00:01
We're given a space u, which is the span of these four polynomials, and we want to determine the dimension of it.
00:10
And the dimension of it will be the number of polynomials or vectors, if you want to get general, the number of vectors in this set that are linearly independent.
00:23
So how do we figure out which ones are linearly independent and which ones are extraneous, not contributing that? well, the first thing i think we can do is let's convert this to a set of column vectors instead of polynomials, because then we could use all of our knowledge of matrices.
00:39
It's going to be the same sort of thing.
00:42
So actually, before we do that, we need to set up an ordered basis.
00:45
So let's set up a basis b for the vector space of polynomials that we're looking at here.
00:51
And we'll have 1x, x squared, x cubed.
00:56
And so that means the vectors, each component of the vector, will be the coefficient on one of those terms.
01:03
So for example, the column vector for the first polynomial will be 1, negative 1, 1, 1, and 1.
01:11
And that represents 1 plus a negative 1x plus 1x squared plus 1x cubed.
01:17
And then we could continue along with the rest of them.
01:21
So that we have a 1, we have a positive 1x, a positive 1x squared, and a negative 1x cubed.
01:30
V3 is a 1 -0 -x's, 1x squared, and 0x cubed.
01:37
And then finally, v4 equals 1.
01:41
Oh, no, no constants.
01:43
Remember the first one's the constant term.
01:45
So 0 constants, 1x, 0 -x squared, and negative 1x cubed.
01:50
So now in terms of column vectors, we want to know, are these vectors linearly independent? it's the same question, but it's written this way.
01:58
And i think the column vectors are a lot nicer, because when we have column vectors, we can use our knowledge of matrices.
02:04
So let's set up a matrix, 1 -1 -1 -1 -1 -1 -1 -1 -negative 1 -1 -1 -0 -0.
02:12
It's nice that the row -zero's and ones, i guess.
02:18
And we can use row reduction.
02:22
And after we do row reduction, the pivot columns will correspond to the original vectors that are independent from each other.
02:31
So what that means or how that applies here, we'll get to.
02:35
But i suppose first, let's do some row reduction.
02:40
Row one is fine because it has the leading one.
02:44
Let's take row two and add row one to it.
02:50
So 1 -1 -1 -0 is fine.
02:55
Row two, we're going to add row one to it...