同济大学：《线性代数》课程教学资源（课件讲稿）Spanning Sets and Independent Sets

Spanning Sets and Independent Sets Li Zhang Department of Mathematics Tongji University 悠 4口+心左4生主9QG

Spanning Sets and Independent Sets Li Zhang Department of Mathematics Tongji University

Span Definition Let v1.v2,....Un be the vectors in a vector space V.A vector uEV is said to be a linear combination of v1.v2.....vn if there exist scalars c1. C2,·,Cn such that u C1v1+c2v2+...+CnUn. The set of all linear combinations of v1,v2.....Un is said to the span of U1.U2,....Un written S(v1,v2,...,Un)or span (v1,v2,...Un) 4口+心左4生主9QG

Span Definition Let v1, v2, . . ., vn be the vectors in a vector space V . A vector u ∈ V is said to be a linear combination of v1, v2, . . ., vn if there exist scalars c1, c2, . . ., cn such that u = c1v1 + c2v2 + . . . + cnvn. The set of all linear combinations of v1, v2, . . ., vn is said to the span of v1, v2, . . ., vn written S (v1, v2, . . . , vn) or span (v1, v2, . . . , vn)

Examples (1)The set of the solutions to the differential equation Py dz2-=0 (*) is a vector space V under the rules of and.for functions, Example ll of the last lecture. Then this vector space is spaned by e2 and e-.If y is a solution of (*)then there are constants(scalars)c1,c2 ER such that y(x)=cle=+ce-=. 4口+++左+4生+定QC

Examples (1) The set of the solutions to the differential equation d 2y dx2 − y = 0 (∗) is a vector space V under the rules of + and • for functions, Example II of the last lecture. Then this vector space is spaned by e x and e −x . If y is a solution of (∗), then there are constants (scalars) c1, c2 ∈ R such that y(x) = c1e x + c2e −x

Examples (continued) (2)V=R3.e1=(1,0,0),e2=(0,1,0),e3=(0,0,1).Then S(e1,e2)=the zy-plane. 4口+心左4生+主9QG

Examples (continued) (2) V = R 3 , e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Then S(e1, e2) = the xy-plane

Subspaces Suppose X is a set with a binary operationo so given any pair 1. x2∈X we have1ox2∈X. Let Y be a subset of X.We say Y is closed under o if for any pair y1,y2∈Y we have1o2∈Y. Definition Let(V,+,.)be a vector space and U C V be a subset.Then U is said to be a subvector space or subspace of V if (i)U is closed under + (ii)U is closed under.. 4口++心++左+4生+定QC

Subspaces Suppose X is a set with a binary operation ◦ so given any pair x1, x2 ∈ X we have x1◦x2 ∈ X. Let Y be a subset of X. We say Y is closed under ◦ if for any pair y1, y2 ∈ Y we have y1◦y2 ∈ Y . Definition Let (V, +, •) be a vector space and U ⊂ V be a subset. Then U is said to be a subvector space or subspace of V if (i) U is closed under +, (ii) U is closed under •

Subspaces Proposition If U is closed under and.,then U equipped with the restrictions of+ and.satisifies the 8 vector space axioms Al,A2,A3,A4 and S1,S2, S3,S4.Hence (U,+,is a vector space.U is said to be a subspace of V Examples (1)The zy-plane CR3. (2)The space of polynomial functions of one variable of degree n Poln(R)∈FR(R) 4口+心左4生主9QG

Subspaces Proposition If U is closed under + and •, then U equipped with the restrictions of + and • satisifies the 8 vector space axioms A1, A2, A3, A4 and S1, S2, S3, S4. Hence (U, +, •) is a vector space. U is said to be a subspace of V . Examples (1) The xy-plane ⊂ R 3 . (2) The space of polynomial functions of one variable of degree n Poln(R) ∈ FR (R)

