Basic Theorems about Independence, Spanning, Basis, and Dimension
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1 Basic Theorems about Independence, Spanning, Basis, and Dimension Theorem 1 If v 1 ; : : : ; v m span V; then every set of vectors in V with more than m vectors must be linearly dependent. Proof. Let w 1 ; : : : ; w n be vectors in V with n > m: We will show that w 1 ; : : : ; w n are linearly dependent. Consider the canonical equation c 1 w 1 + : : : + c n w n = 0 (1) We will show that there exist c 1 ; : : : ; c n not all zero satisfying the canonical equation above. Since v 1 ; : : : ; v m span V we can write each w i in as a linear combination of v 1 ; : : : ; v m : Then equation (1) becomes c 1 (a 11 v 1 + : : : + a m1 v m ) + : : : + c n (a 1n v 1 + : : : + a mn v m ) = 0 (2) Rearranging the terms gives (a 11 c 1 + : : : + a 1n c n )v 1 + : : : + (a m1 c 1 + : : : + a mn c n )v m = 0 (3) Now consider the homogeneous system of equations shown below. a 11 c 1 + : : : + a 1n c n = 0... (4) a m1 c 1 + : : : + a mn c n = 0 A previous theorem implies there exist solutions c 1 ; : : : ; c n not all zero to system because n > m; that is, there are more unknowns than equations. These same c 1 ; : : : ; c n (not all zero) satisfy equations (3) and (2) and (1). Because there exist c 1 ; : : : ; c n not all zero satisfying the canonical equation (1) the vectors w 1 ; : : : ; w n are linearly dependent. Theorem 2 If v 1 ; : : : ; v n are linearly independent in V; then every set of vectors in V with fewer than n vectors cannot span V: Proof. (by contradiction) Assume w 1 ; : : : ; w m (with m < n) span V: Then by Theorem 1 every set of vectors in V with more than m vectors must be linearly dependent. This contradicts our hypothesis that v 1 ; : : : ; v n are linearly independent.
2 Theorem 3 If v 1 ; : : : ; v n is a basis of V and w 1 ; : : : ; w m is a basis of V; then n = m: Proof. By denition of basis v 1 ; : : : ; v n span V and w 1 ; : : : ; w m are linearly independent, so theorem 1 says m n: Similarly, w 1 ; : : : ; w m span V and v 1 ; : : : ; v n are linearly independent, so theorem 1 implies n m: The. only possibility is that n = m: Theorem 3 says that if V has one basis with n elements, then all bases of V have n elements. The number n is intrinsic to the space itself and is independent of which basis you choose. We make the following denition. Denition Let V be a vector space with a basis of n vectors. The number n is called the dimension of V: Theorem 4 Suppose v 1 ; : : : ; v k are linearly independent. If v k+1 =2 spanfv 1 ; : : : ; v k g; then v 1 ; : : : ; v k ; v k+1 are linearly independent. Proof. Consider the canonical equation c 1 v 1 + : : : + c k v k + c k+1 v k+1 = 0 We rst claim that c k+1 = 0: For if c k+1 6= 0; the we can solve the canonical equation for v k+1 to get v k+1 = 1 c k+1 ( c 1 v 1 c k v k ) This contradicts the fact that v k+1 =2 spanfv 1 ; : : : ; v k g: Hence, c k+1 = 0 as claimed. The canonical equation now simplies to c 1 v 1 + : : : + c k v k = 0 Now, since v 1 ; : : : ; v k are linearly independent we have c 1 = 0; : : : ; c k = 0 (in addition to c k+1 = 0): Hence, the vectors v 1 ; : : : ; v k ; v k+1 are linearly independent, which is what we wanted to show.
3 Theorem 5 Suppose v 1 ; : : : ; v k span V: If v k 2 spanfv 1 ; : : : ; v k 1 g; then v 1 ; : : : ; v k 1 span V: Proof. Choose an arbitrary vector v in V: and show that v can be written as a linear combination of v 1 ; : : : ; v k 1 : By hypothesis we can write v as a linear combination of v 1 ; : : : ; v k ; That is, v = c 1 v 1 + : : : + c k 1 v k 1 + c k v k But since v k 2 spanfv 1 ; : : : ; v k 1 g we can write v k = b 1 v 1 + : : : + b k 1 v k 1 Substituting this expression for v k into the previous expression for v gives v = c 1 v 1 + : : : + c k 1 v k 1 + c k (b 1 v 1 + : : : + b k 1 v k 1 ) v = (c 1 + c k b 1 )v 1 + : : : + (c k 1 + c k b k 1 )v k 1 Thus we see that v is a linear combination of v 1 ; : : : ; v k 1 : Since v was an arbitrary vector in V; v 1 ; : : : ; v k 1 span V; as claimed
4 Theorem 6 (The Cut-Your-Work-In-Half Theorem) (Part I) If V is of dimension n and v 1 ; : : : ; v n are linearly independent, then v 1 ; : : : ; v n must span V (and hence form a basis of V ): (Part II) If V is of dimension n and v 1 ; : : : ; v n span V; then v 1 ; : : : ; v n must be linearly independent (and hence form a basis of V ): Proof. (of Part I by contradiction) We want to show v 1 ; : : : ; v n span V: To arrive at a contradiction, suppose they do not span V: That means there is some vector in V which is not in the span of v 1 ; : : : ; v n ; call it v n+1 : By Theorem 4 the vectors v 1 ; : : : ; v n ; v n+1 are linearly independent: Now our hypothesis that V is of dimension n says V has a basis w 1 ; : : : ; w n and these n basis vectors span V: This is now a contradiction to Theorem 1 which says every set with more than n vectors must be linearly dependent. This completes the proof of part I. (of Part II by contradiction) We want to show that v 1 ; : : : ; v n are linearly independent. To arrive at a contradiction suppose they are linearly dependent. This means that one of v 1 ; : : : ; v n is a linear combination of the others. Without loss of generality, suppose v n = c 1 v 1 + : : : + c n 1 v n 1 Then by Theorem 6 it must the case that v 1 ; : : : ; v n 1 span V: Now V has a basis w 1 ; : : : ; w n and these n vectors are linearly independent. This is now a contradiction to theorem 2 which says that fewer than n vectors cannot span V:
5 Theorem 7 Suppose V is of dimension n: Then every linearly independent set with fewer than n vectors can be extended to a basis of V: Proof. Suppose S k is a set of k linearly independent vectors with k < n: By Theorem 2 S k does not span V: So there exists a vector v k+1 =2 span(s k ): By Theorem 4 S k+1 = S k [ fv k+1 g must be linearly independent. In this way, we can continue to append vectors to the original set S k until we arrive at a set S n of n linearly independent vectors. By Theorem 6 (Part I), S n must be a basis of V Theorem 8 Suppose V is of dimension n: Then every spanning set with more than n vectors can be reduced to a basis of V: Proof. Suppose S k is a set of k vectors which spans V with k > n: By Theorem 1 S k must be linearly dependent, meaning one is a linear combination of the others. By Theorem 5 that one vector may be discarded from the set and the remaining set S k 1 of k 1 vectors still spans V: In this way, we can continue to discard vectors from the original set S k until we arrive at a set S n of n vectors which spans V: By Theorem 6 (Part II) S n must be a basis of V:
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