in-exercises-29-and-30-v-is-a-nonzero-finite-dimensional-vector-space-and-the-vectors-listed-belong-

Question

In Exercises 29 and $30, V$ is a nonzero finite-dimensional vector space, and the vectors listed belong to $V$ . Mark each statement True or False. Justify each answer. (These questions are more difficult than those in Exercises 19 and $20 .$ )
a. If there exists a set $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\}$ that spans $V,$ then $\operatorname{dim} V \leq p .$
b. If there exists a linearly independent set $\left\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\}$ in
$V,$ then $\operatorname{dim} V \geq p$
c. If $\operatorname{dim} V=p,$ then there exists a spanning set of $p+1$ vectors in $V .$

Savannah Bogus · Accepted Answer

problem. 29 has a set of givens and gives us three statements and asks whether or not these statements are true or false. The givens Are you a vector Space V, which is non zero in finite dimensional, and all of the vectors listed below belong to the Now let&#x27;s look at a A gives us the vector space of you. One to V. P tells us that this vector space spans me, then says then the dimension a v Slesin or equal to P. Now, in order to solve this, we&#x27;re gonna consider two types of spanning sets. First, we&#x27;re gonna consider a linearly independent, Spanning said for a linearly independent, spanning set. That means none of the actors in this set can be written as a linear combination of any of the out other vectors. Oh, linearly independent spinning set is also known as a basis, and by definition, if we have a basis, the dimension of the equals P. Now the other type of set that we are going to consider is a linear Lee depended spanning set. If we have a linearly dependent spinning. So that would mean that some of the vectors in the set V one to V. P could be written as linear combinations of other vectors in the set. Therefore, those vectors would not contribute to the dimension at all. Therefore, the dimension of the is less than p. So a is true Now we&#x27;re gonna consider be Once I erase all of this, B is gonna be solved fairly similarly, by considering two different types of sets for B, we have the set V one to V. P, and this set is linearly independent. Then we&#x27;re given the dimension of the is greater than or equal to P and we want to know whether or not this is true or false. So let&#x27;s consider our two types of linearly independent sets. There&#x27;s a spanning set which we already know from A That would be a basis which would make the dimension of the equal to P. Now, if we have a not spanning set, we have a not spanning set. That would mean some of the vectors in the set don&#x27;t contribute to the dimension. All therefore to mention of the is greater then p so be is also true. Next we&#x27;re gonna consider, see? All right, So for C the statement. ISS If the dimension of the equals p then there exists a spanning set of P plus one vectors. After thinking through and be, we know that a linearly depended spanning set exists, and in linearly dependent, spanning sets, the dimension of the is less than P. But so if the dimension of the is less than P, there could exist expanding set of P plus one vectors. Now, for this one, we can consider a short example say that we have the equals, our swear that would make the dimension of the equal to the dimension of R squared, which equals two equals p. And then we could also have a set of 30 for one vector 10 for another vector 01 for another factor.

Runpeng Li · Answer

All right. So in this in this problem, we&#x27;re gonna identify where the following statements are true, Boss. So the first statement, we have a bunch of junior dependent specters. Uh, we need to identify whether dimension off B smaller or equal to P. So actually, this statement is false. To provide a counter example, let&#x27;s say we have 100 and 200 so in this case, he will be too. But these two, these two vectors. All right, so in this in this problem, we&#x27;re gonna identify where the following statements are true, boss. So the first statement, we have a bunch of linear dependent specters. Uh, we need to identify whether dimension a, B R they all have dimension three. And they are actually from our three. So say are three will be be in this case. So the dimension not be iss three, it is bigger than p, which is to so this statement is false. The second statement, small early, equal to p. So actually this demon is false. To provide a counter example. Let&#x27;s say we have 100 and 200 So in this case, he will be too But these two These two vectors are, um, says if every every set off P elements in the fails to span V, then dimension V is bigger than P. Well, um, that&#x27;s first consider recalled dimensional to be. It is defined as the element in the set off pieces is number off sectors and, uh, basis and or so are They all have to mention three and they are actually from our three. So say are three will be in this case. So the dimension of B iss three it is bigger than he which is to. So this statement is false. The second statement, um says if every every set off p element as we recall from our textbook, this is what this will be also equal to the people columns, number of people, columns Seeing the fails to span V, then Dimension B s bigger than p. Well, um, that&#x27;s first consider recalled. Dimension not to be. It is defined as the number of element in the set off basis is number, uh, vectors and, uh, basis. And also, as we recall from So what does that mean? So a basis for the column space off upper matrix is equivalent to the number of people Columns in the Matrix. This is what we when we when we read in the textbook. So that means if every set off P elements in B feels to spend be so that implies, by our assumption, our assumption. Then there will be a P boat in every column. Every column as a people are textbook. This is what this will be also equal to the people Columns, number of people, columns. So what does that mean? So a basis for the column Space off for me, shakes is equivalent to the number of people columns in the Matrix. This is what we when we when we read in the textbook. So that means if every set off P elements in beef okay, so that implies. As a result, the dimension will be equal to the number of factors rather than greater than the number of factors. So that means dimension off the is equal to P rather than bigger, bigger than p. So Stillman is false. Okay, feels to spen be so that implies, by our assumption, our assumption. Then there will be a P boat in every column, every column as a people. The third statement we have if P is bigger away for 22 and Dimension V is go to pee than every set off P minus 10 back turns his leaner independent. So again, this stain these falls to see that we can check a Kendra example say s a P equals three we consider are okay, so that implies. As a result, dimension will be equal to the number of factors other than greater than the number of factors. So that means dimension I&#x27;ll be is equal to P rather than bigger, bigger than P. So Stillman is false. Okay, three. So let&#x27;s find a ah sad off P minus B minus one. That is, too. Just find a set off two vectors, that is Eun, you&#x27;re independent. That is meaner dependent because our statement says it. It is Leonard and independent. So we find a counter example you need to find a linear in your defendant set. So it&#x27;s very easy to find. Let&#x27;s say, um, ones. The third statement we have if P is bigger away for 22 and Dimension V is going to be then every set off p minus 10 back turns his leaner independent. So again, this stain these falls to see that we can check a Kendra example say s A P equals three we consider are three. There was zero and 200 See, these two factors are the New York dependent. But we only have like and this is Yeah, we only have two vectors. So dead at exactly sapi minus one. So too So that studies flyover assumption here. And we we did find a union linearly dependent factors. So So let&#x27;s find a ah sad off P minus B minus one. That is, too. Just find a set off two vectors, that is, Eun, you&#x27;re independent. That is very dependent because our statement says it is Leonard and independent. So we find a counter example you need to find a leaner in your defendant set. So it&#x27;s very easy to find. Let&#x27;s say, um one there is 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

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

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.

Problem

In Exercises 29 and $30, V$ is a nonzero finite-d…

Problem 29 Hard Difficulty

Answer

a) True
b) True
c) True.

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