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nt4564vectr2 - Further Properties of Vectors and...

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Further Properties of Vectors and n-Dimensional Vector Spaces Subspaces; Span of a Set of Vectors A subset S of the vector space E n is a subspace of E n if it is closed under the formation of linear combinations of vectors in the set; i.e., if, given X and Y in S and arbitrary scalars α and β , the linear combination α X + β Y also lies in S . Such a subspace is, geometrically, a line, plane or hyperplane, depending on its dimension, passing through the origin. For example, if Z is a non-zero vector, then the set Z ≡ { X E n | Z · X = 0 } is a subspace. For if X and Y lie in Z then Z · X = Z · Y = 0 and then it is clear that for arbitrary scalars α and β we will have Z · ( α X + β Y ) = α Z · X + β Z · Y = 0 . If S is a subspace then S is easily seen, in much the same way, to also be a subspace. It is called the orthogonal complement of S . Example The set of vectors S = { X E n | x 1 + x 2 + · · · + x n = 0 } is easily seen to be a subspace. The orthogonal complement is S = { Y E n | y k = y j , k, j = 1 , 2 , · · · , n } , i.e., vectors all of whose components are equal (“constant” vectors). For if Y S then it is orthogonal to vectors D j = (1 , 0 , 0 , · · · - 1 , 0 , · · · 0) , , j = 2 , 3 , · · · , n , the - 1 appearing in the j -th position. Then 0 = Y · D j = y 1 - y j y j = y 1 . 1
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Suppose X 1 , X 2 , · · · X m is a finite set of vectors in E n . The span of this set, sp { X 1 , X 2 , · · · X m } , is the set of all vectors X which are linear combina- tions of X 1 , X 2 , · · · X m . An easy computation shows that sp { X 1 , X 2 , · · · X m } is a subspace of E n : the subspace spanned by X 1 , X 2 , · · · X m . Example Again let S = { X E n | x 1 + x 2 + · · · + x n = 0 } . We claim this space is the span of the n - 1 vectors D j described in the preceding example. First of all, it is clear that any linear combination of the D j must have the indicated property since the D j themselves have this property. On the other hand, given any vector X such that x 1 + x 2 + · · · + x n = 0 it is easy to see that X = x 1 D 2 + ( x 1 + x 2 ) D 3 + · · · + ( x 1 + x 2 + · · · + x n - 1 ) D n . (Working through the arithmetic the right hand side is seen to be the vector ( x 1 , x 2 , · · · , x n - 1 , ( - x 1 - x 2 · · · - x n - 1 )) and the last component is just x n since x 1 + x 2 + · · · + x n = 0. A set of vectors X 1 , X 2 , · · · , X m in E n is linearly independent if none of these vectors can be written as a linear combination of the others; that is, it is not possible, for any j , to write X j = X k 6 = j d k X k for any scalar coefficients d k , k 6 = j . Another way to say this is that it is not possible to find coefficients c k , k = 1 , 2 , · · · , m , not all zero, such that c 1 X 1 + c 2 X 2 + · · · + c m X m = 0 2
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(for, if this were possible, selecting a non-zero c j , dividing by this quantity and transposing X j to th other side we would have X j = k 6 = j d k X k with d k = - c k c j , k 6 = j . It is not hard to see that this is the same as saying that if any of the vectors in the list X 1 , X 2 , · · · , X m are eliminated, the span of the
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