Lecture 15: 4.7 Row Space, Column Space, and Null space
Wei-Ta Chu
2011/11/9 Elementary Row Operation
¢ Performing an elementary row operation on an augmented matrix does not change the solution set of the corresponding linear system. ¢ It follows that applying an elementary row operation to a matrix A does not change the solution set of the corresponding linear system Ax=0, or stated another way, it does not change the null space of A.
The solution space of the homogeneous system of equation Ax = 0, which is a subspace of Rn, is called the null space of A.
2 Example 2 2− 10 1 −1 − 12 − 31 ¢ Find a basis for the nullspace of A = 11− 20 − 1 0 0 1 1 1 ¢ Solution ß The nullspace of A is the solution space of the homogeneous system
2x1 + 2x2 – x3 + x 5 = 0 -x1 – x2 – 2 x3 – 3x4 + x 5 = 0 x1 + x 2 – 2 x3 – x5 = 0 x3 + x 4 + x 5 = 0 ß In Example 10 of Section 5.4 we showed that the vectors −1 − 1 1 0 v=0 and v = − 1 1 2 0 0 0 1 form a basis for the nullspace.
3 Theorems 4.7.3 and 4.7.4
¢ Theorem 4.7.3 ß Elementary row operations do not change the nullspace of a matrix.
¢ Theorem 4.7.4 ß Elementary row operations do not change the row space of a matrix.
4 Proof of Theorem 4.7.4
¢ Suppose that the row vectors of a matrix A are r1,r2,…, rm, and let B be obtained from A by performing an elementary row operation. (We say that A and B are row equivalent.) ¢ We shall show that every vector in the row space of B is also in that of A, and that every vector in the row space of A is in that of B. ¢ If the row operation is a row interchange, then B and A have the same row vectors and consequently have the same row space.
5 Proof of Theorem 4.7.4
¢ If the row operation is multiplication of a row by a nonzero scalar or a multiple of one row to another, then
the row vector r1’, r2’,…, rm’ of B are linear combination of r1,r2,…, rm; thus they lie in the row space of A. ¢ Since a vector space is closed under addition and scalar
multiplication, all linear combination of r1’, r2’,…, rm’ will also lie in the row space of A. Therefore, each vector in the row space of B is in the row space of A.
6 Proof of Theorem 4.7.4
¢ Since B is obtained from A by performing a row operation, A can be obtained from B by performing the inverse operation (Sec. 1.5). ¢ Thus the argument above shows that the row space of A is contained in the row space of B.
7 Remarks
¢ Do elementary row operations change the column space?
ß Yes! ¢ The second column is a scalar multiple of the first, so the column space of A consists of all scalar multiplies of the first column vector.
Add -2 times the first row to the second
¢ Again, the second column is a scalar multiple of the first, so the column space of B consists of all scalar multiples of the first column vector. This is not the same as the column space of A.
8 Theorem 4.7.5
¢ Theorem 4.7.5 ß If a matrix R is in row echelon form, then the row vectors with the leading 1’s (i.e., the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1’s of the row vectors form a basis for the column space of R.
9 Bases for Row and Column Spaces The matrix 1− 2503 0 1 300 R = 0 0 010 0 0 000 is in row-echelon form. From Theorem 5.5 .6 the vectors = r 1 [1 -2 5 0 3] = r2 [0 1 3 0 0] = r3 [0 0 0 1 0] form a basis for the row space of R, and the vectors 1 − 2 0 0 1 0 c= , c = , c = 10 2 0 4 1 0 0 0 form a basis for the column space of R.
10 Example
¢ Find bases for the row and column spaces of 1− 34 − 25 4 2− 69 − 18 2 A = 2− 69 − 19 7 ¢ Solution: −13 − 42 −− 5 4
ß Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis that of any row-echelon form of A.
ß Reducing A to row-echelon form we obtain
1− 34 − 25 4 0013− 2 − 6 R = 0000 1 5 0000 0 0
11 1− 34 − 25 4 1− 34 − 25 4 Example 2− 69 − 18 2 0013− 2 − 6 A = R = 2− 69 − 19 7 0000 1 5 −13 − 42 −− 5 4 0000 0 0
¢ The basis vectors for the row space of R and A
r1 = [1 -3 4 -2 5 4]
r2 = [0 0 1 3 -2 -6]
r3 = [0 0 0 0 1 5] ¢ Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R.
12 Theorem 4.7.6
¢ Theorem 4.7.6 ß If A and B are row equivalent matrices, then: ¢ A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. ¢ A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B.
