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En Garde, whole no. 16, January 1946
Page 18
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page 18. (1, 0, 2, 0), (-1, 3, -1, -2), & (-1, 6, 0, -4) are linearly dependent because (1,0,2,0) + 2( -1,3,-1,-2) - (-1, y, 0, -4) = = (1,0,2,0) + (-2,6,-2,-4) + (1,-6,0-4) = (0, 0, 0, 0) Any vector is linearly dependent with any multiple of itself, for example 2(9,18,12,-3) + 3(-6, -12,-8,2) = (0, 0, 0, 0) A very simple example of vectors which are not linearly dependent is provided by (1,0,0), (0,1,0), & (0,0,1). Their mutual linear independence is obvious, but taken together with (1,1,1,) or even (69,5,-732) they give a linearly dependent set. Any non-zero vector is linearly independent by itself, that is, you can't multiply it by a number different from zero and get a vector all of whose components are zero; and zero vector (one all of whose components are zero) is linearly independent of any non-zero vector whatever. Using the definition we have just set up, we can get down to the main point of just how a group of vectors may "determine" a space. Remember that so far this determining is something we understand only through geometry and intuition, so that we are unable to extend the concept to space of more than 3 dimensions. We met the dame difficulty in our extending the idea of "vector" to extra dimensions, and we solved it there by setting up a completely abstract and general definition of vectors as ordered number-groups and then defining relations of addition and multiplication between them in such a way that they behaved, when they had 3 or fewer components, like ordinary vectors referred to coordinate systems having 3 or fewer perpendicular axes. Since it made difference in our definitions if our vectors happened to have more than 3 components, we were confident that these definitions would lead to results in these difficult-to imagine cases which would square with what we do know to be true in our universe. So we'll do the same thing here. Set up a definition, test it, and then, having tested it, apply it. First of all, we will define a space as any class of vectors at all. If you will think it over you will see that this really is exactly what we meant by the term anyhow. (Pause while you think it over.) Then, on to the definition: The space determined by a set of vectors includes any vector which forms, with the given set or with any group of vectors from the given set, a linearly dependent group; it also includes (0,0,0) but does not include any further vectors. It doesn't matter whether the original set is linearly independent of itself or not. If I give some examples of this definition, and if you study them carefully, it'll make the testing of it that much easier. All the examples will use 3-vectors, so that you can visualize their significance intuitively by thinking of a coordinate system with 3 axes.
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page 18. (1, 0, 2, 0), (-1, 3, -1, -2), & (-1, 6, 0, -4) are linearly dependent because (1,0,2,0) + 2( -1,3,-1,-2) - (-1, y, 0, -4) = = (1,0,2,0) + (-2,6,-2,-4) + (1,-6,0-4) = (0, 0, 0, 0) Any vector is linearly dependent with any multiple of itself, for example 2(9,18,12,-3) + 3(-6, -12,-8,2) = (0, 0, 0, 0) A very simple example of vectors which are not linearly dependent is provided by (1,0,0), (0,1,0), & (0,0,1). Their mutual linear independence is obvious, but taken together with (1,1,1,) or even (69,5,-732) they give a linearly dependent set. Any non-zero vector is linearly independent by itself, that is, you can't multiply it by a number different from zero and get a vector all of whose components are zero; and zero vector (one all of whose components are zero) is linearly independent of any non-zero vector whatever. Using the definition we have just set up, we can get down to the main point of just how a group of vectors may "determine" a space. Remember that so far this determining is something we understand only through geometry and intuition, so that we are unable to extend the concept to space of more than 3 dimensions. We met the dame difficulty in our extending the idea of "vector" to extra dimensions, and we solved it there by setting up a completely abstract and general definition of vectors as ordered number-groups and then defining relations of addition and multiplication between them in such a way that they behaved, when they had 3 or fewer components, like ordinary vectors referred to coordinate systems having 3 or fewer perpendicular axes. Since it made difference in our definitions if our vectors happened to have more than 3 components, we were confident that these definitions would lead to results in these difficult-to imagine cases which would square with what we do know to be true in our universe. So we'll do the same thing here. Set up a definition, test it, and then, having tested it, apply it. First of all, we will define a space as any class of vectors at all. If you will think it over you will see that this really is exactly what we meant by the term anyhow. (Pause while you think it over.) Then, on to the definition: The space determined by a set of vectors includes any vector which forms, with the given set or with any group of vectors from the given set, a linearly dependent group; it also includes (0,0,0) but does not include any further vectors. It doesn't matter whether the original set is linearly independent of itself or not. If I give some examples of this definition, and if you study them carefully, it'll make the testing of it that much easier. All the examples will use 3-vectors, so that you can visualize their significance intuitively by thinking of a coordinate system with 3 axes.
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