Rank and Nullity MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015
Objectives We have defined and studied the important vector spaces associated with matrices (row space, column space, and null space). We will focus today on the relationships between the dimensions of these vector spaces.
Equal Dimensions for Row and Column Spaces Consider the matrix A and its reduced row echelon form R. 1 3 4 2 5 4 A = 2 6 9 1 8 2 2 6 9 1 9 7 1 3 4 2 5 4 R = 1 3 0 14 0 37 0 0 1 3 0 4 0 0 0 0 1 5 0 0 0 0 0 0 We can see from R that the column space and row space of A are both three-dimensional.
General Result Theorem The row space and column space of a matrix A have the same dimension.
General Result Theorem The row space and column space of a matrix A have the same dimension. Proof. Elementary row operations do not change the dimensions of the row space or column space of a matrix.
General Result Theorem The row space and column space of a matrix A have the same dimension. Proof. Elementary row operations do not change the dimensions of the row space or column space of a matrix. If R is the row echelon form of A then dim(row space of A) = dim(row space of R) dim(column space of A) = dim(column space of R).
General Result Theorem The row space and column space of a matrix A have the same dimension. Proof. Elementary row operations do not change the dimensions of the row space or column space of a matrix. If R is the row echelon form of A then dim(row space of A) = dim(row space of R) dim(column space of A) = dim(column space of R). Dimension of row space of R is the number of nonzero rows which equals the number of leading 1 s which is the dimension of the column space.
Rank and Nullity Definition The dimension of the row and column space of A is called the rank of A, denoted rank(a). The dimension of the null space of A is called the nullity of A, denoted nullity(a).
Example, 5 5 Matrix Find the rank and nullity of A = 1 3 1 1 2 0 3 6 0 3 2 3 2 3 4 3 6 0 6 5 2 9 2 4 6 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 = R
Example 4 6 Matrix Find the rank and nullity of B = 6 5 2 5 9 4 8 1 6 6 3 1 6 1 6 7 4 1 1 9 1 3 9 8 1105 1148 375 574 164 153 164 2029 1 0 0 0 1148 101 0 1 0 0 574 259 0 0 1 0 0 0 0 1 1989 574 43 574
Solution (1 of 2) The reduced row echelon form of the matrix equation B x = 0 is equivalent to the system of equations: x 1 = 2029 1148 x 5 1105 1148 x 6 x 2 = 101 574 x 5 375 574 x 6 x 3 = 259 164 x 5 + 153 164 x 6 x 4 = 1989 574 x 5 + 43 574 x 6
Solution (2 of 2) If we treat x 5 and x 6 as parameters we see that the null space of B is spanned by two vectors and hence is two-dimensional. x 1 x 2 x 3 x 4 x 5 x 6 = s 2029 1148 101 574 259 164 1989 574 1 0 + t 1105 1148 375 574 153 164 43 574 0 1
Maximum Value for Rank Question: if A is an m n matrix, what is the maximum possible rank(a)?
Maximum Value for Rank Question: if A is an m n matrix, what is the maximum possible rank(a)? Answer: rows of A lie in R n and the columns of A lie in R m. Maximum dimension of the row space is n and the maximum dimension of the column space is m, thus rank(a) min(m, n).
Dimension Theorem for Matrices Theorem If A is a matrix with n columns, then rank(a) + nullity(a) = n.
Proof Since A has n columns, the linear system A x = 0 has n unknowns. # of leading variables + # of free variables = n
Proof Since A has n columns, the linear system A x = 0 has n unknowns. # of leading variables + # of free variables = n The number of leading variables is the same as the number of leading 1 s in the reduced row echelon form of A. This is the rank(a).
Proof Since A has n columns, the linear system A x = 0 has n unknowns. # of leading variables + # of free variables = n The number of leading variables is the same as the number of leading 1 s in the reduced row echelon form of A. This is the rank(a). The number of free variables is the number of parameters in the solution to A x = 0. This is the nullity(a).
Example Show that the matrix below obeys the Dimension Theorem for Matrices. 6 5 2 5 9 4 B = 8 1 6 6 3 1 6 1 6 7 4 1 1 9 1 3 9 8
Example Show that the matrix below obeys the Dimension Theorem for Matrices. 6 5 2 5 9 4 B = 8 1 6 6 3 1 6 1 6 7 4 1 1 9 1 3 9 8 We saw earlier that rank(b) = 4 and nullity(b) = 2.
Parameters for a Nonhomogeneous System Theorem If A x = b is a consistent linear system of m equations and n unknowns, and if rank(a) = r, then the general solution of A x = b contains n r parameters.
Parameters for a Nonhomogeneous System Theorem If A x = b is a consistent linear system of m equations and n unknowns, and if rank(a) = r, then the general solution of A x = b contains n r parameters. Example If A is a 7 9 matrix of rank 6, then how many free variables are in the general solution of A x = b?
Fundamental Matrix Spaces Suppose A is an m n matrix, then we can discuss the following vector spaces: row space of A column space of A null space of A column space of A T row space of A T null space of A T
Fundamental Matrix Spaces Suppose A is an m n matrix, then we can discuss the following vector spaces: row space of A column space of A null space of A column space of A T row space of A T null space of A T Question: which of these spaces are the same?
