer

everybody today, we&#x27;re gonna be showing how to prove the consecutive interior angle serum. Converse, Converse here. All right. And what This is saying that if we&#x27;re given angle one the measure of Ingle one plus the measure of angle to equal 180 degrees In other words, of its supplementary, then I can say that these two lines here M and cut by the Trans Mursal are parallel. Now, in order to do a proof, I need to start with a proof table. So I want to draw my proof table. You should draw yours too. And on the left side, I&#x27;m gonna put statements and on the right side, I&#x27;m going to put reasons right. We always need to say why we&#x27;re saying that things were saying okay in life and in geometry. So let&#x27;s start with our givens. The first statement I&#x27;m gonna write is just this up here. I know that the measure of angle one plus the measure of angle to equals 180. And how do I know this? This is just given to me, right? I know that my to consecutive interior angles here are supplementary. That is just something I have been told. The second thing is, I just want you Look at the diagram. Can you make any other statements about angles here? Well, I see angles to an angle. Three. So angle to in Ingle three. What angles hair do they form together? You should see if they form a linear pair. All right. And good. I&#x27;m sorry. I got a little bit wonky. They form a linear pair. And what do I know about linear Paris? I hope you got it right. They add up to 180 degrees. So my measure of angle to plus my measure of angle three is it cool to 180? And why, uh, they form a linear repair. I was gonna write Lynn pair because I wrote linear over there, So hopefully not so I&#x27;m saying so. They are supplementary. All right, Well, what this shows to us is that if one in Tuller 1 80 in two and three or 1 80 This this is the same between the two statements. This is the same between the two statements. Therefore measure Bingle one and measure bangle three must also be congruent. So that is our third line of reasoning that we&#x27;re going to write down. We know that the measure of Ingle one actually weaken just right. Angle one is congrats into angle three. So I know that angle one is concurrent Teoh Angle three. And this is because of substitution between the two statements in line one in line to okay. And if you&#x27;re still confused, look at the blue and look at the green. Right angle one in Ingle three had to have been the same. Well, this is interesting. Angles one in Ingle three. This is where you go back to our geometry caps. What do we know about angles? One in Ingles. Three. What kind of angles are they to each other? That&#x27;s right there. Corresponding. So here I have to corresponding angles that are congruent. And I have a wonderful the&#x27;re, um that This helps me figure out if these two corresponding angles are congruent. Then m is parallel to end. So let&#x27;s write that down. And I know this because of the corresponding angle. I&#x27;m just going to abbreviate angle like this converse serum, and I will brave eight serum as ta drum. So because of the corresponding angles, the&#x27;re, um I know that from this line in number three, that number four should be correct. So to recap I started with if angle one and angle to are supplementary what was right? Step then M must be parallel toe.

Ashley Boni · Answer

so wanna rate the vector V as a combination of a vector in the span of you one and one in the span of you&#x27;re 23 and four. Uh, so first we want to find the one in the span of you won. And we can do this by, uh, using the formula for the orthogonal projection of a vector onto you want. So let&#x27;s call this vector be hat and it&#x27;s given by we thought you want over a year warm. Got you one, um, all supplied by defector you want? So do some scratch work be dot You won. So we have four times one plus five times two minus 3 10 1 plus three times one. So we get four plus 10 and minus three and 30. Cancel. So we get 14. And for the denominator, uh, you want a product with itself, We get one times one plus two times two close one. That&#x27;s one close one times one. So we get four plus three, which gives us seven. So this quality on the outside here is 14 over seven More too. So we have two multiplied by one sort is gonna multiply each component by two, Giving us 24 22 Um, and since all the use for another thought, little basis for horror, for we can just take B minus free hat, let&#x27;s see what we get. And we know that, um, they will be in the span of the other directors in you. So v minus. We had ISS four minus two by minus four. Negative three minus 23 months is too. So we get to one, maybe 51 So that&#x27;s the vector we need to add to be had to get back to be so. Therefore we can write be as a sum of 24 to 2 plus 21 negative 51

Problem

Repeat Exercise 15 for points $\mathbf{v}_{1}=\le…

Problem 15 Easy Difficulty

Answer

$p = \frac { 1 } { 2 } v _ { 1 } + \frac { 1 } { 6 } v _ { 2 } + \frac { 1 } { 3 } v _ { 3 }$

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

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$p = \frac { 1 } { 2 } v _ { 1 } + \frac { 1 } { 6 } v _ { 2 } + \frac { 1 } { 3 } v _ { 3 }$

Linear Algebra and Its Applications

Lily A.

Heather Z.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Heather Z.

Samuel H.

Joseph L.

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

Linear Algebra and Its Applications