Row Space, Column Space, and Nullspace MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015
Introduction Every matrix has associated with it three vector spaces: row space column space nullspace
Definitions Definition If A is an m n matrix, the subspace of R n spanned by the row vectors of A is called the row space of A. The subspace of R m spanned by the column vectors of A is called the column space of A. The vector space of solutions to Ax = 0, which is a subspace of R n is called the nullspace of A. Questions: What relationships exist between the row space, column space, and nullspace? What relationships exist between the solutions of Ax = b and the row space, column space, and nullspace?
Definitions Definition If A is an m n matrix, the subspace of R n spanned by the row vectors of A is called the row space of A. The subspace of R m spanned by the column vectors of A is called the column space of A. The vector space of solutions to Ax = 0, which is a subspace of R n is called the nullspace of A. Questions: What relationships exist between the row space, column space, and nullspace? What relationships exist between the solutions of Ax = b and the row space, column space, and nullspace?
Column Space If A = a 11 a 12 a 1n a 21 a 22 a 2n... a m1 a m2 a mn Ax = x 1 a 11 a 21. a m1 + x 2 and x = a 12 a 22. a m2 x 1 x 2. x n then + + x n a 1n a 2n. a mn which is a linear combination of the columns of A.
Results A system of linear equations Ax = b is consistent if and only if b lies in the column space of A. Example Determine if the following system of equations is consistent. 2 1 3 1 0 1 1 1 2 x 1 x 2 x 3 = 3 2 5
Results A system of linear equations Ax = b is consistent if and only if b lies in the column space of A. Example Determine if the following system of equations is consistent. 2 1 3 1 0 1 1 1 2 x 1 x 2 x 3 = 3 2 5
Results (continued) If x 0 is any solution to a consistent nonhomogeneous linear system Ax = b and if {v 1, v 2,..., v k } form a basis for the nullspace of A, then every solution of Ax = b can be expressed as x = x 0 + c 1 v 1 + c 2 v 2 + + c k v k and conversely for all scalars c 1, c 2,..., c k, the vector x above is a solution to Ax = b. Proof.
Results (continued) If x 0 is any solution to a consistent nonhomogeneous linear system Ax = b and if {v 1, v 2,..., v k } form a basis for the nullspace of A, then every solution of Ax = b can be expressed as x = x 0 + c 1 v 1 + c 2 v 2 + + c k v k and conversely for all scalars c 1, c 2,..., c k, the vector x above is a solution to Ax = b. Proof.
General and Particular Solutions Definition The vector x 0 described previously is called a particular solution to Ax = b. The vector c 1 v 1 + c 2 v 2 + + c k v k is called a general solution to Ax = 0. The general solution to Ax = b is is the sum of any particular solution to Ax = b and the general solution to Ax = 0.
Basis for the Nullspace Recall: elementary row operations do not change the solution space of a homogeneous linear system Ax = 0. Elementary row operations do not change the nullspace of a matrix. Example Find a basis for the nullspace of A where A = 2 0 1 1 2 0 4 0 2 1 0 1 2 1 0 1 4 0 0 0.
Basis for the Nullspace Recall: elementary row operations do not change the solution space of a homogeneous linear system Ax = 0. Elementary row operations do not change the nullspace of a matrix. Example Find a basis for the nullspace of A where A = 2 0 1 1 2 0 4 0 2 1 0 1 2 1 0 1 4 0 0 0.
Basis for the Nullspace Recall: elementary row operations do not change the solution space of a homogeneous linear system Ax = 0. Elementary row operations do not change the nullspace of a matrix. Example Find a basis for the nullspace of A where A = 2 0 1 1 2 0 4 0 2 1 0 1 2 1 0 1 4 0 0 0.
Basis for the Row Space Elementary row operations do not change the row space of a matrix. Proof. Remark: elementary row operations do change the column space of a matrix.
Basis for the Row Space Elementary row operations do not change the row space of a matrix. Proof. Remark: elementary row operations do change the column space of a matrix.
