Lecture 21: 5.6 Rank and Nullity Wei-Ta Chu 2008/12/5 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). 2008/12/5 Elementary Linear Algebra 2 Example (Rank and Nullity) Find the rank and nullity of the matrix 1 2 0 4 5 3 3 7 2 0 1 4 A 2 5 2 4 6 1 4 9 2 4 4 7 Solution: The reduced row-echelon form of A is 1 0 4 28 37 13 0 1 2 12 16 5 0 0 0 0 0 0 0 0 0 0 0 0 Since there are two nonzero rows (two leading 1’s), the row space and column space are both two-dimensional, so rank(A) = 2. 2008/12/5 Elementary Linear Algebra 3 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 x1 4 28 37 13 x 2 12 16 5 2 x3 1 0 0 0 Thus, nullity(A) = 4. r s t u x4 0 1 0 0 x 0 0 1 0 5 x6 0 0 0 1 2008/12/5 Elementary Linear Algebra 4 Theorems Theorem 5.6.2 T If A is any matrix, then rank(A) = rank(A ). T Proof: rank(A) = dim(row space of A) = dim(column space of A ) = rank(AT) Theorem 5.6.3 (Dimension Theorem for Matrices) If A is a matrix with n columns, then rank(A) + nullity(A) = n. 2008/12/5 Elementary Linear Algebra 5 Proof of Theorem 5.6.3 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 2008/12/5 Elementary Linear Algebra 6 Proof of Theorem 5.6.3 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 2008/12/5 Elementary Linear Algebra 7 Theorem 5.6.4 Theorem 5.6.4 If A is an mn 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. 2008/12/5 Elementary Linear Algebra 8 Example (Sum of Rank and Nullity) The matrix 1 2 0 4 5 3 3 7 2 0 1 4 A 2 5 2 4 6 1 4 9 2 4 4 7 has 6 columns, so rank(A) + nullity(A) = 6 This is consistent with the previous example, where we showed that rank(A) = 2 and nullity(A) = 4 2008/12/5 Elementary Linear Algebra 9 Example Find the number of parameters in the general solution of Ax = 0 if A is a 57 matrix of rank 3. Solution: nullity(A) = n –rank(A) = 7 –3 = 4 Thus, there are four parameters. 2008/12/5 Elementary Linear Algebra 10 Dimensions of Fundamental Spaces Suppose that A is an mn matrix of rank r, then T A is an nm matrix of rank r by Theorem 5.6.2 T nullity(A) = n –r, nullity(A ) = m –r by Theorem 5.6.3 Fundamental Space Dimension Row space of A r Column space of A r Nullspace of A n –r Nullspace of AT m –r 2008/12/5 Elementary Linear Algebra 11 Maximum Value for Rank If A is an mn matrix The row vectors lie in Rn and the column vectors lie in Rm. The row space of A is at most n-dimensional and the column space is at most m-dimensional. Since the row and column space have the same dimension (the rank A), we must conclude that if m n, then the rank of A is at most the smaller of the values of m or n. That is, rank(A) min(m, n) 2008/12/5 Elementary Linear Algebra 12 Example If A is a 74 matrix, then the rank of A is at most 4 and, consequently, the seven row vectors must be linearly dependent. If A is a 47 matrix, then again the rank of A is at most 4 and, consequently, the seven column vectors must be linearly dependent. 2008/12/5 Elementary Linear Algebra 13 Theorem 5.6.5 Theorem 5.6.5 (The Consistency Theorem) If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent. Ax = b is consistent. b is in the column space of A. The coefficient matrix A and the augmented matrix [A | b] have the same rank. 2008/12/5 Elementary Linear Algebra 14 Proof of Theorem 5.6.5 (b) → (c) If b is in the column space of A, then the column spaces of A and [A | b] are actually the same, from which it will follow that these two matrices have the same rank. The column spaces of A and [A | b] can be expressed as span{c1, c2, …, cn} and span{c1, c2, …, cn, b} If b is in the column space of A, then each vector in the set {c1,c2,…,cn,b} is a linear combination of the vectors in {c1,c2,…,cn}. Thus, from Theorem 5.2.4, the column spaces of A and [A | b] are the same. 2008/12/5 Elementary Linear Algebra 15 Proof of Theorem 5.6.5 (c) → (b) Assume that A and [A | b] have the same rank r. By Theorem 5.4.6a, there is some subset of the column vectors of A that forms a basis for the column space of A. Suppose that those column vectors are c’1, c’2,…,c’r These r basis vectors also belong to the r-dimensional column space of [A | b]; hence they also form a basis for the column space of [A | b] by Theorem 5.4.6a. This means that b is expressible as a linear combination of c’1, c’2,…,c’r, and consequently b lies in the column space of A. 2008/12/5 Elementary Linear Algebra 16 Example We see from the third row that the system is inconsistent. However, it is also because of this row that the reduced row- echelon form of the augmented matrix has fewer zero rows than the reduced row-echelon form of the coefficient matrix. This forces the coefficient matrix and the augmented matrix for the system to have different ranks. 2008/12/5 Elementary Linear Algebra 17 Theorem 5.6.6 Theorem 5.6.6 If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent. Ax = b is consistent for every m1 matrix b. The column vectors of A span Rm. rank(A) = m. 2008/12/5 Elementary Linear Algebra 18 Proof of Theorem 5.6.6 : The system Ax = b can be expressed as from which we can conclude that Ax=b is consistent for every matrix b if and only if every such b is expressible as a linear combination of the column vectors c1, c2, …, cn, or, equivalently, if and only if these column vectors span Rm. 2008/12/5 Elementary Linear Algebra 19 Proof of Theorem 5.6.6 : From the assumption that Ax=b is consistent for every matrix b, and from (a) and (b) of the Consistency Theorem (5.6.5), it follows that every vector b in Rm lies in the column space of A; that is, the column space of A is all of Rm. Thus rank(A)=dim(Rm)=m. 2008/12/5 Elementary Linear Algebra 20 Proof of Theorem 5.6.6 : From the assumption that rank(A)=m, it follows that the column space of A is a subspace of Rm of dimension m and hence must be all of Rm by Theorem 5.4.7. It now follows from parts (a) and (b) of the Consistency Theorem (5.6.5) that Ax=b is consistent from ever vector b in Rm, since every such b is in the column space of A. 2008/12/5 Elementary Linear Algebra 21 Overdetermined System A linear system with more equations than unknowns is called an overdetermined linear system (超定線性方程 組). If Ax = b is an overdetermined linear system of m equations in n unknowns (so that m > n), then the column vectors of A cannot span Rm. Thus, the overdetermined linear system Ax = b cannot be consistent for every possible b. 2008/12/5 Elementary Linear Algebra 22 Example x12 x 2 b 1 x1 x 2 b 2 The linear system x1 x 2 b 3 x12 x 2 b 4 x13 x 2 b 5 is overdetermined, so it cannot be consistent for all possible values of b1, b2, b3, b4, and b5.