for-the-matrices-in-exercises-17-20a-find-k-such-that-nul-a-is-a-subspace-of-mathbbrk-and-b-find-k-s

Question

For the matrices in Exercises $17-20,(a)$ find $k$ such that Nul $A$ is a subspace of $\mathbb{R}^{k},$ and $(b)$ find $k$ such that $\operatorname{Col} A$ is a subspace of $\mathbb{R}^{k} .$ 
$$
A=\left[\begin{array}{rr}{2} & {-6} \ {-1} & {3} \ {-4} & {12} \ {3} & {-9}\end{array}ight]
$$

Michael Jacobsen · Accepted Answer

in this video, we&#x27;re starting out with a four by two matrix A that&#x27;s indicated here. Now when you know for this particular matrix A. We conform to interesting sub spaces. The first up space is the null space of a Let&#x27;s write out what the null space of a is specifically. It&#x27;s the set of all vectors X such that the matrix equation eight times X equals zero is satisfied. Or you could say it&#x27;s all vectors X that are mapped to zero by the transformation X goes to a X, so we want to determine next the following. We know that the null space of a is a subspace of the following. So no, a is a subspace of our something. So we want to determine what should go here. Will it be? Are for or will it be are too? One way to determine what their appropriate value is is to look very closely up This matrix multiplication, the Matrix A is of size for by two and the zero vector is going to be of size four by one. That tells us something about the vector X. What&#x27;s right down the dimensions here. Well, first it has to be a to, in order for this value to match up and allow for the multiplication to be defined, and it has to be of size one. So we match this outer dimension here. So the Vector X, which is what the null space consists of, has two entries, and that tells us no lays a subspace of our two. So that&#x27;s one very interesting and important subspace that comes from this matrix. Let&#x27;s change gears and talk about the column space of a Indicated this way. Well, the column space of a we Know by words is equal to the span of the columns of a and the span of the columns of a would be linear combinations of this column and this column. Such linear combinations give us new vectors, which still contain four entries, but that employees next that the calm space of A is a subspace of altogether are four since we&#x27;re spanning vectors with four entries, so to pause for a moment before we end this example when we&#x27;re considering the null space and we want to know that dimension, or rather which subspace it belongs to, we look to the number of columns. So are two for this matrix. If you&#x27;re concerned about the column space, we look to the number of rows and that will tell us where the calm space is contained in.

Doruk Isik · Answer

in this problem, we&#x27;re s to find value case like that. No, it is a subspace in our to decay in the column. A subspace in our cake so hard today. No es. So as you can see, our has one roll. And 12345 colds. Right So than a ISS one by five. Um, makings. So since this no way. We&#x27;re looking, uh, percent of recreational perform eggs, is it zero and now is on my eye and no. Yeah, we&#x27;re looking for eggs here. That X should be open for five. Fine. One or five by one reason is the number of columns and a should be cooking number Rose in are, uh, eggs majors. So from this case should be five then l A is the subspace pain or fine part. Be cold, eh? So this time work creation looks. There&#x27;s also perform exit B. So a CZ again one, my fine. Now we know not be a cz fine by one from front, eh? So since the number of Poland&#x27;s and a mixes with the number rose in eggs, we want I have a problem and the resulting medics will be up to size one by one. So from this Okay, Is these what I call a maze of subspace and or

Doruk Isik · Answer

this problem was given the metrics I has to determine. Finally, integer a such that no, a record space of art to decay and told the ladies and the K K K senses. No, actually, we&#x27;re looking for a solution off this form. So a is a four by three mentions to test for Rose and three clothes. So for right? Three. And if you want a most split, this with a vector, director should have the same size, same number. So the columns up they should be called the rolls off. He&#x27;s a writer, so this should be t I want. So it means that hey should be free. So then no, A and R Jew can be. Let&#x27;s say we have another Victor X and let&#x27;s say R B is a eyes column a again, a cz or rectory. So rumors we know that extra be three by one. It&#x27;s a number Collins off a is equal number off rolls of X sign on the metrics. Speed will be up to size for by one. So it means that K should be four and then hold A and were subspace in our before

Brian Ketelobeter · Answer

Okay, Um, let&#x27;s try this again. It&#x27;s still doing the same weird thing. Okay, so I have matrix a here. This is a problem. Number 11 from section 2.8. I&#x27;m given a matrix A, um and I&#x27;m after the null space and the cone space. Okay, so on drily, this is just the null space is just a question of matrix multiplication. So we&#x27;re after all the vectors accession X equals zero and just examining matrix multiplication. You can see that the vectors X would have to be in our four. So that means that the, um uh the, um que ah que would have to equal to four. Um, this is a subspace of our four. I&#x27;m sorry. I don&#x27;t know what&#x27;s going on here all of a sudden, um, this is contained in, um are sore. Okay. On dso Similarly, when we look at the column space So we&#x27;re looking at the span of the columns, and just by looking at the columns, you can see that there three, they have three coordinates. So that means that the column space, um, is contained in our three. So the column space is a subspace of our three And that means that he would have to be, um uh in P would have to be three p. Just being the thes subspace are the dimension of the subspace in which the Colin spaces in. So just by looking at that, you can see that the Colin space is contained in our three. So, uh, yeah, so the null space eyes all the vectors acts such an ax equals zero. And just by looking at the the number of columns in a, you can see that it&#x27;s four. So that means that it would have to be a subspace of our four. And the column space is just the subspace spanned by the columns. And when you look at the columns, you can see that they&#x27;re three dimensional vectors. So that means that it would have to be contained in art. So that&#x27;s it for this problem here. Thank you very much and have a nice

