Math 3191 Applied Linear Algebra

Transcription

1 Math 9 Applied Linear Algebra Lecture : Null and Column Spaces Stephen Billups University of Colorado at Denver Math 9Applied Linear Algebra p./8

2 Announcements Study Guide posted HWK posted Math 9Applied Linear Algebra p./8

3 Subspaces, review A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. For each u and v are in H, u + v is in H. (In this case we say H is closed under vector addition.) c. For each u in H and each scalar c, cu is in H. (In this case we say H is closed under scalar multiplication.) Math 9Applied Linear Algebra p./8

4 Example Is H = 8 < : x x + 9 = : x is real ; a subspace of? I.e., does H satisfy properties a, b and c? x x Graphical Depiction of H Math 9Applied Linear Algebra p./8

5 Solution All three properties must hold in order for H to be a subspace of R. Property (a) is not true because. Therefore H is not a subspace of R. x x Math 9Applied Linear Algebra p./8

6 Another way to show that H is not a subspace of R : Let u = and v =, then u + v = and so u + v =, which is in H. So property (b) fails and so H is not a subspace of R. x x Math 9Applied Linear Algebra p./8

7 A Shortcut for Determining Subspaces THEOREM If v,..., v p are in a vector space V, then Span{v,..., v p } is a subspace of V. Proof: In order to verify this, check properties a, b and c of definition of a subspace. a. is in Span{v,..., v p } since = v + v + + v p. Math 9Applied Linear Algebra p./8

8 b. To show that Span{v,..., v p } is closed under vector addition, we choose two arbitrary vectors in Span{v,..., v p } : u =a v + a v + + a p v p and v =b v + b v + + b p v p. Then u + v = (a v + a v + + a p v p ) + (b v + b v + + b p v p ) = ( v + v ) + ( v + v ) + + ( v p + v p ) = (a + b ) v + (a + b ) v + + (a p + b p ) v p. So u + v is in Span{v,..., v p }. Math 9Applied Linear Algebra p.8/8

9 Proof (cont.) c. To show that Span{v,..., v p } closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span{v,..., v p } : Then v =b v + b v + + b p v p. cv =c (b v + b v + + b p v p ) = v + v + + v p So cv is in Span{v,..., v p }. Since properties a, b and c hold, Span{v,..., v p } is a subspace of V. Math 9Applied Linear Algebra p.9/8

10 Recap. To show that H is a subspace of a vector space, use Theorem. (I.e., rewrite H as the span of a set of vectors).. To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. Math 9Applied Linear Algebra p./8

11 EXAMPLE: Is V = {(a + b, a b) : a and b are real} a subspace of R? Why or why not? Solution: Write vectors in V in column form: a + b = a a b a + b b = + So V =Span{v, v } and therefore V is a subspace of by Theorem. Math 9Applied Linear Algebra p./8

12 Is H = a + b a + a Why or why not? EXAMPLE: : a and b are real a subspace of R? Solution: is not in H since a = b = or any other combination of values for a and b does not produce the zero vector. So property fails to hold and therefore H is not a subspace of R. Math 9Applied Linear Algebra p./8

13 EXAMPLE: [ a Is the set H of all matrices of the form a + b subspace of M? Explain. Solution: Since [ ] [ ] [ a b a b = + a + b b a b b [ ] [ ] Therefore H =Span subspace of M. = a {[ + b ], [. ]} b b ] ] a and so H is a Math 9Applied Linear Algebra p./8

14 Sec.. Null Spaces, Column Spaces, & Linear Transformations The null space of an m n matrix A, written as Nul A, is the set of all solutions to the homogeneous equation Ax =. Nul A = {x : x is in R n and Ax = } (set notation) THEOREM The null space of an m n matrix A is a subspace of R n. Equivalently, the set of all solutions to a system Ax = of m homogeneous linear equations in n unknowns is a subspace of R n. Proof: Nul A is a subset of R n since A has n columns. Must verify properties a, b and c of the definition of a subspace. Property (a) Show that is in Nul A. Since, is in. Math 9Applied Linear Algebra p./8

15 Proof (cont). Property (b) If u and v are in Nul A, show that u + v is in Nul A. Since u and v are in Nul A, and. Therefore A (u + v) = + = + =. Property (c) If u is in Nul A and c is a scalar, show that cu in Nul A: A (cu) = A (u) = c =. Since properties a, b and c hold, A is a subspace of R n. Solving Ax = yields an explicit description of Nul A. Math 9Applied Linear Algebra p./8

