all-vectors-are-in-mathbbrn-mark-each-statement-true-or-false-justify-each-answer-a-not-every-linear

Question

All vectors are in $\mathbb{R}^{n}$ . Mark each statement True or False. Justify each answer.
a. Not every linearly independent set in $\mathbb{R}^{n}$ is an orthogonal set.
b. If y is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.
c. If the vectors in an orthogonal set of nonzero vectors are normalized, then some of the new vectors may not be orthogonal.
d. A matrix with orthonormal columns is an orthogonal matrix.
e. If $L$ is a line through $\mathbf{0}$ and if $\hat{\mathbf{y}}$ is the orthogonal projection of $\mathbf{y}$ onto $L,$ then $\|\hat{\mathbf{y}}\|$ gives the distance from $\mathbf{y}$ to $L .$

Foster Wisusik · Accepted Answer

Okay, This question asks us various true false questions relating to orthogonal matrices. So are a is to false saying that not all linearly independent sets Arthur Ivan all So this is saying, Yeah, a linearly independent set isn&#x27;t necessarily orthogonal. And this is true because if we just think of the vectors 11 and 13 they&#x27;re definitely linearly independent, but 11 dotted with 13 is equal to four. So they&#x27;re not orthogonal. So that&#x27;s a nice little counter example. Then Part B says that if why is a linear combination of vectors from the north agonal sent the weights can be computed without row operations, and this is also true. So if we say that a vector V is equal to a one x one plus a two x two plus dot, dot, dot dot We can calculate these coefficients just by taking the projection of the vector onto each of the basis. So, for example, a one would be equal to the dotted with X one over the magnitude of X one squared, and this could be done for any of the vectors because all we&#x27;re doing since their orthogonal, none of them overlap with their projections. So we just need to calculate the component of the that&#x27;s projected under each of the basis vectors. And we don&#x27;t have to do anything with the Matrix to find that. Then Part C says if vectors in an orthogonal set of non zero vectors air normalized than some of the new vectors may not be orthogonal. Okay, So, for example, let&#x27;s say we have a vector, the one and another vector V two such that their orthogonal so v one dotted with the to is equal to zero. Then it says we normalize each of these factors. So v one over the magnitude of the one started with V two over the magnitude of the two, and we want to see if this dot product is still zero, and we can see that the DOT product allows us to pull out scaler multiples. So in actuality, we can write this as one over the magnitude of the one times one over the magnitude of V two, and then that is just a coefficient outside V one dotted with the to. And we see this is still equal to zero because this thing in brackets zero so the claim is false for C because we can just pull the magnitudes out front. And the only thing affecting or Thorgan ality are the vectors themselves, not their magnitudes, because they they just come along for the ride. Then for D were asked a matrix with Ortho Normal columns is an orthogonal matrix. So this is really tricky because you might want to say true. But actually this is false because the matrix must be. And by n so on, Lee Square matrices could be classified as orthogonal. So even so, the Ortho Normal column claim would be enough if it included the fact that the Matrix had to be square. So if it said there was a square matrix, then do you would be true? And then e says if we have a line through the origin and why hat is the orthogonal projection than the magnitude of why gives the distance from L. Yeah. So we say this is a line and this is a point, and we want the distance between these two points. So what we&#x27;d actually do is we would split this up this blue vector into a component parallel and a component perpendicular which I&#x27;ll actually write this better. So this is the parallel component and this is the perpendicular component. And it is saying that the magnitude of this component right here is the distance to the line. And that&#x27;s false because we want. We want the magnitude of the perpendicular component. So this is the distance we want the perpendicular, not the parallel, so it&#x27;s 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

Viswesh Uppalapati · 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

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.

