Prove the theorem cyclically given Let W,..., W be subspaces of a vector space V. We...

  

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Prove the theorem cyclically given Let W,..., W be subspaces of a vector space V. We say that they are independent if, whenever w W,..., WkE Wk with w + +wk = 0, then w = ... = Wk = 0. In this case, the sum W+...+W is called a direct sum and denoted by WD...OWk. If k = 2 and W W = V, then the subspaces W and W are called complementary subspaces of V. Note: For every w E W 1i k, such that w = w + Wk there exists a unique collection of vectors wi Wi, + Wk. Let V be finite dimensional. Let W,..., Wk be subspaces of V. Then the following are equivalent: (i) W,..., We are independent. (ii) W; ^ (W + + Wi-1 + Wi+1+ + Wk)= {0} for every i (iii) If B, is a basis of W, then B U... UB is a basis of W + ... 1,..., k. + Wk. Prove the theorem cyclically given Let W,..., W be subspaces of a vector space V. We say that they are independent if, whenever w W,..., WkE Wk with w + +wk = 0, then w = ... = Wk = 0. In this case, the sum W+...+W is called a direct sum and denoted by WD...OWk. If k = 2 and W W = V, then the subspaces W and W are called complementary subspaces of V. Note: For every w E W 1i k, such that w = w + Wk there exists a unique collection of vectors wi Wi, + Wk. Let V be finite dimensional. Let W,..., Wk be subspaces of V. Then the following are equivalent: (i) W,..., We are independent. (ii) W; ^ (W + + Wi-1 + Wi+1+ + Wk)= {0} for every i (iii) If B, is a basis of W, then B U... UB is a basis of W + ... 1,..., k. + Wk.