mark-each-statement-true-or-false-justify-each-answer-a-if-mathbff-is-a-function-in-the-vector-space

Question

Mark each statement True or False. Justify each answer.
a. If $\mathbf{f}$ is a function in the vector space $V$ of all real-valued functions on $\mathbb{R}$ and if $\mathbf{f}(t)=0$ for some $t,$ then $\mathbf{f}$ is the zero vector in $V$
b. A vector is an arrow in three-dimensional space.
c. A subset $H$ of a vector space $V$ is a subspace of $V$ if the zero vector is in $H .$
d. A subspace is also a vector space.
e. Analog signals are used in the major control systems for the space shuttle, mentioned in the introduction to the chapter.

Daniel Pezzi · Accepted Answer

Hello. For this question, we just need to answer five basic, true and false questions. Vector spaces. Well for basic, true and false questions about vector spaces on, then, a question about analog control systems on the space shuttle Andi Zehr Just thio test to make sure that, yeah, your your knowledge of definitions and kind of the weird border cases is where it needs to be. So for the first question, it&#x27;s it. It&#x27;s is a function F 50 that is equal to zero for some tea. Is this the zero vector? Where are vector space? Is the set of all real valued functions on the real line and the answer to this is false. Hopefully, this should be pretty intuitive. The zero vector when your vector space is the set of all real value function should be zero function. The names kind of work out. But the reason why is if you eyes actually in the definition of what this zero vector is, zero vector is a vector, such that for any other vector in the vector space. This relationship holds, um, this pretty kind of elementary definition of what a narrative zero is, and a good counter example to this is the sine function. So I think of sine of t and I add it and I add say, uh what za good function say one one is a real valued function and I add it Thio I had sign of tea which is zero at zero So it meets this condition over here. This does not equal one again like ah supposed zero vector should. So this shows that this is in fact false. You need thio be equal to zero everywhere along the rial line. So the second question is is a vector in three dimensional space just a narrow on that is true So this is kind of geometric interpretation. So in three dimensional space we write factors as three arbitrary constants A B and C O r you know, not not arbitrary constants but three particular constants a, B and C on This translates to some vector in three space. If I could get my drawing right, um, where the first coordinate is through the length of the the amount of the vector in the X direction. Second coordinate is the amount of vector in the Y direction and then the third coordinates the amount of actor in the Z direction. Um, so, yes, this is the correct physical interpretation of vectors in three space. So the third true and false question asks about is, uh if I have a subset of a vector space and I know that the zero vector is within this space Does this imply that h is a vector space? Um, and the answer is no. There are three conditions that a subset of a vector space has to meet to be considered a subspace. Um, the first is this condition that it contains the zero vector. The 2nd and 3rd are that they are closed under arbitrary edition of elements in this space and, um, multiplication by arbitrary constants. Eso ah, good counter example to this is if you think about our to and which is clearly a vector space, then if I think of the the unit disk which is the set of all elements, um such that this relationship holds X squared plus y squared less than or equal to one. This contains the zero vector or which is just the point which of the origin written as a vector. But it is not a vector space. Because if I if I take a vector in this space and I multiplied by by a large enough constant, it will escape this space. Eso It&#x27;s not closed under multiplication by arbitrary constants. It&#x27;s also not closed under addition Ah, vectors in space. Like if I take the vectors 10 and add it to the vector 01 this new vector V does not satisfy this relation show. It&#x27;s not in this space. H so h is not necessarily of a subspace. All subspace is of a vector space satisfied this property. But not everything that satisfies this property is a subspace of vector space. So next question is, um, kind of related to that, which is is the subspace of a vector space. So if I have a similar relationship, I know that h is a subspace, and I know that V is a vector space. Does this imply that H is a vector space? And the answer is yes or true on. And this is, um, kind of testing back when you were developing three idea of subspace is is that say, Hey, if I If I have this this thing that sitting inside of, ah, set that I know a lot of information about. Then I should not have to go through and check everything again because there&#x27;s like a dozen or so conditions that something has to meet to be a vector space. There&#x27;s three that it has to meet to be a subspace, but because all of the the other properties of vector space are carried over by this relationship, just the fact that it&#x27;s a simple subset. The only thing stopping, ah subset from being a vector space in and of itself are the three conditions the it has zero closed under arbitrary edition of actors as close under multiplication by arbitrary constants of actors. But yes, if something is a subspace, it is considered a vector space by itself, in fact, that the definition of a subspace is just, ah, vector space contained within another, another vector space. The final question is book dependent, not math dependent, which is, um, our analog systems used in major control systems for the subs for the space shuttle, it&#x27;s mentioned in the introduction of the chapter. If you go back and read it, you&#x27;ll find out that the answer is no. Analog is not used in the space shuttle anymore. That is false, all right, and that&#x27;s it.