16 EXAMPLE Find an explicit description of Nul A where A = 9. Solution: Row reduce augmented matrix corresponding to Ax = : 9 x x x x = x x x x x + x x == x + x + x x x Then Nul A =span{u, v, w}. Question: What are u, v, w? Math 9Applied Linear Algebra p./8

17 Observations:. Spanning set of Nul A, found using the method in the last example, is automatically linearly independent: c + c + c = = c = c = c =. If Nul A {}, the the number of vectors in the spanning set for Nul A equals the number of free variables in Ax =. Math 9Applied Linear Algebra p./8

18 Column Space The column space of an m n matrix A (Col A) is the set of all linear combinations of the columns of A. If A = [a... a n ], then Col A =Span{a,..., a n } THEOREM The column space of an m n matrix A is a subspace of R m. Why? (Theorem, page ) Recall that if Ax = b, then b is a linear combination of the columns of A. Therefore Col A = {b : b =Ax for some x in R n } Math 9Applied Linear Algebra p.8/8

19 EXAMPLE Find a matrix A such that W = Col A where W = 8 >< >: x y y x + y 9 >= : x, y in R. >; Solution: x y y x + y = x + y = x y Therefore A =. (By Theorem, Chapter ), The column space of an m n matrix A is all of R m if and only if the equation Ax = b has a solution for each b in R m. Math 9Applied Linear Algebra p.9/8

20 Contrast between Nul A and Col A EXAMPLE: Let A =. (a) The column space of A is a subspace of R k where k =. (b) The null space of A is a subspace of R k where k =. (c) Find a nonzero vector in Col A. (There are infinitely many possibilities.) + + = Math 9Applied Linear Algebra p./8

21 Contrast between Nul A and Col A (cont). (d) Find a nonzero vector in Nul A. Solve Ax = and pick one solution. = x = x x is free x = Let x = ; then x = x x =. x Contrast Between Nul A and Col A where A is m n (see page ) Math 9Applied Linear Algebra p./8

22 Review A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. For each u and v in H, u + v is in H. (In this case we say H is closed under vector addition.) c. For each u in H and each scalar c, cu is in H. (In this case we say H is closed under scalar multiplication.) If the subset H satisfies these three properties, then H itself is a vector space. Math 9Applied Linear Algebra p./8

23 Review: THEOREM, and (Sections. &.) If v,..., v p are in a vector space V, then Span{v,..., v p } is a subspace of V. The null space of an m n matrix A is a subspace of R n. The column space of an m n matrix A is a subspace of R m. Math 9Applied Linear Algebra p./8

24 EXAMPLES: (a) Determine whether the following set is a vector space or provide a counterexample. {[ ] } x H = : x y = y [ ] Solution: Since = space. is not in H, H is not a vector Math 9Applied Linear Algebra p./8

25 Examples (cont) (b) Determine whether the following set is a vector space or provide a counterexample. 8 >< x V = y >: : x y = y + z = z 9 >= Solution: Rewrite x y = >; y + z = So V =Nul A where A = a vector space. as x y z =.. Since Nul A is a subspace of R, V is Math 9Applied Linear Algebra p./8

26 Examples (c) Determine whether the following set is a vector space or provide a counterexample. 8 >< x + y S = x y >: y 9 >= : x, y, z are real >; One Solution: Since x + y x y y = x + y, 8 >< S =span >:, 9 >= ; therefore S is a vector space by Theorem. >; Math 9Applied Linear Algebra p./8

27 Examples Another Solution: Since x + y x y y = x + y, S =Col A where A = space is a vector space. ; therefore S is a vector space, since a column Math 9Applied Linear Algebra p./8

28 Kernel and Range of a Linear Transformation A linear transformation T from a vector space V into a vector space W is a rule that assigns to each vector x in V a unique vector T (x) in W, such that i. T (u + v) = T (u) +T (v) for all u, v in V ; ii. T (cu) =ct (u) for all u in tin V and all scalars c. The kernel (or null space) of T is the set of all vectors u in V such that T (u) =. The range of T is the set of all vectors in W of the form T (u) where u is in V. So if T (x) =Ax, col A =range of T. Math 9Applied Linear Algebra p.8/8