Sean Thrasher · Answer

to true or false were given. That w is the span of X one x two x tree where x one x two extra linearly independence and then V one V two V three is orthogonal and double Do you okay? Does this mean that V one b two B three is a basis for W? Now always make sure you check your definitions. So let&#x27;s go to page 340. What is north organist set a set of vectors you want up to U P. They say in Oran is orthogonal means that that each pair of distinct vectors are Thorgan all so you I don&#x27;t you j is equal to zero for every eye not equal j that&#x27;s between one and p. Okay, So please notice the very important fact that this doesn&#x27;t rule out the case where all the vectors in my set of zero okay, so two vectors could be orthogonal to each other simply because at least one of the vectors are zero. So this is where the meaning of war Thorgan ality slightly departs from our intuition. Perhaps I mean is 00 orthogonal 210 the pictures, you know 00 is just the origin right here. And then one zero&#x27;s this vector. The answer is yes, there are dog. No, Because if you take the dot product, you get zero. So the set of vectors where the set, where every single vector is the zero vector is still a North organo set. So you have to be careful. Okay? Just because you have a North organo set doesn&#x27;t mean that you can declare it is linearly Independence. Okay. And then the linearly dependent said cannot be a basis. So of course. So this statement is false. Okay, this statement is false. I don&#x27;t need to come up with a counter example. So, some advice for thick for thinking of counter examples. It&#x27;s usually best to choose the simplest on the most extreme counter example. We need to come up. Where? With a statement. Where? Um, so we need to come up with a case where this statement right here fails. Okay, this is what it means to have a counter example. Okay, well, let&#x27;s take our three. Okay. And let&#x27;s w b So let x one x two extra b x will be the standard basis. It&#x27;s too. It&#x27;s a lot extra. Next to extreme be the stunt basis, then w is the entire are three is the entire vector space of our three. And there&#x27;s no there&#x27;s nothing wrong with doing that. Okay, so and then also, what will we choose for V one? V two v three. So let&#x27;s take the one b two, b 32 equal zero vector. Okay, so this is the simplest counter example I can think of. And then the most extreme case where the or Thorgan ality condition fails to guarantee linearly in the linear independence is when you know each of the vectors are the zero vector. Okay, so so x one x two extra or certainly linearly independence. Okay, on dhe V one v two V three. Well, they&#x27;re definitely orthogonal in w right. There are Fogell, are three all right, because I mean, this condition right here certainly satisfied. Right? V one dot ve to zero. We wound up the 30 and vi tue dot be 300 So that&#x27;s fine. Then we should check that the conclusion actually fails, and it certainly does. Right? So in this case, even though x one x two extra linearly independence. And even though the one V two V three are Fogell and W V one V two V three cannot be a basis for W. This fails right here simply because, you know, this is a linearly dependent Sets. Okay. Now on to question two are were asked if X is not in the subspace w than X minus the projection of excellent w is not zero. So this is true, Okay? It&#x27;s pretty obvious. It should be pretty obvious, too. That is true, right? But we can&#x27;t just, you know, we can&#x27;t just wave our hands and justify it. We&#x27;re gonna actually have to come up with the proof. No, this comes from obviously the orthogonal decomposition Terram. Right? So you can take any vector X and split. It decomposes into projections w plus some factors that where that is in, you&#x27;re Thorgan of compliments. Okay, so we&#x27;re going to use something to do with this. All right, So let let me show you my proof. Well, it&#x27;s a little bit awkward to work with. You know, this ex not in a subspace. And then this is not zero. Right? So let me show you a little magic mathematical magic trick, if you like. So saying a implies b is completely equivalent to saying that not be implies, not a okay. If I live in London, then I live in England is equivalent to saying if I don&#x27;t live in England, then I certainly cannot live in London, right? And so this right here this fact, we use quite a law in my all right, I&#x27;m gonna use this here. So instead of showing that X not in the subspace w implies that this is not zero I&#x27;m going to say, OK, well, suppose that this guy right here is zero. Then I&#x27;m going to prove to you that X must be in the subspace. W shouldn&#x27;t. This is a little bit confusing. Sorry. I&#x27;m talking about the same Max here. Okay, so here&#x27;s what I&#x27;m going to do. I want to show that X minus. This equals zero, implies that X is in W. So let&#x27;s do that. Well, let&#x27;s call the skies ET from here, right, It&#x27;s called the C. So Z is equal zero, right? And when I want to show, the ex is in W, Right. So So, um, this is how I&#x27;m gonna do it. X is equal to this plus Zed rights. And I want to show the exes and w So Zed is equal to zero. Right? If you rearrange us, you find that set is equal zero. So is that is certainly in w. Hey, why is that? Well, w is a subspace, right? And any subspace is a vector space in his own right. And vector spaces must contain the zero vector. And so w must contain the zero vector as well. Okay, this guy right here is in W as well. Why is that? That&#x27;s from the or so Colonel Decomposition theory. Right? We say that you can split X into a vector that&#x27;s in w and effectiveness in the orthogonal complement of w. So this guy started in W No. Eggs is the sum of two vectors that