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Lect15

# Lect15 - Chapter 3 Vector Spaces Math1111 Basis and...

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Theorem 3.4.4 Theorem 3.4.4 Let V be a vector space and dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . ii) If v 1 , ··· , v k are linearly independent where k < n , we can add suitably vectors x k + 1 , ··· , x n so that { v 1 , ··· , v k , x k + 1 , ··· , x n } is a basis. iii) Suppose v 1 , ··· , v m where m > n span V . We may drop suitably m - n vectors to get a basis.

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 i) Theorem 3.4.4 i) Let V be a vector space, dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . Proof.
Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 i) Theorem 3.4.4 i) Let V be a vector space, dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . Proof. Suppose Span ( v 1 , ··· , v k ) = V , where k < n .

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 i) Theorem 3.4.4 i) Let V be a vector space, dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . Proof. Suppose Span ( v 1 , ··· , v k ) = V , where k < n . If v 1 , ··· , v k are linearly independent, then { v 1 , ··· , v k } is a basis for V . So dim V = k ( < n ) . Contradict to dim V = n .
Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 i) Theorem 3.4.4 i) Let V be a vector space, dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . Proof. Suppose Span ( v 1 , ··· , v k ) = V , where k < n . If v 1 , ··· , v k are linearly independent, then { v 1 , ··· , v k } is a basis for V . So dim V = k ( < n ) . Contradict to dim V = n . If v 1 , ··· , v k are linearly dependent, then one of v 1 , ··· , v k is a linear combination of the other k - 1 vectors. Repeat the argument in the proof of Theorem 3.4.3 II). We conclude that dim V = r < k < n .

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 i) Theorem 3.4.4 i) Let V be a vector space, dim V = n 1 . i) For any v 1 , ··· , v k V with k < n , Span ( v 1 , ··· , v k ) ( V . Proof. Suppose Span ( v 1 , ··· , v k ) = V , where k < n . If v 1 , ··· , v k are linearly independent, then { v 1 , ··· , v k } is a basis for V . So dim V = k ( < n ) . Contradict to dim V = n . If v 1 , ··· , v k are linearly dependent, then one of v 1 , ··· , v k is a linear combination of the other k - 1 vectors. Repeat the argument in the proof of Theorem 3.4.3 II). We conclude that dim V = r < k < n . Contradict to dim V = n again.
Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 ii) Theorem 3.4.4 ii) Let V be a vector space, dim V = n 1 . ii) If v 1 , ··· , v k are linearly independent where k < n , we can add suitable vectors x k + 1 , ··· , x n so that { v 1 , ··· , v k , x k + 1 , ··· , x n } is a basis. Proof.

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Chapter 3. Vector Spaces Math1111 Basis and Dimension Proof of Thm 3.4.4 ii) Theorem 3.4.4 ii) Let V be a vector space, dim V = n 1 . ii) If v 1 , ··· , v k are linearly independent where k < n , we can add suitable vectors x k + 1 , ··· , x n so that { v 1 , ··· , v k , x k + 1 , ··· , x n } is a basis.
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Lect15 - Chapter 3 Vector Spaces Math1111 Basis and...

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