mark-each-statement-true-or-false-justify-each-answer-a-if-mathbfyc_1-mathbfv_1c_2-mathbfv_2c_3-math

Question

Mark each statement True or False. Justify each answer.
a. If $\mathbf{y}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3}$ and $c_{1}+c_{2}+c_{3}=1,$ then $\mathbf{y}$ is a convex combination of $\mathbf{v}_{1}, \mathbf{v}_{2},$ and $\mathbf{v}_{3}$ .
b. If $S$ is a nonempty set, then conv $S$ contains some points that are not in $S$ .
c. If $S$ and $T$ are convex sets, then $S \cup T$ is also convex.

Lucía Guerrero · Accepted Answer

and this exercise statement a is false. We also need that every c I is a non negative number. Yeah, a statement fee is also false In the case when we Onley have one vector and its convex hole is itself a statement See, he&#x27;s also false Consider the following example and are too Every line segment is a convex set. Thio is anti in this case our comeback sets on their union is not a comeback set. If it were comebacks, the it won&#x27;t contain the following green segment, but it doesn&#x27;t.

Gideon Idumah · Answer

so we have some show of force. Question. We&#x27;ll see that the A part is true because this is a symptom of US definition given on pains. 4 57 The be pot is also true. This is you can see Sarah AIDS on page 4 50 seats. The Sea Potts is also true because this is exactly this sister insane. That&#x27;s his car, it&#x27;s your dog returns.

Chris Trentman · Answer

forgiven a statement and were asked to market tour false and then to just fire answer in part a. The statement is Cube in. Our three has exactly five facets. The statement is false. A number of facets of a cube in our three is six because in our three, a facet is the same as a square face. Now, in part B, the statement is a point. P is an extreme point of a poly tope if and only if p is a vertex of the politics. This statement is true. This is a direct consequence of the&#x27;re um 14. In the book, the serum says that a point of a poly tope is an extreme point if and only if it&#x27;s a vertex and definitely if it lies in the profile now in part C were given the statement if s is a non empty, compact convex set and a linear functional attains its maximum at a point p then P is an extreme point of s. This is false. So while the maximum is always attained at some extreme point maybe other points that are not extreme points I wish the maximum is attained. In fact, there are examples given in the book. I would say, for example, look, at example three part C. After the&#x27;re, um 16. We see that the linear functional attendance maximum value at to extreme points, but then also at every point in the complex whole of these extreme points and these points in the comics whole are not. Extreme points now in Part D were given the statement a two dimensional Polito always has the same number of Vergis is and edges. This statement is true to see why. Consider Oilers formula so by Oilers formula checking in to be to We get that for a two dimensional polito f zero of call this P and then minus If one of P is equal to one plus negative one to the first, which is zero So it follows F zero p is equal to F one of p. Therefore, the number of Vergis is for this play. Tope is always the same as the number of edges

Viswesh Uppalapati · Answer

in this video, we&#x27;re gonna be selling problem number 24 of section four point would. And, um, this is basically a five part question where were in each part of us? US. Approved, whether given same is true or false. So, um, the first statement says that it for a it&#x27;s 24 aces, that a vector is any element of the vector space. And this is true because it is the definition of a vector. And, um, it is listed under the definition of the vector space in the book. So we know that a vector is any element of electric space. Um, which basically just means that, um, which basically just means that if a vector contains, like, let&#x27;s, say, three countries, any vector in, um the vector space that is three dimensional because of three column entries would would also be a vector. So this thing that is true B says that if you if you is a vector in a vector space V than negative one, you is the same as negative you. So it&#x27;s basically just saying that negative u equals negative one times you. If you look at Axiom number 10 on the first page of the chapter, Section 4.1. It says that one U equals you. So if you multiply this by a negative one, we got negative one U equals negative. You. So this is true. Um, see, Asa&#x27;s states that a vector space is also a subspace. And this is true because we know that we know that, um, every set is a subset of itself. So seeing the definition of a vector space, you know that every vector spaces subspace of itself, and it is listed in the second paragraph. Um, after the definition of a vector space in this section. Okay, Um, uh, d states that are two of the subspace of our three. And this is false, because if you look at example eight. Um, we know that our two is not not even a subset of our three because there were talking about two different dimensions. Um, and because elements of our two have two entries in them, 12 and elements of our three have three column entries. We know that this demon is false, and the last one states that, um e a subset age of vector space via subspace of we um, if the following conditions are satisfied, the zero backer of E is an h u v e com movie, and you plus a v e r N h um, See is a scaler and see you as an H. So this is tricky as this is Basie how we test if something is a subspace of another vector space. So I stayed in the conditions two and three. Um, the conditions don&#x27;t make sense because it doesn&#x27;t say that the vectors u n Veer actually, um, conditions for this statement to be true. So to be correct, you have toe you have toe say that for every U. N V in H in a couple of places. So this definition is incomplete. It&#x27;s missing some words. Um, so it is not so here. Um, the definition of two which says you come Avi and duplicity yarn h and sees a scaler and see you as an h, um are false because, um, because these vectors also have to be an age when there&#x27;s killed by any value are added thio Any other vector? Um, that that is from the subspace of age that exists in the vector space of age. So this is also false

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

Gideon Idumah · Answer

cinemas. All we have that the A part is force because I find off s equal toe s and they can see uh, the park of below the definition on page for 59 number B parts. This is true. Um, scenes, The fine all off s always contents s. We know these. So if, as contains, it&#x27;s fine all it&#x27;s fine. All So we have a friend all always convinced ISS. So if s contents are fine, all this would imply that s with after Beatie quote So they find all off a fine of s on DDE bites urine Sue serum sue because s a course to a fine of s. This implies that s is our friend Number C is also true. You can see the Parker off below the definition in page for 42. Question is force This statements the statement is only true is all the true when floods he&#x27;s in our three on then is true because floods is a trust leads off subspace. So this means that so if we translate it back if we translates, it&#x27;s Bach to the origin. It becomes a subspace ce good

Problem

Mark each statement True or False. Justify each a…

Problem 11 Easy Difficulty

Answer

F, F, F

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

Linear Algebra and Its Applications

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Kayleah T.

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F, F, F

Linear Algebra and Its Applications

Lily A.

Heather Z.

Kayleah T.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Heather Z.

Kayleah T.

Michael J.

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

Linear Algebra and Its Applications