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5.4 Basis And Dimension

5.4 Basis And Dimension. Nonrectangular Coordinate Systems (1/4).

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5.4 Basis And Dimension

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  1. 5.4 Basis And Dimension

  2. Nonrectangular Coordinate Systems (1/4) • We describe this by saying that the coordinate system establishes a one-to-one correspondence between points in the plane and ordered pairs of real numbers. Although perpendicular coordinate axes are the most common, any two nonparallel lines can be used to define a coordinate system in the plane.

  3. Nonrectangular Coordinate Systems (2/4) • Our first objective in this section is to extend the concept of a coordinate system to general vector. In Figure 5.4.2a, for example, v1 and v2 are vectors of length 1 that points in the positive direction of the axis. As illustrated in the figure Similarly, the coordinates (a, b, c) of the point P in Figure 5.4.1c can be obtained by expressing as a linear combination of the vectors shown in Figure 5.4.2b.

  4. Nonrectangular Coordinate Systems (3/4) • Informally stated, vectors that specify a coordinate system are called “basis vectors” for that system. Although we used basis vectors of length 1 in the preceding discussion, we shall see in a moment that this is not essential – nonzero vectors of any length will suffice.

  5. Nonrectangular Coordinate Systems (4/4)

  6. Definition • If V is any vector space and S={v1, v2, …,vn} is a set of vectors in V, then S is called a basis for V if the following two conditions hold: • S is linearly independent. • S spans V.

  7. Theorem 5.4.1 Uniqueness of Basis Representation • If S={v1, v2, …,vn} is a basis for a vector space V, then every vector v in V can be expressed in the form v=c1v1+c2v2+…+cnvn in exactly one way.

  8. Coordinates Relative to a Basis • If S={v1, v2, …,vn} is a basis for a vector space V, and v=c1v1+c2v2+…+cnvn is the expression for a vector v in terms of the basis S, then the scalars c1, c2, …, cn, are called the coordinates of v relative to the basis S. The vector (c1, c2, …, cn) in Rn constructed from these coordinates is called the coordinate vector of v relative to S; it is denoted by (v)s=(c1, c2, …, cn)

  9. Remark • It should be noted that coordinate vectors depend not only on the basis S but also on the order in which the basis vectors are written; a change in the order of the basis vectors results in a corresponding change of order for the entries in the coordinate vector.

  10. Example 1Standard Basis for R3 • Suppose i=(1, 0, 0), j=(0, 1, 0),and k=(0, 0, 1), S={i, j, k} is a linearly independent set in R3. This set also spans R3 since any vector v=(a, b, c) in R3 can be written as v=(a, b, c)=a(1, 0, 0)+b(0, 1, 0)+c(0, 0, 1)=ai+bj+ck (1) Thus, S is a basis for R3; it is called the standard basis for R3.Looking at the coefficients of i, j, and k in (1), if follows that the coordinates of v relative to the standard basis are a, b, and c, so (v)s=(a, b, c) Comparing this result to (1) we see that v=(v)s

  11. Example 2Standard Basis for Rn • If e1=(1, 0, 0, …, 0), e2=(0, 1, 0, …, 0), …, en=(0, 0, 0, …, 1) then S={e1, e2, …, en} is a linearly independent set in Rn. Moreover, this set also spans Rn since any vector v=(v1, v2, …, vn) in Rn can be written as V=v1e1+v2e2+…+vnen (2) Thus, S is a basis for Rn; it is called the standard basis for Rn. It follows from (2) that the coordinates of v=(v1, v2, …, vn) relative to the standard basis are v1, v2, …, vn, so (v)s=(v1, v2, …, vn) As in Example 1, we have v= (v)s, so a vector v and its coordinate vector relative to the standard basis for Rn are the same.

  12. Example 3Demonstrating That a Set of Vectors Is a Basis (1/3) • Let v1=(1, 2, 1), v2=(2, 9, 0), and v3=(3, 3, 4). Show that the set S={v1, v2, v3} is a basis for R3. Solution. To show that the set S spans R3, we must show that an arbitrary vector b=(b1, b2, b3) can be expressed as a linear combination b=c1v1+c2v2+c3v3 of the vectors in S. Expressing this equation in terms of components gives (b1, b2, b3)=c1(1, 2, 1)+c2(2, 9, 0)+c3(3, 3, 4) or, on equating corresponding components, c1+2c2+3c3=b1 2c1+9c2+3c3=b2 (3) c1 +4c3=b3

  13. Example 3Demonstrating That a Set of Vectors Is a Basis (2/3) Thus, to show that S spans R3, we must demonstrate that system (3) has a solution for all choices of b=(b1, b2, b3). To prove that S is linearly dependent, we must show that the only solution of c1v1+c2v2+c3v3=0 (4) is c1=c2=c3=0. As above, if (4) is expressed in terms of components, the verification of independence reduces to showing that the homogeneous system c1+2c2+3c3=0 2c1+9c2+3c3=0(5) c1 +4c3=0 has only the trivial solution. Observe that system (3) and (5) have the same coefficient matrix.

