in-exercises-21-and-22-mark-each-statement-true-or-false-justify-each-answer-a-if-d-is-a-real-number

Question

In Exercises 21 and $22,$ mark each statement True or False. Justify each answer.
a. If $d$ is a real number and $f$ is a nonzero linear functional defined on $\mathbb{R}^{n},$ then $[f : d]$ is a hyperplane in $\mathbb{R}^{n}$ .
b. Given any vector $\mathbf{n}$ and any real number $d,$ the set $\{\mathbf{x} : \mathbf{n} \cdot \mathbf{x}=d\}$ is a hyperplane.
c. If $A$ and $B$ are nonempty disjoint sets such that $A$ is compact and $B$ is closed, then there exists a hyperplane that strictly separates $A$ and $B .$
d. If there exists a hyperplane $H$ such that $H$ does not strictly separate two sets $A$ and $B,$ then $(\operatorname{conv} A) \cap$ $(\operatorname{conv} B) 
eq \varnothing$

James Chok · Accepted Answer

okay, In this question, we will answer with a Siri&#x27;s off truth and false is so a hey, if Israel number and F is a non zero linear functional than FT is a hard time in our So this diamond is true If you go to the textbook on page 4644464 The statement after the question three say this. Okay, given any vector in on any rule number, date, set, end, times X and secretive Gate is a high plane. This is false. Um, the victor in must be nausea. So Paige, six for six number Our Question three s a c if a and B a non empty. I&#x27;m not empty district sets such that a is compact and being close. Then there exists a high plane that&#x27;s strictly separates A and B. This is false. The sets must also be convex. Well, they must be convex. So the scene Why, that&#x27;s just crap. Very simple example. So well, over this set as a so this is both close and compact. Then we can create a set B, which is just a line. So this line B, this is a B is close So now we have a set which is strictly coast and compact. But there is no line. There&#x27;s a hyper thing which week, which, with we can separate these two, there&#x27;s no way do you if there exists 100 times such that taste of usted please separate two sets A and B in convicts of eight, capped by convicts of these non empty. This is false because some other hard thing, um, on the I could play cancer right then can&#x27;t sit right them.

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.

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

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.

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

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

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Let $\mathbf{v}_{1}=\left[\begin{array}{l}{1} \\ …

Problem 22 Hard Difficulty

Answer

a) $\mathrm{T}$
b) $\mathrm{F}$
c) $\mathrm{F}$
d) $\mathrm{F}$

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

Linear Algebra and Its Applications

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a) $\mathrm{T}$b) $\mathrm{F}$c) $\mathrm{F}$d) $\mathrm{F}$

Linear Algebra and Its Applications

Lily A.

Catherine R.

Anna Marie V.

Kristen K.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Anna Marie V.

Kristen K.

a) $\mathrm{T}$
b) $\mathrm{F}$
c) $\mathrm{F}$
d) $\mathrm{F}$

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

Linear Algebra and Its Applications