Objective: Introduction of vector spaces, subspaces, and bases. Linear Algebra: Section

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1 Objective: Introduction of vector spaces, subspaces, and bases.

2 Vector space Vector space Examples: R n, subsets of R n, the set of polynomials (up to degree n), the set of (continuous, differentiable) real valued functions, etc. Definition A vector space is a (non-empty) set with elements called vectors equipped with an addition of vectors u + v and a scalar multiplication of vectors by (real) scalars cu such that for any vectors u, v, w and scalars c and d: 1. u + v V. 2. u + v = v + u. 3. (u + v) + w = u + (v + w). 4. There is a zero vector 0 such that 0 + v = v. 5. For every v there is v V such that v + ( v) = cv V. 7. c(u + v) = cu + cv. 8. (c + d)v = cv + dv. 9. c(dv) = (cd)v. 10. The number 1 satifies 1v = v. Properties. For any u V and scalar c: 0u = 0, c 0 = 0, u = ( 1)u.

3 Examples R n, R m n, R 1 n, P n(t), P(t), Set of continuous function on R or on an interval, Set of differentiable function on R or on an interval. Let A be m n. The solution set of Ax = 0 in R n. Let A be m n. The set of linear combinations of the columns of A.

4 Subspaces Definition A subset H of a vector space V is a subspace if H is a vector space under the same operations. Remark One needs only check three things: (a) 0 H, (b) u + v H, and (c) cu H u, v H. Subspaces related to a matrix for any scalar c, and Let A be m n. The null space Nul A and the column space of A are: Nul A = {x R n : Ax = 0} R n and Col A = span{a 1,..., a n} = {Ax R m : x R n } R m. Kernel and Range of a Linear Transformation Suppose T : R n R m is defined by T (x) = Ax. Then the kernel of T is kernel(t ) = {x R n : T (x) = 0} = {x R n : Ax = 0}. The range space of T is range(t ) = {T (x) R m : x R n } = {Ax R m : x R n }.

5 Linearly independent sets Definition A set of vectors {v 1,..., v p} V is linearly independent if the vector equation c 1v c pv p = 0 only has trivial solution c 1 = = c p = 0. Theorem 4 Let p 2. Then the set {v 1,..., v p} V of vectors is linearly independent if and only if v j is not a linear combination of v 1,..., v j 1, for every j > 1. Proof. If v j = a 1v a pv p without the v j term on the right, then 0 = a 1v a pv p where a j = 1. So, {v 1,..., v p} is linearly dependent. If {v 1,..., v p} is linearly dependent, then 0 = a 1v a pv p with at least one j such that a j 0. Then a jv j = a 1v a pv p without the v j term on the right. Dividing both side by the scalar a j, we see that v j is a linear combination of the other vectors.

6 Bases Definition Let H be a subspace of V. A basis for H is a linearly independent subset of H that spans H. Theorem 6 The pivot columns of an m n matrix A form a basis for Col A. Proof. Let à be the submatrix of A using the pivot columns of A. Then Ax = b has solution precisely when Ãx = b has solution. Theorem 5 Suppose S = {v 1,..., v p} V that spans a subspace H. (a) If v k is a linearly combination of the other vectors in S, then H is also spanned by S {v k }. (b) If H {0}, then a subset of S forms a basis for H. Proof. (a) If v is a linear combination of S, then it is a linear combination of vectors in S {v k }. (b) The smallest spanning set chosen from S will be linearly independent and spanning. Remark Let S V. Then S is a basis if any one of the following is true. (a) S is a maximal linearly independent set. (b) S is minimal spanning set.