Aaron W. But I just said that w is a vector space. I take two vectors and w I add them. I must stay in w. And so ex isn&#x27;t w. And that&#x27;s it. We&#x27;re done. Okay, now, right? What? I said out clearly line by line. Okay. In the queue, our factories ation ese click. You times are where a has linearly independent columns were told. Okay. Is it true? Is it true that the columns of Q. Forman Ortho normal basis for the common space of a Well, if you go to page 359 where they discussed the cure factories ation, they say the following If a is an n by N matrix with linearly independent columns. All right, then, lips then they can be factored as que times are where Q is an n by N matrix. Whose columns? Right, listen up. Those columns, Forman. Ortho. Normal basis for the column. Space of A They literally just say that. And so this statement is true. Okay, this statement is true, provided that are you know, it provided that we are actually factoring this in this way, right? But that they tell us that we&#x27;re in a queue, our factories a shin right when a cure factories ation and so the cure factories ation. When you do that, you get a Lynn, you know, matrix that as linearly independent columns split into a matrix whose columns from the north a normal basis for the column space of a times some, you know, upper triangular in vertical matrix. All right, so this is certainly true. You know, it&#x27;s basically a direct rip off from what they say in page through 59. All right, so we&#x27;re done now with the questions, but I want to go back to question one to discuss some interesting subtlety. Okay, so, you know, feel free to and the video here. But if you&#x27;re interested, then listen up. So let&#x27;s go back to this question yet. The very start X one x two x tree. And then we said x one x two x three are linearly independence and we want 3 to 3 or four, Colonel. Okay, so we found that this does not imply that B one b two B three must be a basis. All right, now, let me change a statement a little bit. Okay? So instead of making this fourth, Organon and w let me make the condition on this set a bit stronger. I a little bit more stringent. Okay, so let&#x27;s say that I say that this is a set of non zero orthogonal vector&#x27;s right. The counter example used for this was when we had, you know, least 10 At least a zero vector in here. What if I completely disallowed that? What would happen with this statement? Still be false, Or would it be true, Paul&#x27;s And ponder this, and then if you think it&#x27;s true, come up with a proof. All right? The answer is yes. I think this would be true. So this would imply this. Okay, Now, why is that? Okay, So why is that? Well, you know, you can&#x27;t just you can&#x27;t just wave your hands and, you know, kind of argue by your intuition. You have to actually come up with a proof using the theorems and stuff that has been derived in the book. Okay, well, why is this so I&#x27;m gonna kind of sketched this out to you. So page 340 they&#x27;re in four this time. Says that if you have an orthogonal set of non zero vectors in RN, then the set is linearly independence. Okay. No, we know therefore, that this sets must be linearly independence. Now, the question is, does it span? Well, the final ingredient you need is theory 12 on page 229. Where it says that Well, okay, this staring right here. Go to it and check it. Right. And in this case, W is a three dimensional vector space. Okay, the dimensional W is three, and we have a linear, independent set of exactly three elements. And w so by the room 12 on page 229 this set right here must automatically be a basis for W. Okay, so since we&#x27;ve realized that B one B two B three is a linear in independent sets, theorem 12 from Page Turner, 29 again guarantees that this must be a basis for W. So I hope this helps you appreciate why we spend so much time carefully proving these serums. Because without it, it is impossible to convincingly argue anything in the New year. Algebra, even if it feels obvious. Okay, so this is it. That&#x27;s all for now. Thanks for watching

James Chok · Answer

Okay, so this question will be answering with a serious off a tree and forces. So it was a if b is a basis for subspace hated in each. Better in hate can be ridden in only one way as linear combination of factors in B. This fact is true, which is, quite literally, the definition of a basis. And if you need more help on this question, if you go to the text book page 1 55 the beginning off section to 2.9 on the second paragraph, it says, is basically verbatim. Question be if B is a basis for suspicion of hate in the correspondence, makes hatred look and act the same as our p. So this is also true, which is literally told bourbon in page 1 57 See the dimensions off mo off a number of variables in the equation X because is there this is false is equal to the number off free variables in a executed the dimension off The column Space off A is rank A. This is true because this is that that this is the definition off, Frank. Hey, if head is a p dimensional subspace, there linearly independent set off the vectors in hedge is a basis for Paige. This is true by the base system.

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All vectors are in $\mathbb{R}^{n} .$ Mark each s…

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

Linear Algebra and Its Applications

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A. TrueB. TrueC. FalseD. FalseE. False

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Caleb E.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

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Anna Marie V.

Caleb E.

Kristen K.

A. True
B. True
C. False
D. False
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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications