Math 2331 Linear Algebra

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1 4.3 Linearly Independent Sets; Bases Math 233 Linear Algebra 4.3 Linearly Independent Sets; Bases Jiwen He Department of Mathematics, University of Houston math.uh.edu/ jiwenhe/math233 Jiwen He, University of Houston Math 233, Linear Algebra / 2

2 4.3 Linearly Independent Sets; Bases Linearly Independent Sets Definition Facts s A Basis Set: Definition A Basis Set: s Nul A: s and Theorem Col A: s and Theorem The Spanning Set Theorem Jiwen He, University of Houston Math 233, Linear Algebra 2 / 2

3 Linearly Independent Sets Linearly Independent Sets A set of vectors {v, v 2,..., v p } in a vector space V is said to be linearly independent if the vector equation c v + c 2 v c p v p = has only the trivial solution c =,..., c p =. The set {v, v 2,..., v p } is said to be linearly dependent if there exists weights c,..., c p,not all, such that c v + c 2 v c p v p =. Jiwen He, University of Houston Math 233, Linear Algebra 3 / 2

4 Linearly Independent Sets: Facts The following results from Section.7 are still true for more general vectors spaces. Fact A set containing the zero vector is linearly dependent. Fact 2 Fact 3 A set of two vectors is linearly dependent if and only if one is a multiple of the other. A set containing the zero vector is linearly independent. Jiwen He, University of Houston Math 233, Linear Algebra 4 / 2

5 Linearly Independent Sets: s {[ ] [, ] [ 3 2, 3 ]} is a linearly set. {[ since ] [ 3 6, 9 [ ]} is a linearly ] is not a multiple of [ ]. set Jiwen He, University of Houston Math 233, Linear Algebra 5 / 2

6 Linearly Independent Sets: s Theorem (4) An indexed set {v, v 2,..., v p } of two or more vectors, with v, is linearly dependent if and only if some vector v j (j > ) is a linear combination of the preceding vectors v,..., v j. Let {p, p 2, p 3 } be a set of vectors in P 2 where p (t) = t, p 2 (t) = t 2, and p 3 (t) = 4t + 2t 2. Is this a linearly dependent set? Solution: Since p 3 = p + p 2, {p, p 2, p 3 } is a linearly set. Jiwen He, University of Houston Math 233, Linear Algebra 6 / 2

7 A Basis Set Let H be the plane illustrated below. Which of the following are valid descriptions of H? (a) H =Span{v, v 2 } (b) H =Span{v, v 3 } (c) H =Span{v 2, v 3 } (d) H =Span{v, v 2, v 3 } A basis set is an efficient spanning set containing no unnecessary vectors. In this case, we would consider the linearly independent sets {v, v 2 } and {v, v 3 } to both be examples of basis sets or bases (plural for basis) for H. Jiwen He, University of Houston Math 233, Linear Algebra 7 / 2

8 A Basis Set: Definition and s A Basis Set Let H be a subspace of a vector space V. An indexed set of vectors β = {b,..., b p } in V is a basis for H if i. β is a linearly independent set, and ii. H = Span{b,..., b p }. Let e =, e 2 =, e 3 =. Show that {e, e 2, e 3 } is a basis for R 3. The set {e, e 2, e 3 } is called a standard basis for R 3. Solutions: (Review the IMT, page 2) Let A = [ ] e e 2 e 3 =. Jiwen He, University of Houston Math 233, Linear Algebra 8 / 2

9 A Basis Set: Definition and s Since A has 3 pivots, the columns of A are linearly and the columns of A therefore, {e, e 2, e 3 } is a basis for R 3. by IMT; Let S = {, t, t 2,..., t n}. Show that S is a basis for P n. by the IMT, Solution: Any polynomial in P n is in span of S.To show that S is linearly independent, assume c + c t + + c n t n =. Then c = c = = c n =. Hence S is a basis for P n. Jiwen He, University of Houston Math 233, Linear Algebra 9 / 2

10 A Basis Set: Let v = 2, v 2 = Is {v, v 2, v 3 } a basis for R 3?, v 3 = Solution: Let A = [v v 2 v 3 ] = By row reduction, 2 5 and since there are 3 pivots, the columns of A are linearly independent and they span R 3 by the IMT. Therefore {v, v 2, v 3 } is a basis for R 3. Jiwen He, University of Houston Math 233, Linear Algebra / 2

