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. Exact conditions under which the system is consistent can be obtained by solving the linear system by Gauss-Jordan elimination.
1 0 2b2 b 1 0 1 b b 2 1
0 0b3 3 b 2 2 b 1 0 0b4 4 b 2 3 b 1 0 0b5 5 b 2 4 b 1
2008/12/5 Elementary Linear Algebra 23 Example
Thus, the system is consistent if and only if b1, b2, b3, b4, and b5 satisfy the conditions
2b1 3 b 2 b 3 =0
2b1 4 b 2 b 4 =0
4b1 5 b 2 b 5 =0
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.
2008/12/5 Elementary Linear Algebra 24 Theorem 5.6.7
Theorem 5.6.7
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 isa matrixwithrank4,andif 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 Theorem 5.6.8
Theorem 5.6.8
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 Proof of Theorem 5.6.8
: If c1, c2, …, cn are the column vectors of A, then the linear system Ax=0 can be written as
If c1, c2, …, cn are linearly independent vectors, then this equation is satisfied only by x1=x2=…=xn=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 x1=x2=…=xn=0, which means that c1, c2, …, cn are linearly independent.
2008/12/5 Elementary Linear Algebra 27 Proof of Theorem 5.6.8
: 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 x0 by any solution. From Theorem 5.5.2, we conclude that the general solution of
Ax=b is x0+0=x0. Thus the only solution of Ax=b is x0.
2008/12/5 Elementary Linear Algebra 28 Proof of Theorem 5.6.8
: 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 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 Theorem 5.6.9 (Equivalent Statements) n n If A is an nn matrix, and if TA : R R 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 n1 matrix b. Ax = b has exactly one solution for every n1 matrix b. det(A)≠0. n The range of TA is R . TA is one-to-one. The column vectors of A are linearly independent. The row vectors of A are linearly independent. n 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 . A has rank n. A has nullity 0.
2008/12/5 Elementary Linear Algebra 31