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4.7 Row Space, Column Space, and Null Space

4.7 Row Space, Column Space, and Null Space. Row Space and Column Space. Definition

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4.7 Row Space, Column Space, and Null Space

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  1. 4.7Row Space, Column Space, and Null Space

  2. Row Space and Column Space • Definition • If A is an mn matrix, then the subspace of Rnspanned by the row vectors of A is called the row space (列空間) of A, and the subspace of Rmspanned by the column vectors is called the column space (行空間) of A. • The solution space of the homogeneous system of equation Ax = 0, which is a subspace of Rn, is called the null space (零核空間) of A.

  3. Remarks • In this section we will be concerned with two questions • What relationships exist between the solutions of a linear system Ax=b and the row space, column space, and null space of A. • What relationships exist among the row space, column space, and null space of a matrix.

  4. Remarks • It follows from Formula(10) of Section 1.3 • We conclude that Ax=b is consistent if and only if b is expressible as a linear combination of the column vectors of A or, equivalently, if and only if b is in the column space of A.

  5. Theorem 4.7.1 • Theorem 4.7.1 • A system of linear equations Ax = b is consistent if and only if b is in the column space of A.

  6. Example • Let Ax = b be the linear systemShow 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

  7. General and Particular Solutions • Theorem 4.7.2 • If x0 denotes any single solution of a consistent linear system Ax = b, and if v1, v2, …, vkform a basis for the null space 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 Conversely, for all choices of scalars c1, c2, …, ck, the vector x in this formula is a solution of Ax = b.

  8. Proof of Theorem 4.7.2 • Assume that x0 is any fixed solution of Ax=b and that x is an arbitrary solution. Then Ax0 = b and Ax = b. • Subtracting these equations yields Ax – Ax0 = 0 or A(x-x0)=0 • Which shows that x-x0 is a solution of the homogeneous system Ax = 0. • Since v1, v2, …, vk is a basis for the solution space of this system, we can express x-x0 as a linear combination of these vectors, say x-x0 = c1v1+c2v2+…+ckvk. Thus, x=x0+c1v1+c2v2+…+ckvk.

  9. Proof of Theorem 4.7.2 • Conversely, for all choices of the scalars c1,c2,…,ck, we have Ax = A(x0+c1v1+c2v2+…+ckvk) Ax = Ax0 + c1(Av1) + c2(Av2) + … + ck(Avk) • But x0 is a solution of the nonhomogeneous system, and v1, v2, …, vk are solutions of the homogeneous system, so the last equation implies that Ax = b + 0 + 0 + … + 0 = b • Which shows that x is a solution of Ax = b.

  10. Remark • Remark • The vector x0 is called a particular solution (特解) of Ax = b. • The expression x0 + c1v1 + · · · + ckvkis called the general solution (通解) of Ax = b, the expression c1v1 + · · · + ckvkis called the general solution of Ax = 0. • The general solution of Ax = b is the sum of any particular solution of Ax = b and the general solution of Ax = 0.

  11. The solution to the nonhomogeneous system x1 + 3x2 – 2x3 + 2x5 = 0 2x1 + 6x2 – 5x3 – 2x4 + 4x5 – 3x6 = -1 5x3 + 10x4 + 15x6 = 5 2x1 + 5x2 + 8x4 + 4x5 + 18x6 = 6 is x1 = -3r - 4s - 2t, x2 = r, x3 = -2s, x4 = s, x5 = t, x6 = 1/3 The result can be written in vector form as which is the general solution. The vector x0 is a particular solution of nonhomogeneous system, and the linear combination x is the general solution of the homogeneous system. Example (General Solution of Ax = b)

  12. Elementary Row Operation • Performing an elementary row operation on an augmented matrix does not change the solution set of the corresponding linear system. • It follows that applying an elementary row operation to a matrix A does not change the solution set of the corresponding linear system Ax=0, or stated another way, it does not change the null space of A. • The solution space of the homogeneous system of equation Ax = 0, which is a subspace of Rn, is called the null space of A.