11 A Basis Set: Explain why each of the following sets is not a basis for R 3. 4 (a) 2, 5,, (b) 2, Jiwen He, University of Houston Math 233, Linear Algebra / 2

12 Bases for Nul A: Find a basis for Nul A where [ A = ]. Solution: Row reduce [ A ] : [ ] = x = 2x 2 3x 4 33x 5 x 3 = 6x 4 + 5x 5 x 2, x 4 and x 5 are free Jiwen He, University of Houston Math 233, Linear Algebra 2 / 2

13 Bases for Nul A: (cont.) x 2 x x 2 x 3 x 4 x 5 2 u = + x 4 2x 2 3x 4 33x 5 x 2 6x 4 + 5x 5 x v x 5 + x 5 = 33 5 w Therefore {u, v, w} is a spanning set for Nul A. In the last section we observed that this set is linearly independent. Therefore {u, v, w} is a basis for Nul A. The technique used here always provides a linearly independent set. Jiwen He, University of Houston Math 233, Linear Algebra 3 / 2

14 The Spanning Set Theorem A basis can be constructed from a spanning set of vectors by discarding vectors which are linear combinations of preceding vectors in the indexed set. Suppose v = [ ] [, v 2 = ] and v 3 = [ 2 3 ]. Solution: If x is in Span{v, v 2, v 3 }, then x =c v + c 2 v 2 + c 3 v 3 = c v + c 2 v 2 + c 3 ( v + v 2 ) = v + v 2 Therefore, Span{v, v 2, v 3 } =Span{v, v 2 }. Jiwen He, University of Houston Math 233, Linear Algebra 4 / 2

15 The Spanning Set Theorem Theorem (5 The Spanning Set Theorem) Let S = {v,..., v p } be a set in V and let H = Span {v,..., v p }. a. If one of the vectors in S - say v k - is a linear combination of the remaining vectors in S, then the set formed from S by removing v k still spans H. b. If H {}, some subset of S is a basis for H. Jiwen He, University of Houston Math 233, Linear Algebra 5 / 2

16 Bases for Col A: s Find a basis for Col A, where Solution: A = [a a 2 a 3 a 4 ] = Row reduce: [a a 2 a 3 a 4 ] = [b b 2 b 3 b 4 ] Jiwen He, University of Houston Math 233, Linear Algebra 6 / 2

17 Bases for Col A: s (cont.) Note that b 2 = b and a 2 = a b 4 = 4b + 5b 3 and a 4 = 4a + 5a 3 b and b 3 are not multiples of each other a and a 3 are not multiples of each other Elementary row operations on a matrix do not affect the linear dependence relations among the columns of the matrix. Therefore Span {a, a 2, a 3, a 4 } = Span {a, a 3 } and {a, a 3 } is a basis for Col A. Jiwen He, University of Houston Math 233, Linear Algebra 7 / 2

18 Bases for Col A: Theorem and Theorem (6) The pivot columns of a matrix A form a basis for Col A. Let v = 2 3 Span{v, v 2, v 3 }., v 2 = 2 4 6, v 3 = Find a basis for Solution: Let and note that A = ColA = Span {v, v 2, v 3 }. Jiwen He, University of Houston Math 233, Linear Algebra 8 / 2

19 Bases for Col A: Theorem and (cont.) By row reduction, A 2. Therefore a basis for Span{v, v 2, v 3 } is,. Jiwen He, University of Houston Math 233, Linear Algebra 9 / 2

20 Bases for Nul A & Col A: Review Review. To find a basis for Nul A, use elementary row operations to transform [A ] to an equivalent reduced row echelon form [B ]. Use the reduced row echelon form to find parametric form of the general solution to Ax =. The vectors found in this parametric form of the general solution form a basis for Nul A. 2. A basis for Col A is formed from the pivot columns of A. Warning: Use the pivot columns of A, not the pivot columns of B, where B is in reduced echelon form and is row equivalent to A. Jiwen He, University of Houston Math 233, Linear Algebra 2 / 2