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Two such are W1 = [{e1 , e2 }] and W2 = [{e3 , e4 }], and also U1 = [{e1 }] and U2 = [{e2 , e3 , e4 }]. ) In contrast, any partition of R1 ’s single-vector basis will give one basis with no elements and another with a single element. Thus any decomposition involves R1 and its trivial subspace. 29 Set inclusion one way is easy: {w1 + · · · + wk wi ∈ Wi } is a subset of [W1 ∪ . . ∪ Wk ] because each w1 + · · · + wk is a sum of vectors from the union. For the other inclusion, to any linear combination of vectors from the union apply commutativity of vector addition to put vectors from W1 first, followed by vectors from W2 , etc.

RepB (c · p + d · q) = RepB ( (cp1 + dq1 )β1 + (cp2 + dq2 )β2 + (cp3 + dq3 )β3 )   cp1 + dq1 = cp2 + dq2  cp3 + dq3     q1 p1 = c · p2  + d · q2  p3 q3 = RepB (p) + RepB (q) (d) Use any basis B for P2 whose first two members are x + x2 and 1 − x, say B = x + x2 , 1 − x, 1 . 34 See the next subsection. 35 (a) Most of the conditions in the definition of a vector space are routine. We here sketch the verification of part (1) of that definition. For closure of U ×W , note that because U and W are closed, we have that u1 +u2 ∈ U and w1 + w2 ∈ W and so (u1 + u2 , w1 + w2 ) ∈ U × W .

That’s simply an assertion that the three-element set is linearly independent. 12, that’s equivalent to an assertion that the set is a basis for R3 . 29 Let the space V be finite dimensional. Let S be a subspace of V . (a) The empty set is a linearly independent subset of S. 10, it can be expanded to a basis for the vector space S. (b) Any basis for the subspace S is a linearly independent set in the superspace V . Hence it can be expanded to a basis for the superspace, which is finite dimensional.

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A short linear algebra book (answers) by Hefferon J.

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Categories: Algebra