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En Garde, whole no. 16, January 1946

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moved head. But in moving the vector we kept it parallel to its original position (otherwise it would have become a different vector), and in its new position it passes through a point--the other vector's head--which is in the plane of the two original vectors. Therefore it is still in this plane, its head is still in the plane, and the sum vector, having both tail and head in the plane, is also in it, and our assumption is proved correct. Now in addition it is self-evident that any multiple of any of the group of coplanar vectors with which we started is in their plane. From these two facts it follows that any sum of multiples of the coplanar vectors is in the plane also! Knowing this makes our job very easy. For it means that any vector in the plane of our determining set is expressible as a sum of multiples of the set. So now to test our definition we have only to show that it includes in the determined space all sums of multiples of the set and excludes all other vectors.
Suppose the set is vector A, vector B, and vector C, and let vector D be any vector in the plane, so that some numbers a, b, & c can be found so that
vector D = vector aA + vector bB + vector cC
Then it's obvious that
vector aA + vector bB + vector cC - vector D = (0,0,0)
so that the definition checks here too. And it's equally obvious that if vector D were not expressible as in the first of these equations then no equation such as the second would be found to hold. So we are all set.
The next part of our test I'll let you work out for yourself.
If vector A, vector B, vector C, and vector D are 3-vectors which are not coplanar, they should determine the space of all 3-vectors. If you can prove that any 3-vector can be expressed as a sum of multiples of these arbitrary non-coplanar vectors, then you can complete the testing of the definition much as I have just done for the case where the set is coplanar. If you're good at algebra the proof is a snap; if you are not, content yourself with verifying it intuitively.
So now the definition has been weighed in the balance and, found not wanting. By now you should have an inkling of how we're going to go about obtaining our main result. We were trying, you recall, to define the "dimension" of an arbitrary space in a way which would depend only on the components of the determining vectors.
We have already seen that many different sets may determine the same space. It should be plausible to you that in order to define the dimension of a space we would like to find a determining set for it which is as simple as possible. Specifically, we would like to get as few vectors in the set as possible. OK, let's do.
It is a fact that if any group of vectors from the set is linearly sependent, then one of this group may be eliminated from the set and the same space will still be determined. If one of the

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