VECTOR SPACES & SUBSPACES

Transcription

1 VECTOR SPACES & SUBSPACES Definition: The real Vector Space " V " is the set of all entities called "vectors" with real entries that satisfy two closure properties and obey a set of eight rules. If "x" and "y" are in "V" and " α" is a scalar (real number), then those vectors must satisfy these closure properties. (.) x + y is also in "V". (.) α x is also in "V". If "x", "y", & "z" are vectors in "V" and "α" & "β" are scalars (real numbers), then all of the vectors in "V" must obey all of the following rules for "V" to be a Vector Space. (.) x + y y + x (.)( x + y) + z x + ( y + z) (.) x + x (4.) x + ( x) (5.)α ( x + y) α x + (6.)( α + β) x α x + (7.)( αβ ) x α β x (8.)I x x ( ) α y β x Example : Show that R mxn with the usual addition and scalar multiplication of matrices, is a vector space. Page of

2 Let "A", "B " and " C " be in R mxn, but otherwise arbitrary matrices. To show that R mxn is a vector space, we capitalize on the properties of real numbers, which are the entries of the matrices. Note that a ij + b ij c ij. Therefore, the sum is in R mxn. Also, α a ij c ij. Therefore, a scalar multiple is in R mxn. The closure property is thus satisfied. (.) a ij + b ij b ij + a ij. Thus, A + B B + A. ( ) c ij ( ) (.) a ij + b ij + a ij + b ij + c ij. Thus, ( A + B) + C A + ( B + C). (.) a ij + a ij. Thus, A + O A. ( ) (4.) a ij + a ij. Thus, A + ( A) O. ( ) (5.) α a ij + b ij α a ij + α b ij. Thus, α ( A + B) α A + α B. (6.) ( α + β) ( a ij ) ( α + β) ( A) ( ) ( a ij ) (7.) αβ α a ij + β a ij. Thus, α A + β A. ( ) ( ) A α β a ij. Thus, αβ ( ) ( ) α β A. ( ) (8.) ( ) a ij a ij. Thus, I A A. Page of

3 (Note that the Identity Element " I " is a matrix that is in R nxm and not R mxn. Otherwise, the matrix product, " I A", would not be defined.) R mxn satisfies all of the requirements for a vector space and thus constitutes a vector space. Definition: If " S " is a non-empty subset of a vector space " V ", and " S " satisfies the closure property, then " S " is said to be a subspace of " V ". Example a: Determine whether the set: { x x such that x + x } is a subspace of R. Let x x x, and y y y. x + y z x x y + y z z x + y x + y x + y ( ) x + y z z Thus, the given set is closed under addition. Page of

4 α x z α x x z z α x α x ( α x ) ( ) α x z z Thus, the given set is closed under scalar multiplication. Because the given set satisfies the closure property in R, it is a subspace of R. Example b: Determine whether the set: { x x such that x x } is a subspace of R. Let x x x, and y y y. x + y z x x y + y z z x + y x + y ( ) ( x y ) z z x + y + x x + y y + x y + x y Page 4 of

5 x x y y z z x y + x y Thus, the given set is NOT closed under addition. Because the given set does not satisfy the closure property in R, it is NOT a subspace of R. Example c: Let " A " be a particular vector in R x & " B " be any vector in R x. Determine whether the set: { Β ε R x I such that A B R x. BA } is a subspace of A ( α B) ( α B) A α A α B B A AB BA Thus, the given set is closed under scalar multiplication. Page 5 of

6 AB BA AC CA ( B + C) A BA + CA AB + AC A ( B + C) Thus, the given set is closed under addition. Because the given set satisfies the closure property in R x, it is a subspace of R x. Example d: Let " A " be a particular vector in R x & " B " be any vector in R x. Determine whether the set: { Β ε R x I such that A B R x. BA } is a subspace of AB AC BA CA ( B + C) A BA + CA AB + AC A ( B + C) Let D B + C. Then D ε R x. But A D DA. Thus, the given set is NOT closed under addition. Because the given set does not satisfy the closure property in R x, it is NOT a subspace of R x. Page 6 of

7 Definition: If {v... v p } is a spanning set for the vector space " V ", then the set of all linear combinations of v... v p is denoted by Span{v... v p } and every vector, x, in " V " can be written as a Linear Combination of v... v p. Those vectors, thus, have this form, where the c k are scalars. x c v + c v c k v k c p v p p c k v k k Example # : Give a geometric description of Span{v, v }. v v These vectors reside in R. But they do not span R. Instead they span a plane in R that passes thru the origin. More specifically, that plane is the xz-plane. Page 7 of

8 Vectors lie in plane z y x Theorem:The equation, Ax b, has a solution if and only if b is a linear combination of the columns of "A". Let "A" have pxn dimensions with columns, a k. Ax x a + x a x n a n Ax b x a + x a x n a n Thus, b must be expressible as a linear combination of the columns of "A" for Ax b to have at least one solution. Page 8 of

9 Example # 4a: Does the set {,, } span R? α β + γ + a b c α β γ a b c Since the matrix is non-singular, a linear combination of its columns can always be found to produce any vector in. Thus, the given set spans R. Page 9 of

10 Example # 4b: Does the set {,,, } span R? We have shown in the previous example that the first three vectors in this set span R. The fourth vector is superfluous. The given set spans R. Example # 4c: Does the set {,, } span R The matrix is singular because. Therefore, a linear combination of the columns of the matrix can not always be found for every vector in R. Accordingly, the given set does NOT span R. Page of