Basis for the Row Space Elementary row operations do not change the row space of a matrix. Proof. Remark: elementary row operations do change the column space of a matrix.
Results (continued) If A and B are row equivalent matrices then: 1. A given set of column vectors of A is linearly independent if and only if the corresponding column vectors from B are linearly independent. 2. A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding vectors of B form a basis for the column space of B. Proof.
Results (continued) If A and B are row equivalent matrices then: 1. A given set of column vectors of A is linearly independent if and only if the corresponding column vectors from B are linearly independent. 2. A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding vectors of B form a basis for the column space of B. Proof.
Results (continued) If a matrix R is in row-echelon form, then the vectors containing the leading 1 s form a basis for the row space and the columns containing the leading 1 s of the row vectors form a basis for the column space. Example Find bases for the row space and column space of A where A = 1 2 1 2 1 2 4 2 4 2 1 2 1 3 1 3 6 3 6 0 1 2 1 2 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
Results (continued) If a matrix R is in row-echelon form, then the vectors containing the leading 1 s form a basis for the row space and the columns containing the leading 1 s of the row vectors form a basis for the column space. Example Find bases for the row space and column space of A where A = 1 2 1 2 1 2 4 2 4 2 1 2 1 3 1 3 6 3 6 0 1 2 1 2 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
Example Example Find a basis for the space spanned by v 1 = (1, 2, 0, 1) v 2 = (2, 4, 0, 2) v 3 = ( 1, 1, 2, 0) v 4 = (0, 1, 2, 3) Using the vectors as the rows of the following matrix: 1 2 0 1 1 0 4 0 2 4 0 2 1 1 2 0 0 1 2 0 0 0 0 1 0 1 2 3 0 0 0 0
Example Example Find a basis for the space spanned by v 1 = (1, 2, 0, 1) v 2 = (2, 4, 0, 2) v 3 = ( 1, 1, 2, 0) v 4 = (0, 1, 2, 3) Using the vectors as the rows of the following matrix: 1 2 0 1 1 0 4 0 2 4 0 2 1 1 2 0 0 1 2 0 0 0 0 1 0 1 2 3 0 0 0 0
Algorithm for Finding a Basis Given a set of vectors S = {v 1, v 2,..., v k } in R n we can find a basis for span(s) by using the following. Steps: 1. Form a matrix A having columns v 1, v 2,..., v k. 2. Reduce A to its row-echelon form R and let w 1, w 2,..., w k be the columns of R. 3. Identify the columns of R containing leading 1 s. The corresponding columns vectors of A are the basis vectors for the column space of A. 4. Each column vector of R that does not contain a leading 1 is a linear combination of the columns to its left.
Example Example Find subset of the vectors below that forms a basis for the space spanned by v 1 = (1, 1, 5, 2) v 2 = ( 2, 3, 1, 0) v 3 = (4, 5, 9, 4) v 4 = (0, 4, 2, 3) v 5 = ( 7, 18, 2, 8) and express each vector that is not in the basis as a linear combination of the basis vectors. 1 2 4 0 7 1 0 2 0 1 1 3 5 4 18 5 1 9 2 2 0 1 1 0 3 0 0 0 1 2 2 0 4 3 8 0 0 0 0 0
Example Example Find subset of the vectors below that forms a basis for the space spanned by v 1 = (1, 1, 5, 2) v 2 = ( 2, 3, 1, 0) v 3 = (4, 5, 9, 4) v 4 = (0, 4, 2, 3) v 5 = ( 7, 18, 2, 8) and express each vector that is not in the basis as a linear combination of the basis vectors. 1 2 4 0 7 1 0 2 0 1 1 3 5 4 18 5 1 9 2 2 0 1 1 0 3 0 0 0 1 2 2 0 4 3 8 0 0 0 0 0
Homework Read Section 5.5 and work exercises 1, 2ac, 3acd, 4, 6cd, 7, 9cd, 11, 13.