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For each subspace in Exercises 1–8, (a) find a basis, and (b) state the dimension.$$\left\{\left[\begin{array}{c}{a-4 b-2 c} \\ {2 a+5 b-4 c} \\ {-a+2 c} \\ {-3 a+7 b+6 c}\end{array}\right] : a, b, c \text { in } \mathbb{R}\right\}$$

a. Since $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ are not multiples of each other, $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\}$ is linearly independent and hence, a basis for $S .$b. 2

Calculus 3

Chapter 4

Vector Spaces

Section 5

The Dimension of a Vector Space

Vectors

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Harvey Mudd College

Baylor University

University of Nottingham

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Hello. For this question, we'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'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'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'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'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'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'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're gonna be considering next. We have vector that's multiplying are arbitrary, constant B, which we can just read off the the components of it. We have negative for 50 There'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'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'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't matter which two of these vectors I pick. I'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'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'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're done.

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