Lecture 21: 5.6 Rank and Nullity
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1 Lecture 21: 5.6 Rank and Nullity Wei-Ta Chu 2008/12/5
2 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
3 Example (Rank and Nullity) Find the rank and nullity of the matrix A 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) = /12/5 Elementary Linear Algebra 3
4 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 x 1 4x 3 28x 4 37x x 6 = 0 x 2 2x 3 12x 4 16 x x 6 = 0 It follows that the general solution of the system is x 1 = 4r + 28s + 37t 13u, x 2 = 2r + 12s + 16t 5u, x 3 = r, x 4 = s, x 5 = t, x 6 = u or x x x r s t u x x x Thus, nullity(a) = /12/5 Elementary Linear Algebra 4
5 Theorems Theorem If A is any matrix, then rank(a) = rank(a T ). Proof: rank(a) = dim(row space of A) = dim(column space of A T ) = rank(a T ) Theorem (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
6 Proof of Theorem 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
7 Proof of Theorem 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
8 Theorem Theorem 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 = /12/5 Elementary Linear Algebra 8
9 Example (Sum of Rank and Nullity) The matrix has 6 columns, so A rank(a) + nullity(a) = 6 This is consistent with the previous example, where we showed that rank(a) = 2 and nullity(a) = /12/5 Elementary Linear Algebra 9
10 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
11 Dimensions of Fundamental Spaces Suppose that A is an mn matrix of rank r, then A T is an nm matrix of rank r by Theorem nullity(a) = n r, nullity(a T ) = m r by Theorem Fundamental Space Row space of A Column space of A Nullspace of A Nullspace of A T Dimension r r n r m r 2008/12/5 Elementary Linear Algebra 11
12 Maximum Value for Rank If A is an mn matrix The row vectors lie in R n and the column vectors lie in R m. 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
13 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
14 Theorem Theorem (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
15 Proof of Theorem (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{c 1, c 2,, c n } and span{c 1, c 2,, c n, b} If b is in the column space of A, then each vector in the set {c 1,c 2,,c n,b} is a linear combination of the vectors in {c 1,c 2,,c n }. Thus, from Theorem 5.2.4, the column spaces of A and [A b] are the same. 2008/12/5 Elementary Linear Algebra 15
16 Proof of Theorem (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
17 Example We see from the third row that the system is inconsistent. However, it is also because of this row that the reduced rowechelon 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
18 Theorem Theorem 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 R m. rank(a) = m. 2008/12/5 Elementary Linear Algebra 18
19 Proof of Theorem : 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 c 1, c 2,, c n, or, equivalently, if and only if these column vectors span R m. 2008/12/5 Elementary Linear Algebra 19
20 Proof of Theorem : 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 R m lies in the column space of A; that is, the column space of A is all of R m. Thus rank(a)=dim(r m )=m. 2008/12/5 Elementary Linear Algebra 20
21 Proof of Theorem : From the assumption that rank(a)=m, it follows that the column space of A is a subspace of R m of dimension m and hence must be all of R m by Theorem 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 R m, since every such b is in the column space of A. 2008/12/5 Elementary Linear Algebra 21
22 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 R m. Thus, the overdetermined linear system Ax = b cannot be consistent for every possible b. 2008/12/5 Elementary Linear Algebra 22
23 Example The linear system x 2 x b x x b x x b x 2 x b x 3 x b is overdetermined, so it cannot be consistent for all possible values of b 1, b 2, b 3, b 4, and b 5. Exact conditions under which the system is consistent can be obtained by solving the linear system by Gauss-Jordan elimination b2 b b2 b b3 3 b2 2 b b4 4 b2 3 b1 0 0 b5 5 b2 4 b /12/5 Elementary Linear Algebra 23
24 Example Thus, the system is consistent if and only if b 1, b 2, b 3, b 4, and b 5 satisfy the conditions 2b 3 b b = b 4 b b = b 5 b b = or, on solving this homogeneous linear system, b 1 =5r-4s, b 2 =4r-3s, b 3 =2r-s, b 4 =r, b 5 =s where r and s are arbitrary. 2008/12/5 Elementary Linear Algebra 24
25 Theorem Theorem 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. 2008/12/5 Elementary Linear Algebra 25
26 Theorem Theorem If A is an mn matrix, then the following are equivalent. Ax = 0 has only the trivial solution. The column vectors of A are linearly independent. Ax = b has at most one solution (0 or 1) for every m1 matrix b. 2008/12/5 Elementary Linear Algebra 26
27 Proof of Theorem : If c 1, c 2,, c n are the column vectors of A, then the linear system Ax=0 can be written as If c 1, c 2,, c n are linearly independent vectors, then this equation is satisfied only by x 1 =x 2 = =x n =0, which means that Ax=0 has only the trivial solution. Conversely, if Ax=0 has only the trivial solution, then it is satisfied only by x 1 =x 2 = =x n =0, which means that c 1, c 2,, c n are linearly independent. 2008/12/5 Elementary Linear Algebra 27
28 Proof of Theorem : Assume that Ax=0 has only the trivial solution. Either Ax=b is consistent or it is not. If it is not consistent, then there are no solutions of Ax=b, and we are done. If Ax=b is consistent, let x 0 by any solution. From Theorem 5.5.2, we conclude that the general solution of Ax=b is x 0 +0=x 0. Thus the only solution of Ax=b is x /12/5 Elementary Linear Algebra 28
29 Proof of Theorem : Assume that Ax=b has at most one solution for every matrix b. Then, in particular, Ax=0 has at most one solution. Thus Ax=0 has only trivial solution. 2008/12/5 Elementary Linear Algebra 29
30 Example A linear system with more unknowns than equations is called an underdetermined linear system. A consistent underdetermined linear system must have infinitely many solutions. An Undetermined System If A is a 57 matrix, then for every 71 matrix b, the linear system Ax = b is undetermined. Thus, Ax = b must be consistent for some b, and for each such b the general solution must have 7 r parameters, where r is the rank of A. 2008/12/5 Elementary Linear Algebra 30
31 Theorem (Equivalent Statements) If A is an nn matrix, and if T A : R n R n 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 I n. A is expressible as a product of elementary matrices. Ax = b is consistent for every n1 matrix b. Ax = b has exactly one solution for every n1 matrix b. det(a) 0. The range of T A is R n. T A is one-to-one. The column vectors of A are linearly independent. The row vectors of A are linearly independent. The column vectors of A span R n. The row vectors of A span R n. The column vectors of A form a basis for R n. The row vectors of A form a basis for R n. A has rank n. A has nullity /12/5 Elementary Linear Algebra 31
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