for-each-subspace-in-exercises-18-a-find-a-basis-and-b-state-the-dimension-leftleftbeginarrayc3-a6-b

Question

For each subspace in Exercises 1–8, (a) find a basis, and (b) state the dimension.
$$
\left\{\left[\begin{array}{c}{3 a+6 b-c} \ {6 a-2 b-2 c} \ {-9 a+5 b+3 c} \ {-3 a+b+c}\end{array}ight] : a, b, c 	ext { in } \mathbb{R}ight\}
$$

Runpeng Li · Accepted Answer

were given a linear system which is, uh, three a US six. B minus C and six a minus to be minus to see to see. And also we have negative nine a us five e and plus three say and elected three a plus B plus C where ABC are all real numbers. Okay, So to find the basis off this system, um, we first need to separate those, um, the scale er&#x27;s ABC. So to do that, we have first scaler a and we can&#x27;t those those turns that contain that those come that contain eight. So that&#x27;s 36 negative. 89 and 10 to 3. So that&#x27;s three six connective 9 to 3. And also for Ms Cater be We checked the coefficients. That&#x27;s Ah, six negative 25 and one six connected to buy one. And for skaters. See you. We have a negative 12 and three and one. So that&#x27;s next one Naked too. And three and one. So that&#x27;s for skater. See? So the vectors generated by, uh, the system is will be 36 elective nigh and elective, three for the one say, and to be two, we have six negative too. Five and a one. And for the three, we have negative one. Next, two and three and one. Okay, so the first thing we need to do is to write here. We need to check the win your independence off these three vectors. So, um, one thing to observe dead is be one is scuse me is negative. Three times B three. Right? Because if we times to be three by a skater, negative three, then the first term will be three. And second term will be six. The third term will be negative. Die and fourth term is negative. Three. So that means a V one is dependent on the B three. So the only reason you&#x27;re independent vectors in this system is vey to MP three. There are only too linearly independent vectors B two and B three. Sorry. Oh, revised is you too. And B three linearly independent independent. Okay, so that means that dimension of this you mention off this subspace is two. And this is the basis which is formed by 3 to 1 of these three

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.

Runpeng Li · Answer

we&#x27;ll get something. This problem were given subspace. That is a B C D. Where? A minus three B. Let&#x27;s see. Zero. So by this equation, we know that see can be represented as freebie minus a So that means C can be determined by being a so our vector a B c D can be re read. Could be right. We can rewrite this vector. Um, terms off, eh? Hey, freebie mind, say, and the so. I noticed that A, B and D R V variables here. Well, c can be represented as, uh, A and B. So keep going. We can separate the specter so that it will be 10 91. I&#x27;m a skater. Eight and zero. 13 zero. Time Specter beak. And for the vector d off. Sorry. Should be scaler instead of Victor. Um, times of skated be And poor skater d we have There was 00 times skater d All right, so our pieces will be these three vectors. 10 negative. 10 013 and 0001 All right, So this is our faces and our dimension of this subspace. Well, the three

Runpeng Li · Answer

So in this problem still, we are given a subspace. We need to find the basis and the dimension. That&#x27;s a plus B to a hey minus B and negative. Okay, well, which may be our real numbers. Okay. Um, so again, we need to separate this vector purse. So we could, Although all the interests that contain a burst, that is a to a Sorry how? Just rewrite this too, eh? Three a zero. Right. Because in the purse, entry would contain that contains one A and, uh, second African things to a they&#x27;re entering. Continue contains three a. So we couldn&#x27;t plus find some e and nothing. Nothing in second entries that zero on. Don&#x27;t be on connected. Okay, so this gives two vectors. 12 three. Sorry. Just rewrite the three 30 and the second vector one. You negative 11 So these two vectors are linearly independent. So this is the basis for the subspace. And you mentioned be two

Runpeng Li · Answer

Okay, So in this problem were given subspace saying or as meant, if three s and teeth for which s anti our real numbers. Okay, so in this case, we first need to separate this vector. So we could we put all the entries that contain, as in one vector and all the interested contained t another vector and find some. So that is or ask connective three as and zero plus t zero through. So it turns out we the space is spent by these two factors and we can observe that these two vectors are you nearly independent. So dimension of this subspace is too. And the basis is just these two vectors, which is for next 30 and activities and 001

Michael Jacobsen · 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.

Problem

For each subspace in Exercises 1–8, (a) find a ba…

Problem 6 Medium Difficulty

For each subspace in Exercises 1–8, (a) find a basis, and (b) state the dimension.
$$
\left\{\left[\begin{array}{c}{3 a+6 b-c} \\ {6 a-2 b-2 c} \\ {-9 a+5 b+3 c} \\ {-3 a+b+c}\end{array}\right] : a, b, c \text { in } \mathbb{R}\right\}
$$

Answer

a. See solution
b. 2

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

Linear Algebra and Its Applications

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a. See solutionb. 2

Linear Algebra and Its Applications

Catherine R.

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Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Heather Z.

Caleb E.

Samuel H.

a. See solution
b. 2

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

Linear Algebra and Its Applications