mark-each-statement-true-or-false-justify-each-answer-on-the-basis-of-a-careful-reading-of-the-tex-2

Question

Mark each statement True or False. Justify each answer on the basis of a careful reading of the text.
a. Two vectors are linearly dependent if and only if they lie on a line through the origin.
b. If a set contains fewer vectors than there are entries in the vectors, then the set is linearly independent.
c. If $\mathbf{x}$ and $\mathbf{y}$ are linearly independent, and if $\mathbf{z}$ is in Span $\{\mathbf{x}, \mathbf{y}\},$ then $\{\mathbf{x}, \mathbf{y}, \mathbf{z}\}$ is linearly dependent.
d. If a set in $\mathbb{R}^{n}$ is linearly dependent, then the set contains more vectors than there are entries in each vector.

Viswesh Uppalapati · Accepted Answer

in this video, we&#x27;re gonna be solving problem number 22 of section 1.7, which is a true falls four part question. I didn&#x27;t wantto right all the words on the screen, So if you have the book in front of you, please follow on. So ace us if, ah, to back a states that two vectors air linearly dependent if and only if they lie on a line through the origin. And, um, that is true, because when a set of vectors air linearly independent, that means they&#x27;re not linear combinations of each other. So they&#x27;re gonna be, um, in separate planes, and they&#x27;re gonna intersect the origin at different times, and they won&#x27;t intersect it in a line. But if a set of vectors are literally dependent, then they&#x27;re gonna be, ah, linear combination of each other. So they should, uh, they, like, extended within the same lines. So they passed through the origin of the same point. Simple enough for more. This is the exact same figure as Figure one on page 59 of the book. So it check it out for to make it easier, which is a zit looks cleaner in the book anyway, uh, the states that if a set of if a set of vectors contains fewer vectors than there are entries in the vectors, um, then the set is linearly independent. So they&#x27;re made space that, um, saythe opposite of this where it says if you have more vectors, Ah, number of vectors is greater than a number of entries of the victors, then the set is linearly dependent, not literally independent, because that would result in free variables which had caused winner of the near dependency. So the opposite of this is not always not always true, because, um, just cause a set contains fewer vectors than their entries does not mean it is always linearly independent, and they&#x27;re always examples to disprove it. Ah, see, states that if x and wire literally independent and if Z is in the span of X and y the net, then the set X y Z is linearly dependent. This is true, because if V three is in the span of the one and uh V to, they&#x27;re gonna be in the same plane, Um, so they&#x27;re gonna be literally dependent. And if we three is not in the span of you want me to. That means it&#x27;s in its own plane. And if you draw the planet something like that, then it is literally independent that then the set is literally independent. Because, uh, no set of values would, uh, make the X equals zero equipped equation be just have the trivial solution. And finally, d is if a set is in rn. If a set in Oran is linearly dependent than set, contains more records than their entries in the In the Vector, it is not always, always true. As here, there&#x27;s a counter example are, ah, the one of you to have the same entries. And there&#x27;s two. So there&#x27;s two entries into vectors, so there&#x27;s not always more vectors than their entries. So here the number of actors and entries are equal. So these are linearly and dependent as one. Becker is a linear combination of the other as V two equals to be one. So by here, um, um Bye, dear. Uh, but he runs real quick by the time seven, um, in the book, this set is linearly dependent. So? So Stephen B is 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.

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

Doruk Isik · Answer

in this problem. We&#x27;re given a few statements and were asked for work. It&#x27;s that where it is true or false. So the first statement, uh, schools, Because the set of backers not only should be linearly dependent, it should also coincide which age? Hey, you could check the definition off base base. Second statement is true, and I&#x27;m gonna report you refer to the, uh, spending said here. Now, the statement is statement, you see, is also true. And this time I would refer to the section named two years off a basis. Uh, Indy, begin the statement. This falls because this standard method of producing a spend settle now a is actually always Cordish iss a re nearly Bennett sent. And it actually always produces a basis for no hey and less Damon. ISS also pulls because we need Thio look at D or use the different elements or give its Collins off matrix A not the magics B, which is the Rebecca Formal, eh? Because some columns of B mike nights be in the column space. Oh, okay. So we should also we should always use the pivot columns off our orginal matrix

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.

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

Describe the possible echelon forms of the matrix…

Problem 22 Hard Difficulty

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a. True
b. False
c. True
d. False

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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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a. Trueb. Falsec. Trued. False

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Kayleah T.

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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

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a. True
b. False
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d. False

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

Linear Algebra and Its Applications