  13. Example • Find a basis for the nullspace of • Solution • The nullspace of A is the solution space of the homogeneous system 2x1 + 2x2 – x3+ x5 = 0 -x1 – x2 – 2 x3 – 3x4 + x5 = 0 x1 + x2 – 2 x3– x5 = 0 x3 + x4 + x5 = 0 • In Example 10 of Section 4.5 we showed that the vectorsform a basis for the nullspace.

  14. Theorems 4.7.3 and 4.7.4 • Theorem 4.7.3 • Elementary row operations do not change the nullspace of a matrix. • Theorem 4.7.4 • Elementary row operations do not change the row space of a matrix.

  15. Proof of Theorem 4.7.4 • Suppose that the row vectors of a matrix A are r1,r2,…,rm, and let B be obtained from A by performing an elementary row operation. (We say that A and B are row equivalent.) • We shall show that every vector in the row space of B is also in that of A, and that every vector in the row space of A is in that of B. • If the row operation is a row interchange, then B and A have the same row vectors and consequently have the same row space.

  16. Proof of Theorem 4.7.4 • If the row operation is multiplication of a row by a nonzero scalar or a multiple of one row to another, then the row vector r1’,r2’,…,rm’ of B are linear combination of r1,r2,…,rm; thus they lie in the row space of A. • Since a vector space is closed under addition and scalar multiplication, all linear combination of r1’,r2’,…,rm’ will also lie in the row space of A. Therefore, each vector in the row space of B is in the row space of A.

  17. Proof of Theorem 4.7.4 • Since B is obtained from A by performing a row operation, A can be obtained from B by performing the inverse operation (Sec. 1.5). • Thus the argument above shows that the row space of A is contained in the row space of B.

  18. Remarks • Do elementary row operations change the column space? • Yes! • The second column is a scalar multiple of the first, so the column space of A consists of all scalar multiplies of the first column vector. • Again, the second column is a scalar multiple of the first, so the column space of B consists of all scalar multiples of the first column vector. This is not the same as the column space of A. Add -2 times the first row to the second

  19. Theorem 4.7.5 • Theorem 4.7.5 • 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.

  20. Bases for Row and Column Spaces

  21. Example • Find bases for the row and column spaces of • Solution: • Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis that of any row-echelon form of A. • Reducing A to row-echelon form we obtain

  22. Example • The basis vectors for the row space of R and A r1 = [1 -3 4 -2 5 4] r2 = [0 0 1 3 -2 -6] r3 = [0 0 0 0 1 5] • Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R.

  23. Theorem 4.7.6 • Theorem 4.7.6 • 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.

  24. Example • We can find the basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A. • Basis for R’s column space • Basis for A’s column space

  25. Example (Basis for a Vector Space Using Row Operations ) • Find a basis for the space spanned by the row vectors v1= (1, -2, 0, 0, 3), v2 = (2, -5, -3, -2, 6), v3 = (0, 5, 15, 10, 0), v4 = (2, 6, 18, 8, 6). • Except for a variation in notation, the space spanned by these vectors is the row space of the matrix • The nonzero row vectors in this matrix are w1= (1, -2, 0, 0, 3), w2 = (0, 1, 3, 2, 0), w3 = (0, 0, 1, 1, 0) • These vectors form a basis for the row space and consequently form a basis for the subspace of R5 spanned by v1, v2, v3, and v4.

  26. Remarks • Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of Adirectly from the column vectors of R. • However, if we can find a set of column vectors of R that forms a basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A. • The basis vectors obtained for the column space of A consisted of column vectors of A, but the basis vectors obtained for the row space of A were not all vectors of A. • Transpose of the matrix can be used to solve this problem.