Four Fundamental Spaces There are four fundamental matrix spaces: row space of A ( R n ) column space of A ( R m ) null space of A ( R n ) null space of A T ( R m )
Results Theorem If A is a matrix, then rank(a) = rank(a T ).
Results Theorem If A is a matrix, then rank(a) = rank(a T ). Proof. rank(a) = dim(row space of A) = dim(column space of A T ) = rank(a T )
Implications We can express the dimensions of all four of the fundamental spaces in terms of the rank(a). Suppose A is an n m matrix with rank(a) = r
Implications We can express the dimensions of all four of the fundamental spaces in terms of the rank(a). Suppose A is an n m matrix with rank(a) = r nullity(a) = n rank(a) = n r rank(a T ) = rank(a) = r nullity(a T ) = m rank(a) = m r.
Summary Suppose A is an m n matrix of rank r, then Fundamental Space row space of A column space of A null space of A null space of A T Dimension r r n r m r Remark: rank(a) min(m, n)
Equivalent Statements (1 of 2) Theorem If A is an n n matrix, then the following are equivalent. 1. A is invertible. 2. Ax = 0 has only the trivial solution. 3. The reduced row echelon form of A is I n. 4. A is expressible as the product of elementary matrices. 5. Ax = b is consistent for every n 1 matrix b. 6. Ax = b has exactly one solution for every n 1 matrix b. 7. det(a) 0.
Equivalent Statements (2 of 2) Theorem If A is an n n matrix, then in addition to statements (1) (7) the following are equivalent. 8. The column vectors of A are linearly independent. 9. The row vectors of A are linearly independent. 10. The column vectors of A span R n. 11. The row vectors of A span R n. 12. The column vectors of A form a basis for R n. 13. The row vectors of A form a basis for R n. 14. A has rank n. 15. A has nullity 0.
Proof Many of these equivalences have already been proved. Others are homework exercises. Some are proved here. ((2) = (15)) If A x = 0 has only the trivial solution, then there are no parameters in the solution. Thus nullity(a) = 0.
Proof Many of these equivalences have already been proved. Others are homework exercises. Some are proved here. ((2) = (15)) If A x = 0 has only the trivial solution, then there are no parameters in the solution. Thus nullity(a) = 0. ((15) = (14)) If nullity(a) = 0 then rank(a) = n 0 = n.
Proof Many of these equivalences have already been proved. Others are homework exercises. Some are proved here. ((2) = (15)) If A x = 0 has only the trivial solution, then there are no parameters in the solution. Thus nullity(a) = 0. ((15) = (14)) If nullity(a) = 0 then rank(a) = n 0 = n. ((14) = (2)) If rank(a) = n then there are n leading variables and no free variables in the solution of A x = 0. Thus the only solution is the trivial solution.
Overdetermined Systems Remark: If a linear system has more equations than unknowns, it is said to be overdetermined. Consider Ax = b where A is an m n matrix with m > n. rank(a) < m,
Overdetermined Systems Remark: If a linear system has more equations than unknowns, it is said to be overdetermined. Consider Ax = b where A is an m n matrix with m > n. rank(a) < m, column vectors of A cannot span R m,
Overdetermined Systems Remark: If a linear system has more equations than unknowns, it is said to be overdetermined. Consider Ax = b where A is an m n matrix with m > n. rank(a) < m, column vectors of A cannot span R m, Ax = b cannot be consistent for an arbitrary b R m.
Underdetermined Systems Remark: If a linear system has more unknowns than equations, it is said to be underdetermined. Consider Ax = b where A is an m n matrix with m < n. rank(a) < n,
Underdetermined Systems Remark: If a linear system has more unknowns than equations, it is said to be underdetermined. Consider Ax = b where A is an m n matrix with m < n. rank(a) < n, if Ax = b is consistent then the general solution has at least one parameter (free variable),
Underdetermined Systems Remark: If a linear system has more unknowns than equations, it is said to be underdetermined. Consider Ax = b where A is an m n matrix with m < n. rank(a) < n, if Ax = b is consistent then the general solution has at least one parameter (free variable), a consistent underdetermined system has infinitely many solutions.
Example (1 of 2) Question: What can be said about the solutions of an overdetermined system A x = b of 7 equations with 5 unknowns in which rank(a) = 4?
Example (1 of 2) Question: What can be said about the solutions of an overdetermined system A x = b of 7 equations with 5 unknowns in which rank(a) = 4? Answer: there is some vector b R 7 for which the system is consistent. For any such b the number of parameters in the general solution is n rank(a) = 5 4 = 1.
Example (2 of 2) Question: What can be said about the solutions of an underdetermined system A x = b of 5 equations with 7 unknowns in which rank(a) = 4?
Example (2 of 2) Question: What can be said about the solutions of an underdetermined system A x = b of 5 equations with 7 unknowns in which rank(a) = 4? Answer: the system may be consistent or inconsistent. If it is consistent for vector b R 5, then the general solution has a number of parameters equal to n rank(a) = 7 4 = 3.
Homework Read Section 4.8 Exercises: 1, 3, 5, 7, 10, 11, 15, 19, 27