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

Well, it&#x27;s like a different album. Were given a statement and were asked to mark each statement True or false and justify our answer. V is a vector space. Mm. In part a, the statement is Yeah, our square. It is a two dimensional subspace of are acute dominance of Yes, So they&#x27;re in there. They&#x27;re like, you know, podcast. They got a very dialectic. This is kind of a tricky question, but we know there are two is not a subset of our three because the vectors in R two of two entries and vectors in R three of three entries, however, are too. Is I so more thick to a subset of R three? Therefore, since our two is not even a subset, the statement is false. Then in part B Mm. The statement is the number of variables in the equation. X equals zero equals the dimension of the null space of a Mm. Oh, well, we know that the dimension of the null space of a is actually equal to the number of three variables. Therefore, if the number of free burials variables is not the same as the number of variables which is possible, the statement is false. So the statement is false. So get dick kebab. Pray in part C. Something like statement is a vector. Space is infinite dimensional. If it is spanned by an infinite set, read it. Suck dick Just legal out Grace. Well, in fact, I actually want it&#x27;s true that any and finite character doesn&#x27;t make any sense. Vector space is spanned, and it&#x27;s not like I don&#x27;t I don&#x27;t bad job by an infinite set as well. Green Any non empty finance sector space, I should say. Therefore, to solve your requirement is that an infinite vector space cannot be spanned said by a finite set. And in this way, an infinite vector. Space is different from a finite vector space because finite victor spaces must be spanned by a finite set. Therefore, the statement is once again false. What happens to it finally in part? Mm. Actually, that was part C in part d. The statement is if the dimension of V equals n and s spans V then s is a basis of the Yeah, but you say types that we&#x27;re talking about fucking face. Yeah, this is actually mhm. Not necessarily true. Well, if the dimension of V equals n and s is equal in s spans, V then it follows. That s is a basis for V if and only if the number of vectors and s is also in. Therefore the statement could be false. We could choose a set s which spans V, which has more than or less than n vectors. In fact, it has to have more than n vectors. Finally, in part E were given the statement. The only three dimensional subspace of r three is ours. Three itself. Mhm. All the well, we know that the dimension of our three is three. Yeah, therefore span of our three must contain three. Mm, but nearly independent factors. And therefore he didn&#x27;t finish mhm r three. It&#x27;s happening right now, right? Okay, that&#x27;s true. All right is the only three dimensional subspace of our three. Therefore, the statement is true. Would have