  27. Find a basis for the row space of consisting entirely of row vectors from A. Solution: The column space of AT are Thus, the basis vectors for the row space of A are r1 = [1 -2 0 0 3] r2 = [2 -5 -3 -2 6] r3 = [2 6 18 8 6] Example (Basis for the Row Space of a Matrix )

  28. Example (Basis and Linear Combinations) • (a) Find a subset of the vectors v1 = (1, -2, 0, 3), v2 = (2, -5, -3, 6), v3 = (0, 1, 3, 0), v4 = (2, -1, 4, -7), v5 = (5, -8, 1, 2) that forms a basis for the space spanned by these vectors. • (b) Express each vector not in the basis as a linear combination of the basis vectors. • Solution (a): • Thus, {v1, v2, v4} is a basis for the column space of the matrix.

  29. Example • Solution (b): • We can express w3 as a linear combination of w1 and w2, express w5 as a linear combination of w1, w2, and w4 (Why?). By inspection, these linear combination are w3 = 2w1 – w2 w5 = w1 + w2 + w4 • We call these the dependency equations. The corresponding relationships in the original vectors are v3 = 2v1 – v2 v5 = v1 + v2 + v4

  30. 4.8Rank, Nullity, and the Fundamental Matrix Spaces

  31. Dimension and Rank • Theorem 4.8.1 • If A is any matrix, then the row space and column space of A have the same dimension. • Proof: Let R be any row-echelon form of A. It follows from Theorem 4.7.4 and 4.7.6b that dim(row space of A) = dim(row space of R). dim(column space of A) = dim(column space of R) • The dimension of the row space of R is the number of nonzero rows = number of leading 1’s = dimension of the column space of R

  32. Rank and Nullity • Definition • The common dimension of the row and column space of a matrix A is called the rank (秩) of A and is denoted by rank(A); the dimension of the nullspace of a is called the nullity (零核維數) of A and is denoted by nullity(A).

  33. Example (Rank and Nullity) • Find the rank and nullity of the matrix • Solution: • The reduced row-echelon form of A is • Since there are two nonzero rows (two leading 1’s), the row space and column space are both two-dimensional, so rank(A) = 2.

  34. Example (Rank and Nullity) • To find the nullity of A, we must find the dimension of the solution space of the linear system Ax=0. • The corresponding system of equations will be x1 – 4x3 – 28x4 – 37x5 + 13x6 = 0 x2 – 2x3 – 12x4 – 16 x5+ 5 x6 = 0 • It follows that the general solution of the system is x1 = 4r + 28s + 37t – 13u, x2 = 2r + 12s + 16t – 5u, x3 = r, x4 = s, x5 = t, x6 = u or • Thus, nullity(A) = 4.

  35. Example • What is the maximum possible rank of an matrix A that is not square? • Solution: The row space of A is at most n-dimensional and the column space is at most m-dimensional. Since the rank of A is the common dimension of its row and column space, it follows that the rank is at most the smaller of m and n.

  36. Theorem 4.8.2 • Theorem 4.8.2 (Dimension Theorem for Matrices) • If A is a matrix with n columns, then rank(A) + nullity(A) = n. • Proof: • Since A has n columns, Ax = 0 has n unknowns. These fall into two categories: the leading variables and the free variables. • The number of leading 1’s in the reduced row-echelon form of A is the rank of A

  37. Theorem 4.8.2 • The number of free variables is equal to the nullity of A. This is so because the nullity of A is the dimension of the solution space of Ax=0, which is the same as the number of parameters in the general solution, which is the same as the number of free variables. Thus rank(A) + nullity(A) = n

  38. Example • This matrix has 6 columns, so rank(A) + nullity(A) = 6 • In previous example, we know rank(A) = 4 and nullity(A) = 2

  39. Theorem 4.8.3 • Theorem 4.8.3 • If A is an mn matrix, then: • rank(A) = Number of leading variables in the solution of Ax = 0. • nullity(A) = Number of parameters in the general solution of Ax = 0.

  40. Example • Find the number of parameters in the general solution of Ax = 0 if A is a 57 matrix of rank 3. • Solution: • nullity(A) = n – rank(A) = 7 – 3 = 4 • Thus, there are four parameters.