Spanning Sets Proposition Suppose v1,v2.....Un are vectors in V.Then S(v1,v2,...,Un)is a subspace of V. Proof. S is closed under c1v1+c22+..+cnm+d11+d22+..+dnn= =(c1+d)v+(c2+d2)2+.+(cn+dn)vn. S is closed under. c(c1v1+c2v2+...+cnUn)=(cc1)v1 +(cc2)v2+...+(ccn)Un. S(v1,v2,...,Un)is said to be the subspace of V spanned by U1,U2,·,Un

Spanning Sets Proposition Suppose v1, v2, . . ., vn are vectors in V . Then S (v1, v2, . . . , vn) is a subspace of V . Proof. S is closed under + c1v1 + c2v2 + . . . + cnvn + d1v1 + d2v2 + . . . + dnvn = = (c1 + d1)v1 + (c2 + d2)v2 + . . . + (cn + dn)vn. S is closed under • c(c1v1 + c2v2 + . . . + cnvn) = (cc1)v1 + (cc2)v2 + . . . + (ccn)vn. S (v1, v2, . . . , vn) is said to be the subspace of V spanned by v1, v2, . . . , vn

Properties of+and。 Proposition S(v1,v2,...,Un)is a subspace of V is the smallest subspace of V containing v1,v2,....Un. Proof. Suppose U contains v1,v2....,Un.Then since U is closed under and any linear combination c1v1+c2v2+...+cnun must be in U.Hence S(v1,v2,,vn)≤U But U is the smallest subspace of V containing v1.v2.....Un.so US(1,2,,un), so U=S(v1,2,,vn).□ 4口+++左+4生+定QC

Properties of + and • Proposition S (v1, v2, . . . , vn) is a subspace of V is the smallest subspace of V containing v1, v2, . . ., vn. Proof. Suppose U contains v1, v2, . . ., vn. Then since U is closed under + and •, any linear combination c1v1 + c2v2 + . . . + cnvn must be in U. Hence S (v1, v2, . . . , vn) ⊆ U. But U is the smallest subspace of V containing v1, v2, . . ., vn, so U ⊆ S (v1, v2, . . . , vn), so U = S (v1, v2, . . . , vn)

Spanning Sets Definition Let V be a vector space.Then a collection of vectors v1,v2,...,Un}c V is said to be a spanning set for V if V=S(v1,2,,vn). Examples (1)(e1,e2,e3)is a spanning set for R3. (2)(e1,e2,...,en)is a spanning set for R". (3)(1,,2,....)is a spanning set for Poln(R). 4口++左4生+定QC

Spanning Sets Definition Let V be a vector space. Then a collection of vectors {v1, v2, . . . , vn} ⊂ V is said to be a spanning set for V if V = S (v1, v2, . . . , vn). Examples (1) (e1, e2, e3) is a spanning set for R 3 . (2) (e1, e2, . . . , en) is a spanning set for R n. (3) ￾ 1, x, x2 , . . . , xn
is a spanning set for Poln(R)

Linear Independence There are inefficient(too big)spanning sets for a vector space V.For example {(1,0),(0,1),(1,1)}is a spanning set for R2 but any two of the three vectors still spans. Dependence Relation Let v1,v2,...,UnE V.Then a dependence relation between V1,U2,...,Un is an equation c1v1+c2v2+...+CnUn =0,C1,C2,...Cn ER. The dependence relation is said to the trivial dependence relation if all the cs are zero.So in the above 1(1,0)+1(0,1)-1(1,1)=0. is a dependence relation. 4口+++左+4生+定QC

Linear Independence There are inefficient (too big) spanning sets for a vector space V . For example {(1, 0), (0, 1), (1, 1)} is a spanning set for R 2 but any two of the three vectors still spans. Dependence Relation Let v1, v2, . . . , vn ∈ V . Then a dependence relation between v1, v2, . . . , vn is an equation c1v1 + c2v2 + . . . + cnvn = 0, c1, c2, . . . cn ∈ R. The dependence relation is said to the trivial dependence relation if all the c 0 i s are zero. So in the above 1•(1, 0) + 1•(0, 1) − 1•(1, 1) = 0. is a dependence relation