Bobby Barnes · Answer

right. So they want us to determine if these statements are going to be true or falls. Um, so this first one, it says essentially, that we have the zero vector and H. So this is a subspace of our in and this is going to be false because it doesn&#x27;t say anything about the other two properties that we need to have, which is it Closed under sums and closed under scaler multiplication. So false. Because we do not No. If actually, let me beat the Patriots. Bigger. Do not know if closed under some and scaler multiplication. Okay, so we have a there now for B. They say we have this set of factors, and the set of all linear combinations is a subspace. And so this one is going to be true, because if we just think about, um, what this is so it would be something like see one V one plus c two, b, two plus dot, dot dot all the way up to C p v p. So we can get the, um zero vector by saying all see, I equal to zero for every eye. So that&#x27;s fine. Um, we can get any scaler out of this by saying, see, hi equals zero except for some value. Um, but I is equal to and where it is, like the vector that we want. So that gives us the scale of property and three, some wolf. I mean, some just kind of comes from the definition. So some is, um, true by definition of linear combination linear combo. So that&#x27;s how we end up getting true for B. See, I believe this one is also true, but let&#x27;s go ahead and kind of draw it out. So if we have over here some Mbai in Matrix, remember what the knoll space really is? It&#x27;s saying, Well, what collections of vectors here output the zero vector. And, um, in order for us to multiply these we needed in by one matrix, which means we would have in entries here and then this is an element of our in. So then the null space has to be a subspace of our because it will have the zero vector on. We&#x27;ll also have where if we add two things by the linearity of matrix, that should be fine. And then if we just multiply this, that should also be fine. Um, well, actually, let&#x27;s go ahead and write all those out. So eso let&#x27;s say just so we can kind of be a little bit mawr exact right? So Ah, we have some matrix A with the vectors. Let&#x27;s just call it X Y elements of the knoll of a All right. So if we want to do the some property So let&#x27;s look at a Times X plus y Well, then this is going to be a X plus a y and by definition of both of these being in the null space, this is the zero vector plus zero vector which is zero vector. And so then that implies X Plus y is also an element of nor a Okay, so we have that one and then first scale er&#x27;s. So this is some. So that&#x27;s good. I guess we should have also did the zero vector. But that one&#x27;s kind of true just by definition. So this is zero element of Norway. Alright, so that&#x27;s good so far and then lastly we need these scaler. So we have a times c of X, remember axes so in the north space. But we can go ahead. And do you see? Actually me, Right This over here. So this is going to be C a X like that so we can apply community since he is just some constant. And then we&#x27;ll that c times zero vector, which is equal to zero vector. So then that implies C X is also in the knoll. So then the constant multiple or the scaler also holds. So, um, I don&#x27;t know if you really need to be that exact, but, I mean, it doesn&#x27;t really hurt now for D the column space of Matrix A is the set of solutions. A X is equal to be all right. And so this is actually going to be, uh, false. And the reason why this is false is because it&#x27;s not necessarily just for the solution of one. It&#x27;s supposed to be the span of all of the columns I&#x27;m sorry of. Ah, yeah, the span of all of the columns. Um, so, yeah, that&#x27;s why this one isn&#x27;t necessarily true. Because it&#x27;s not talking about a solution set and then for E. So this is saying that B is the echelon form of a matrix A in the pivot columns of B form a basis for a And so this one, um, is also false Because it&#x27;s not the pivots of B, but the pivots of call a Arab the pivots of a instead. So this should actually say a here on beacon, actually kind of show that off on the side really fast. Eso Let&#x27;s come down here and do that, right? So let&#x27;s say we have some matrix here. And so this is kind of like you can use, like, a counter example as to why this wouldn&#x27;t be the case. Um, so we have, like, 123 uh, then 567 13, 14, 15. And then, I don&#x27;t know, negative 112 So hopefully if I reproduce this year Ah, this gives me 111 000 And if it doesn&#x27;t just kind of tweak this so it does. But I believe this will give us that. Okay, well, if we look at it, this is a and then this is the echelon form. Be over here. Well, if we were to look at this, though, over here, this is saying if we take each of these are fourth position of our vector is always going to be zero. But over here, notice There are numbers here. So we wouldn&#x27;t even be able to get out the, like, original columns here. So because of that, um, we would need to use the columns for a as opposed to the columns of being So Yeah. So that&#x27;s kind of like the rationale behind. Why that? Why that ISS? Yeah. So s it looks like the first statements. False second statement. True, They&#x27;re true. And then the last two are going to be false.