Daniel Pezzi · Answer

Hello. For this question, we&#x27;re asked to find two things we need to find the find a basis for this subspace of our four. And we need to figure out what the dimension of this subspace is. Now we&#x27;re given that this is a subspace and finding a basis just gives us a nice, uh, finite way of representing a set. Infinite things which will greatly cut down on the number of computations we need to dio and finding, uh, the dimension. Um, what&#x27;s a specifically kind of verbalize? How big quote unquote, uh, this subspace actually is. This is a subspace of are four. And we know that because we have four slots in this factor now, before doing any work, um, we can see that we have three arbitrary constants A, B and C. So the largest the dimension of this subspace the dimension of this subspace of our four can be is three. So if we do a bunch of work and we get that the dimensions for we know we&#x27;ve messed up somewhere. But it could be less than three, depending on how these constants are related. So recalling the definition of a basis, Um, if I called the set. The essentially V is the set of all vectors that could be written as, um Constance, the one plus the to plus the three. We&#x27;re V one V two and V three are some vectors. And then, uh, we just way have Thio determine if these three vectors are living early, independent or not, to get our our final answer for the basis. So our first step is to break up, um, this representation of an arbitrary element of this space into this form by factoring out constants and breaking it up over edition. So doing that, I&#x27;m gonna put in a here. And if we look at the first slot, we have won a So that means this vector, which is multiplying the arbitrary constants A and its first lot has a one. We have two A&#x27;s in the second slot, so I can put it to here. We have minus one a in the third slot and minus three a in the fourth slot. So this is the first vector that we&#x27;re gonna be considering next. We have vector that&#x27;s multiplying are arbitrary, constant B, which we can just read off the the components of it. We have negative for 50 There&#x27;s an implicit plus zero be in here and then seven So I could write out minus four, 50 and seven. And now we have the vector multiplying or arbitrary constants C Which again I can just read off is negative two minus four to and, uh, six. Now, I would It is tempting to just say that. Okay, well, we found out what our basis is. It&#x27;s these three vectors, but we have to. But we have to remember that in the definition of basis, all the vectors in that basis have to be linearly independent. And if you notice this vector 12 negative one negative three is a constant multiple off this factor. Negative two minus four, two and six. If I multiply this by negative too, I get this vector. So instead of having, um, three arbitrary constants, we actually only have to arbitrary constants because this constant is not giving us any more geometric information. Because this factor is secretly a multiple of this factor. And if you aren&#x27;t convinced, just factor out a negative to here. Um, you would get minus two c times this factor over here on. Then you would have You could declare a new, arbitrary, constant a minus two C and you would see that the actual true basis is the set, uh, minus four, 507 And remember, bases are only determined upto arbitrary constants. So it doesn&#x27;t matter which two of these vectors I pick. I&#x27;m just gonna pick this first one. Ah, because I like it, Or I guess so. Our basis is these two factors. Um, and we can quickly check that they are, in fact, linearly independent. It could be another situation where all three of these factors of constant multiples of each other we immediately see that&#x27;s not the case. Um, just because this one has a zero in one of the slots and this one, uh, does not have a zero in any of the slot. So there&#x27;s no way that these two could be constant multiples of each other. Um and so they are linearly independent. So our basis Excuse me. So our basis for the space is this set here. So we have a nice finite representation of this kind of Messi arbitrary set on. Then we could immediately read off with the dimension is because the dimension of a a vector space is the same thing as the number of elements in any basis of that space. So the dimension of the we can just quickly write down as to which is in fact less than or equal to four, which is a nice sanity check, given that this space is supposed to be a subset of our four and we&#x27;re done.

James Chok · Answer

Okay, so we want to find the column space and null space off this matrix here to find the column space. We&#x27;re all All we need to do is to find a pivot columns off this matrix and the pivot column off this matrix is going to be the same as a pivot column off. It&#x27;s rational for now, The pivot column. All the right echelon forms are simply one, two and three thes $3. So, in the same sense, the column space is a space spanned by these vectors to the cold of space over A. He&#x27;s even to this man 11 minus two or minus two minus 101 and 551 and one To find a null space, we need to solve the equation. B X is equal to zero. What is all this? We ride up like this. You&#x27;re on the street there. Seven. There is no one wants to. So you&#x27;re you know, and nobody&#x27;s Yes. So what this implies is that x one x two eggs, three x or expired. So what is this? Begins here. Well, we know that X one is going to be if we move this to decide. This is obviously the service of this side. It&#x27;s going to be too. Plans x two plus lying times X minus line Time Victory three minus by times Export minus four times X by. Okay, on dhe X two is going to be three x three and plus 7 +65 Thanks. Bye now x three people call for X three extra is not increasing. Three x three Explore is going to be just to expire and once again, doesn&#x27;t people column for expire? So excellent is just ex wife. So you&#x27;re no space. A little space Well, a is the span Oh 20000 minus 93 100 Wanna spy 0000 Finally bars full said. And two, one who is zero here?

Problem

For the matrices in Exercises $17-20,(a)$ find $k…

Problem 17 Hard Difficulty

For the matrices in Exercises $17-20,(a)$ find $k$ such that Nul $A$ is a subspace of $\mathbb{R}^{k},$ and $(b)$ find $k$ such that $\operatorname{Col} A$ is a subspace of $\mathbb{R}^{k} .$
$$
A=\left[\begin{array}{rr}{2} & {-6} \\ {-1} & {3} \\ {-4} & {12} \\ {3} & {-9}\end{array}\right]
$$

Answer

a. 2
b. 4

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Kayleah T.

Vectors Intro

Vector Basics - Example 1

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a. 2 b. 4

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Kayleah T.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Anna Marie V.

Heather Z.

Kayleah T.

a. 2
b. 4

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

Linear Algebra and Its Applications