  14. Example 3Demonstrating That a Set of Vectors Is a Basis (3/3) Thus, by parts (b), (e), and (g) of Theorem 4.3.4, we can simultaneously prove that S is linearly independent and spans R3 by demonstrating that in systems (3) and (5) the matrix of coefficients has a nonzero determinant. From and so S is a basis for R3.

  15. Example 4Representing a Vector Using Two Bases (1/2) • Let S={v1, v2, v3} be the basis for R3 in the preceding example. • Find the coordinate vector of v=(5, -1, 9) with respect to S. • Find the vector v in R3 whose coordinate vector with respect to the basis S is (v)s=(-1, 3, 2). Solution (a). We must find scalars c1, c2, c3 such that v=c1v1+c2v2+c3v3 or, in terms of components, (5, -1, 9)=c1(1, 2, 1)+c2(2, 9, 0)+c3(3, 3, 4) Solving this, we obtaining c1=1, c2=-1, c3=2. Therefore, (v)s=(-1, 3, 2).

  16. Example 4Representing a Vector Using Two Bases (2/2) Solution (b). Using the definition of the coordinate vector (v)s, we obtain v=(-1)v1+3v2+2v3 =(11, 31, 7).

  17. Example 5Standard Basis for Pn • Show that S={1, x, x2, …, xn} is a basis for the vector space Pn of polynomials of the form a0+a1x+…+anxn. • Find the coordinate vector of the polynomial p=a0+a1x+a2x2 relative to the basis S={1, x, x2}for P2. Solution (a). We showed that S spans Pn in Example11 of Section5.2, and we showed that S is a linearly independent set in Example 5.3. Thus, S is a basis for Pn; it is called the standard basis for Pn. Solution (b). The coordinates of p=a0+a1x+a2x2 are the scalar coefficients of the basis vectors 1, x, and x2, so (p)s=(a0, a1, a2).

  18. Example 6Standard Basis for Mmn (1/2) • Let The set S={M1, M2, M3, M4} is a basis for the vector space M22 of 2×2 matrices. To see that S spans M22, note that an arbitrary vector (matrix) can be written as To see that S is linearly independent, assume aM1+bM2+cM3+dM4=0

  19. Example 6Standard Basis for Mmn (2/2) It follows that Thus, a=b=c=d=0, so S is linearly independent. The basis S in this example is called the standard basis for M22. More generally, the standard basis for Mmnconsists of the mn different matrices with a single 1 and zeros for the remaining entries.

  20. Example 7Basis for the Subspace span(S) • If S={v1, v2, …,vn} is a linearly independent set in a vector space V, then S is a basis for the subspace span(S) since the set S span span(S) by definition of span(S).

  21. Definition • A nonzero vector V is called finite-dimensional if it contains a finite set of vector {v1, v2, …,vn} that forms a basis. If no such set exists, V is called infinite-dimensional. In addition, we shall regard the zero vector space to be finite-dimensional.

  22. Example 8Some Finite- and Infinite- Dimensional Spaces • By Example 2, 5, and 6, the vector spaces Rn, Pn, and Mmn are finite-dimensional. The vector spaces F(-∞, ∞), C(-∞, ∞), Cm(-∞, ∞), and C∞(-∞, ∞) are infinite-dimensional.

  23. Theorem 5.4.2 • Let V be a finite-dimensional vector space and {v1, v2, …,vn} any basis. • If a set has more than n vector, then it is linearly dependent. • If a set has fewer than n vector, then it does not span V.

  24. Theorem 5.4.3 • All bases for a finite-dimensional vector space have the same number of vectors.

  25. Definition • The dimension of a finite-dimensional vector space V, denoted by dim(V), is defined to be the number of vectors in a basis for V. In addition, we define the zero vector space to have dimension zero.

  26. Remark • From here on we shall follow a common convention of regarding the empty set to be a basis for the zero vector space. This is consistent with the preceding definition, since the empty set has no vectors and the zero space has dimension zero.

  27. Example 9Dimensions of Some Vector Spaces dim(Rn)=n [The standard basis has n vectors (Example 2).] dim(Pn)=n+1 [The standard basis has n+1 vectors (Example 5).] dim(Mmn)=mn [The standard basis has mn vectors (Example 6).]

  28. Example 10Dimension of a Solution Space (1/2) • Determine a basis for and the dimension of the solution space of the homogeneous system 2x1+2x2- x3 +x5=0 -x1+ x2+2x3 -3x4+x5=0 x1+ x2- 2x3 -x5=0 x3 + x4+ x5=0 In Example 7 of Section 1.2 it was shown that the general solution of the given system is x1=-s-t, x2=s, x3=-t, x4=0, x5=t Therefore, the solution vectors can be written as

  29. Example 10Dimension of a Solution Space (2/2) span the solution space. Since they are also linearly independent, {v1, v2} is a basis , and the solution space is two-dimensional.