Runpeng Li · Answer

Okay, so for problem 12 we need to identify statement a first to call himself the columns of the change. According the Matrix from B to see Arlene Odie independent, whether this is short balls. So, uh, so this statement is true. The reason this is stated in our textbook and excess. So to call himself off the change of corn and a matrix frumpy to see our linearly independent because they are the coordinates vectors off the we nearly independent set off. So how just marked the page number so you can check this out. Page page 2 28 Okay. So Steam and B says if the d take the view to be are too And, uh, the basis is we won&#x27;t be too. And, uh see wants you to. So the real reduction Upsy once you do you want me to to I and P produces it to reduce its ah metrics. P that satisfy said it&#x27;s 56 b p x c. So one thing to notice here is that misty mate says for all specter eggs in the But this is impossible. So why it is impossible. I&#x27;ll try now. Explain the eight and the next page. So first, by our equation, giving the textbook we have can t the change of corn The matrix to BC to be the change of quantum matrix from Peter to see Then we have, ah record reckon X under B is equal to universe uh, Bracket C So he in first it&#x27;s the Patriots that satisfied is he this equation? And by overthere, um 16 in a textbook that implies, Yeah, p it&#x27;s not always the case. He is equal to P. Ah, yeah, this Now it&#x27;s not always the case that the inverse of peace you got to pee. So this statement is wrong.

Ashley Boni · Answer

eso part, eh? You don&#x27;t be minus b dot You equal zero. Well, we know that the dot product is commuted s. So that means that you dot b is equal to read about you. And these two things are part B for any scaler. See, uh, the norm of CVI is equal to see times the normal b Well, so we want to see if, uh, we can pull a scaler out of the norm. And if this holds true, So let&#x27;s take c t equal Negative one on our vector Be to just be, say, 11 Let&#x27;s compute uh, these two values So the norm of CV is equal to the norm of negative one negative one which is equal to the square root. Um, negative one squared postnegative one squared, which is a squared. But now, if we take, uh, c times the normal V, we get negative one times the norm of 11 which is negative one times the square root of one square post one squared, which is negative, squared or two. So these things aren&#x27;t the same once he&#x27;s negative. So this one iss who else? This person is true. Uh, next for part C. If X is or thought inal toe every vector in a subspace W then X is in, um, the Ortho Wagnalls base to W. So that&#x27;s just the definition. Um, the space here, This just means that all the vectors that or thought minal to the space w closes just true. By definition, party says, if the normal view squared plus the norm of these squares is equal to the norm of you plus b squared then you movie or are orthogonal. So if we know that this is true, let&#x27;s write out what this means. So the normal view squared is you, don&#x27;t you? Normal v squared is we Dock B now on the right We have you close to me dot You cost me So we have you, don&#x27;t you? Plus B, that me is equal to, um this very hand side we can essentially like foil it out So we have you got you Plus to you don&#x27;t be plus me dot B um, so you don&#x27;t use could cancel be Darby&#x27;s can cancel. So we get zero equals two times you don&#x27;t me. So this means that you don&#x27;t be has to equal zero, which means that you and v our orthogonal. So this is true. And lastly, part e ah, for an m by n matrix A vectors in the null space of a our earth Ogle two factors in the roast base of it. Well, if we look in the section on here, um, three. It states that everything orthogonal to row space of a is equal to mall space of a. So this is just another way of saying exactly what the statement says. Um, so this is true because this is just stating they&#x27;re on three, which is on page 3 55

Problem

Mark each statement True or False. Justify each a…

Problem 23 Easy Difficulty

Answer

a. false.
b. true.
c. false.
d. true.
s. false.

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Anna Marie V.

Heather Z.

Caleb E.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Caleb E.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

a. false.b. true.c. false.d. true.s. false.

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Caleb E.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Heather Z.

Caleb E.

a. false.
b. true.
c. false.
d. true.
s. false.

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

Linear Algebra and Its Applications