Linear Algebra Linearly Independent Sets; Bases

Linear Algebra Linearly Independent Sets; Bases Dr. Bisher M. Iqelan biqelan@iugaza.edu.ps Department of Mathematics The Islamic University of Gaza 27-28, Semester 2 Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 / 2

Definition Let V be a vector space and let { v. v 2,..., v p } V. If the only solution to the equation x v + x 2 v 2 + + x p v p = is the trivial solution x = x 2 = = x p = then { v, v 2,..., v p } is said to be linearly independent. 2 If there is a non-trivial solution to the equation then the set of vectors is said to be linearly dependent. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 2 / 2

The following results from Section.7 are still true for more general vectors spaces. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 3 / 2

The following results from Section.7 are still true for more general vectors spaces. Result A set containing the zero vector is linearly dependent. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 3 / 2

The following results from Section.7 are still true for more general vectors spaces. Result A set containing the zero vector is linearly dependent. Result 2 A set of two vectors is linearly dependent if and only if one is a multiple of the other. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 3 / 2

The following results from Section.7 are still true for more general vectors spaces. Result A set containing the zero vector is linearly dependent. Result 2 A set of two vectors is linearly dependent if and only if one is a multiple of the other. Result 3 A set containing a single non-zero vector is linearly independent. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 3 / 2

Example {[ ] [ 2, 3 4 ] [ 3 2, 3 ]} is a linearly set. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 4 / 2

Example {[ ] [ 2, 3 4 ] [ 3 2, 3 ]} is a linearly set. Example {[ ] [ ]} 2 3 6, is a linearly 3 4 9 [ ] [ 3 6 2 is not a multiple of 9 3 4 ]. set since Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 4 / 2

Theorem Let V be a vector space. Suppose that { v, v 2,... v p } V, p 2, and that v =. Then { v, v 2,..., v p } is linearly dependent if and only if there is a j p so that v j is a linear combination of the vectors v, v 2,..., v j. Example Let S = {p, p 2, p 3 } be a set of vectors in P 3 where p (x) = x 2 + 2x + 3, p 2 (x) = x 3 +, and p 3 (x) = x 3 + 2x 2 + 4x + 7. S P 3. Is it linearly dependent or linearly independent? Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 5 / 2

Example Consider the vector space V = {f : [, ] R so that f is continuous} and let S = {sin(x), cos(x)} V. Is this set linearly dependent or linearly independent? Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 6 / 2

Example Consider the vector space V = {f : [, ] R so that f is continuous} and let S = {sin(x), cos(x)} V. Is this set linearly dependent or linearly independent? What about the set S = {sin(x) cos(x), sin(2x)} V? Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 6 / 2

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. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 7 / 2

Definition Suppose that V is a vector space with W V, and that B = { v, v 2,..., v p } W. This ordered set of vectors is a basis for W provided that B is linearly independent, and 2 Span(B) = W. Example (A standard basis for R 3 ) 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. Solution: (Review the IMT, page 4) A = [e, e 2, e 3 ] =. Since A has 3 pivots, the columns of A are linearly independent by the IMT and the columns of A spans R 3. Therefore, {e, e 2, e 3 } is a basis for R 3. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 8 / 2

Example Let v = 2, v 2 =, v 3 = Solution: Let A = [ v, v 2, v 3 ] = 2 3 2R +R 2 2 3 3 2 3 Is { v, v 2, v 3 } a basis for R 3? R 2+R 3. Using 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. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 9 / 2

Example Let S = {, x, x 2,..., x n }. Show that S is a basis for P n. Solution: Any polynomial in P n is in span of S.To show that S is linearly independent, assume c. + c.x + + c n.x n = Then c = c = = c n =. Hence S is a basis for P n. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 / 2

Example Let S = {, x, x 2,..., x n }. Show that S is a basis for P n. Solution: Any polynomial in P n is in span of S.To show that S is linearly independent, assume c. + c.x + + c n.x n = Then c = c = = c n =. Hence S is a basis for P n. Example Explain why each of thefollowing sets isnot a basis for R 3. 4 (a) 2, 5,, 3 (b) 2, 3 7 7 3 4 5 6 Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 / 2

Bases for Nul A Note We already have seen how to find a basis for Nul A: row reduce A to obtain a matrix in reduced row echelon form and use this to express the null space in vector parametric form. The vectors appearing will be the basis for Nul A. Example ( ) 3 6 6 3 9 Find a basis for Nul A where A = 6 2 3 3 ( ) 3 6 6 3 9 Solution: Row reduce [A ] = 6 2 3 3 ( 2 3 33 6 5 ), x = 2x 2 3x 4 33x 5 x 3 = 6x 4 + 5x 5 x 2, x 4, and x 5 are free Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 / 2

Bases for Nul A Example (Continue...) x 2x 2 3x 4 33x 5 x 2 x 3 = x 2 6x 4 + 5x 5 x 4 x 4 x 5 x 5 where, v = 2, v 2 = 3 6 = x 2. v + x 4. v 2 + x 5. v 3, and v 3 = 33 5 Therefore { v, v 2, v 3 } is a spanning set for Nul A. One can easily show that this set is linearly independent. Therefore { v, v 2, v 3 } is a basis for Nul A. The technique used here always provides a linearly independent set. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 2 / 2

Bases for Nul A Example Let A = 3 2 9 5. Find a basis for Nul A. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 3 / 2

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. Example Let v =, v 2 =, v 3 = 2. Show that Span( v, v 2, v 3 ) = Span( v, v 2 ). 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 ) Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 4 / 2

The Spanning Set Theorem Theorem Let V be a vector space, and S = { v, v 2,..., v p } V, and let H = Span( v, v 2,..., v p ). If there exists k so that v k is a linear combination of the other vectors in S, then H = Span( v, v 2,..., v k, v k+,..., v p ) 2 If H = { } then some subset of S is a basis for H. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 5 / 2

Example (Bases for Col A) Find a basis for Col(A), where A = [ a, a 2, a 3, a 4 ] = Solution: Note that: Row reduce: [ a, a 2, a 3, a 4 ] = 2 4 5 b2 = b and a 2 = a b4 = 4 b + 5 b3 and a 4 = 4 a + 5 a 3 2 4 2 4 3 3 6 2 22 4 8 6 = [ b, b2, b3, b4 ] Also, b and b3 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. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 6 / 2

Bases for Col A Example Again, let A = 3 2 9 5. Find a basis for Col A. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 7 / 2

Review Note 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 A x =. 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. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 8 / 2

Note A basis is A spanning set which is as small as possible A linearly independent set which is as big as possible Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 9 / 2

Note A basis is A spanning set which is as small as possible A linearly independent set which is as big as possible Example Which of the following sets of vectors form a basis for R 3., 2 3 2, 2 3, 3, 2 3,, Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 9 / 2

H.W HW Sec. 4.2, p25: # 3,4,9,,3,5,23,24. Dr. Bisher M. Iqelan (IUG) Sec.3.2: Lin. Indep. Sets; Bases 2nd Semester, 27-28 2 / 2