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In Exercises $15-18,$ find a basis for the space spanned by the given vectors, $\mathbf{v}_{1}, \ldots, \mathbf{v}_{5}$$$[\mathbf{M}]\left[\begin{array}{r}{8} \\ {9} \\ {-3} \\ {-6} \\ {0}\end{array}\right],\left[\begin{array}{r}{4} \\ {5} \\ {1} \\ {-4} \\ {4}\end{array}\right],\left[\begin{array}{r}{-1} \\ {-4} \\ {-9} \\ {-9} \\ {-7}\end{array}\right],\left[\begin{array}{r}{6} \\ {8} \\ {4} \\ {-7} \\ {10}\end{array}\right],\left[\begin{array}{r}{-1} \\ {4} \\ {11} \\ {-8} \\ {-7}\end{array}\right]$$

$\left\{\left[\begin{array}{c}{8} \\ {9} \\ {-3} \\ {-6} \\ {0}\end{array}\right],\left[\begin{array}{c}{4} \\ {5} \\ {1} \\ {-4} \\ {4}\end{array}\right],\left[\begin{array}{c}{-1} \\ {-4} \\ {-9} \\ {6} \\ {-7}\end{array}\right]\right\}$

Calculus 3

Chapter 4

Vector Spaces

Section 3

Linearly Independent Sets; Bases

Vectors

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in this example were provided with five different vectors and each of them come from the set are five. What we're going to do with this set of vectors is let H be equal to the span of B one B two, b three, V four and V five. So now h is a subspace of the larger space are five. Merkel here is to form a basis for this speech. The space age in forming a basis. We have two requirements. One requirement being that we spend the space we require. And so that's already met. We already know the one through the five span H because that's how we defined age. So now we're looking for a set of vectors from this which are linearly independent. They may all be linearly independent or just a collection of them. And to determine this, we're going to use the following method. Let a matrix a be formed from columns being the one B two, b three, V four and V five. Now our method is to row, reduce and find which columns are pivot columns, but unfortunately reducing a five by five matrix with not the most friendly numbers like we have is a tremendous amount of effort. So what we're going to be using is a calculator to do the job for us. So that way we can save some time and get back to linear algebra ideas. So in a calculator, I found that this matrix a ro reduces to So now with this matrix, what we need to do our analyze the pivots. But first, let's recall that column. One came from V one, then B two, b three, B four and V five. Also, we have a pivot and comes 12 and three, but nowhere else. This tells us that if we want a basis for H V 13 V three will span and be linearly independent. So that tells us the vectors before and the five can be effectively discarded since V one v two V three generate those two as well. So it's day our conclusion. We know that the set consisting V one v two v three, is linearly independent. Now recall our Cole. Here's to find a basis for the Space H, which is a subset of our five, and we already knew it spans. We found a set that's linearly independent, and now we're ready to stay a conclusion. This implies that the set the one V to V three now forms a basis. Four. H recall. It's not basis for our five as not basis from our three might think with three vectors. It's a basis for H because that's the set that it spans.

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