  41. Theorem 4.8.4 (Equivalent Statements) • If A is an nn matrix, and if TA : Rn  Rn is multiplication by A, then the following are equivalent: • A is invertible. • Ax = 0 has only the trivial solution. • The reduced row-echelon form of A is In. • A is expressible as a product of elementary matrices. • Ax = b is consistent for every n1 matrix b. • Ax = b has exactly one solution for every n1matrix b. • det(A)≠0. • The column vectors of A are linearly independent. • The row vectors of A are linearly independent. • The column vectors of A span Rn. • The row vectors of A span Rn. • The column vectors of A form a basis for Rn. • The row vectors of A form a basis for Rn. • A has rank n. • A has nullity 0.

  42. Overdetermined System • A linear system with more equations than unknowns is called an overdetermined linear system (超定線性方程組). With fewer unknowns than equations, it’s called an underdetermined system. • Theorem 4.8.5 • If Ax = b is a consistent linear system of m equations in n unknowns, and if A has rank r, then the general solution of the system contains n – r parameters. • If A is a matrix with rank 4, and if Ax=b is a consistent linear system, then the general solution of the system contains 7-4=3 parameters.

  43. Theorem 4.8.6 • Let A be an matrix • (a) (Overdetemined Case) If m> n, then the linear system Ax=b is inconsistent for at least one vector b in Rn. • (b) (Underdetermined Case) If m < n, then for each vector b in Rm the linear system Ax=b is either inconsistent or has infinitely many solutions.

  44. Proof of Theorem 4.8.6 (a) • Assume that m>n, in which case the column vectors of A cannot span Rm (fewer vectors than the dimension of Rm). Thus, there is at least one vector b in Rm that is not in the column space of A, and for that b the system Ax=b is inconsistent by Theorem 4.7.1.

  45. Proof of Theorem 4.8.6 (b) • Assume that m<n. For each vector b in Rn there are two possibilities: either the system Ax=b is consistent or it is inconsistent. • If it is inconsistent, then the proof is complete. • If it is consistent, then Theorem 4.8.5 implies that the general solution has n-r parameters, where r=rank(A). • But rank(A) is the smaller of m and n, so n-r = n-m > 0 • This means that the general solution has at least one parameter and hence there are infinitely many solutions.

  46. Example • What can you say about the solutions of an overdetermined system Ax=b of 7 equations in 5 unknowns in which A has rank = 4? • What can you say about the solutions of an underdetermined system Ax=b of 5 equations in 7 unknowns in which A has rank = 4? • Solution: • (a) the system is consistent for some vector b in R7, and for any such b the number of parameters in the general solution is n-r=5-4=1 • (b) the system may be consistent or inconsistent, but if it is consistent for the vector b in R5, then the general solution has n-r=7-4=3 parameters.

  47. Example • The linear system is overdetermined, so it cannot be consistent for all possible values of b1, b2, b3, b4, and b5. Exact conditions under which the system is consistent can be obtained by solving the linear system by Gauss-Jordan elimination.

  48. Example • Thus, the system is consistent if and only if b1, b2, b3, b4, and b5 satisfy the conditions or, on solving this homogeneous linear system, b1=5r-4s, b2=4r-3s, b3=2r-s, b4=r, b5=s where r and s are arbitrary.

  49. Fundamental Spaces of a Matrix • Six important vector spaces associated with a matrix A • Row space of A, row space of AT • Column space of A, column space of AT • Null space of A, null space of AT • Transposing a matrix converts row vectors into column vectors • Row space of AT = column space of A • Column space of AT = row space of A • These are called the fundamental spaces of a matrix A

  50. Theorem 4.8.7 • if A is any matrix, then rank(A) = rank(AT) • Proof: • Rank(A) = dim(row space of A) = dim(column space of AT) = rank(AT) • If A is an matrix, then rank(A)+nullity(A)=n. rank(AT)+nullity(AT) = m • The dimensions of fundamental spaces

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