  30. Theorem 5.4.4Plus/Minus Theorem • Let S be a nonempty set of vectors in a vector space V. • If S is a linearly independent set, and if v is a vector in V that is outside of span(S), then the set S∪{v} that results by inserting v into S is still linearly independent. • If v is a vector in S that is expressible as a linear combination of other vectors in S, and if S-{v} denotes the set obtained by removing v from S, then S and S-{v} span the same space; that is , span(S)=span(S-{v})

  31. Theorem 5.4.5 • If V is an n-dimensional vector space, and if S is a set in V with exactly n vectors, then S is a basis for V if either S spans V or S is linearly independent.

  32. Example 11Checking for a Basis • Show that v1=(-3, 7) and v2=(5, 5) form a basis for R2 by inspection. • Show that v1=(2, 0, 1) , v2=(4, 0, 7), v3=(-1, 1, 4) form a basis for R3 by inspection. Solution (a). Since neither vector is a scalar multiple of the other, the two vectors form a linear independent set in the two-dimensional space R2, and hence form a basis by Theorem 5.4.5. Solution (b). The vectors v1 and v2 form a linearly independent set in the xy-plane. The vector v3 is outside of the xy-plane, so the set {v1, v2 , v3} is also linearly independent. Since R3 is three- dimensional, Theorem 5.4.5 implies that {v1, v2 , v3} is a basis for R3.

  33. Theorem 5.4.6 • Let S be finite set of vectors in a finite-dimensional vector space V. • If S spans V but is not a basis for V, then S can be reduced to a basis for V by removing appropriate vectors from S. • If S is a linearly independent set that is not already a basis for V, then S can be enlarged to a basis for V by inserting appropriate vectors into S.

  34. Theorem 5.4.7 • If W is a subspace of a finite-dimensional vector space V, then dim(W)≦dim(V); moreover, if dim(W)=dim(V), then W=V.

  35. 5.5 Row Space, Column, Space, And Nullspace

  36. Definition

  37. Example 1Row and Column Vectors in a 2×3 Matrix • Let The row vectors of A are and the column vectors of A are

  38. Definition • If A is an m×n matrix, then the subspace of Rn spanned by the row vectors of A is called the row space of A, and the subspace of Rm spanned by the column vectors is called the column space of A. The solution space of the homogeneous system of equation Ax=0, which is a subspace of Rn, is called the nullsapce of A.

  39. Theorem 5.5.1 • A system of linear equations Ax=b is consistent if and only if b is in the column space of A.

  40. Example 2A Vector b in the Column Space of A • Let Ax=b be the linear system Show that b is in the column space of A, and express b as a linear combination of the column vectors of A. Solution. Solving the system by Gaussian elimination yields x1=2, x2=-1, x3=3 Since the system is consistent, b is in the column space of A. Moreover, it follows that

  41. Theorem 5.5.2 • If x0 denotes any single solution of a consistent linear system Ax=b, and if v1, v2, …, vk form a basis for the nullspace of A, that is , the solution space of the homogeneous system Ax=0, then every solution of Ax=b can be expressed in the form x= x0+c1v1+c2v2+…+ckvk (3) And, conversely, for all choices of scalars c1, c2, …, ck the vector x in this formula is a solution of Ax=b.

  42. General and Particular Solutions • There is some terminology associated with Formula (3). The vector x0 is called a particular solution of Ax=b. The expression x0+c1v1+c2v2+…+ckvk is called the general solution of Ax=b, and the expression is called the general solution of Ax=0. With this terminology, Formula (3) states, the general solution of Ax=b is the sum of any particular solution of Ax=b and the general solution of Ax=0.

  43. Example 3General Solution of a Linear System Ax=b (1/2)

  44. Example 3General Solution of a Linear System Ax=b (2/2)

  45. Theorem 5.5.3 • Elementary row operations do not change the nullspace of a matrix.

  46. Example 4 Basis for Nullspace • Find a basis for the nullspace of Solution. The nullspace of A is the solution space of the homogeneous system In Example 10 of Section 5.4 we showed that the vectors Form a basis for this space.

  47. Theorem 5.5.4 • Elementary row operations do not change the row space of a matrix.

  48. Theorem 5.5.5 • If A and B are row equivalent matrices, then: • A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. • A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B.

  49. Theorem 5.5.6 • If a matrix R is in row echelon form, then the row vectors with the leading 1’s (i.e., the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1’s of the row vectors form a basis for the column space of R.

  50. Example 5Bases for Row and Column Spaces

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