13 1− 34 − 25 4 1− 34 − 25 4 Example 2− 69 − 18 2 0013− 2 − 6 A = R = 2− 69 − 19 7 0000 1 5 −13 − 42 −− 5 4 0000 0 0
¢ We can find the basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A. ¢ Basis for R’s column space
¢ Basis for A’s column space 1 4 5 2 9 8 c=, c = , c = 12 3 9 5 9 −1 − 4 − 5
14 Example (Basis for a Vector Space Using Row Operations )
¢ Find a basis for the space spanned by the row vectors
v1= (1, -2, 0, 0, 3), v2 = (2, -5, -3, -2, 6), v3 = (0, 5, 15, 10, 0), v4 = (2, 6, 18, 8, 6). ¢ Except for a variation in notation, the space spanned by these vectors is the row space of the matrix 1− 20 03 1− 2003 − − − 2 5 3 26 0 1 320 0 5 15 10 0 0 0 110 2 6 18 8 6 0 0 000 ß The nonzero row vectors in this matrix are
w1= (1, -2, 0, 0, 3), w2 = (0, 1, 3, 2, 0), w3 = (0, 0, 1, 1, 0) ß These vectors form a basis for the row space and consequently form a 5 basis for the subspace of R spanned by v1, v2, v3, and v4.
15 Remarks
¢ Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R. ¢ However, if we can find a set of column vectors of R that forms a basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A.
¢ The basis vectors obtained for the column space of A consisted of column vectors of A, but the basis vectors obtained for the row space of A were not all vectors of A. ¢ Transpose of the matrix can be used to solve this problem.
16 Example (Basis for the Row Space of a Matrix )
T ¢ Find a basis for the row space of ß The column space of A are 1− 20 03 1 2 2 − − − 2 5 3 26 − − A = 2 5 6 0 5 15 10 0 c=0, c = − 3 , and c = 18 1 2 4 2 6 18 8 6 0 − 2 8 consisting entirely of row vectors 3 6 6 from A. ß Thus, the basis vectors for the row space of A are ¢ Solution: r1 = [1 -2 0 0 3] 1 2 0 2 1 20 2 r = [2 -5 -3 -2 6] −2 − 5 5 6 0 1 5− 10 2 r = [2 6 18 8 6] AT = 0− 3 15 18 000 1 3 0− 2 10 8 000 0 3 6 0 6 000 0
17 Example (Basis and Linear Combinations)
¢ (a) Find a subset of the vectors v1 = (1, -2, 0, 3), v2 = (2, -5, -3, 6), v3 = (0, 1, 3, 0), v4 = (2, -1, 4, -7), v5 = (5, -8, 1, 2) that forms a basis for the space spanned by these vectors. ¢ (b) Express each vector not in the basis as a linear combination of the basis vectors.
¢ Solution (a): 1 2 0 2 5 1 0 2 0 1 − − − − − 2 5 1 1 8 0 1 1 0 1 0 − 3 3 4 1 0 0 0 1 1 3 6 0 − 7 2 0 0 0 0 0 ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑
v1 v 2 v 3 v 4 v 5 w1 w 2 w3 w 4 w5
ß Thus, { v1, v2, v4} is a basis for the column space of the matrix.
18 Example
¢ Solution (b):
ß We can express w3 as a linear combination of w1 and w2, express w5 as a linear combination of w1, w2, and w4 (Why?). By inspection, these linear combination are
w3 = 2 w1 – w2
w5 = w1 + w2 + w4 ß We call these the dependency equations . The corresponding relationships in the original vectors are
v3 = 2 v1 – v2
v5 = v1 + v2 + v4
19 Lecture 15: 4.8 Rank, Nullity, and the Fundamental Matrix Spaces
Wei-Ta Chu
2011/11/9 Dimension and Rank
¢ Theorem 4.8.1
ß If A is any matrix, then the row space and column space of A have the same dimension . ¢ Proof: Let R be any row-echelon form of A. It follows from Theorem 4.7.4 and 4.7.6b that dim(row space of A) = dim(row space of R). dim(column space of A) = dim(column space of R) ¢ The dimension of the row space of R is the number of nonzero rows = number of leading 1’s = dimension of the column space of R
21 Rank and Nullity
¢ Definition
ß 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).
22 Example (Rank and Nullity)
¢ Find the rank and nullity of the matrix −1204 5 − 3 3− 720 1 4 A = 2− 524 6 1 4− 92 − 4 − 47
¢ Solution: ß The reduced row-echelon form of A is 1 0− 4 − 28 − 3713 − − − 01 2 12 16 5 000 0 0 0 000 0 0 0 ß Since there are two nonzero rows (two leading 1’s), the row space and column space are both two-dimensional, so rank( A) = 2.
23 Example (Rank and Nullity)
ß To find the nullity of A, we must find the dimension of the solution space of the linear system Ax=0. ß The corresponding system of equations will be
x1 – 4x3 – 28 x4 – 37 x5 + 13 x6 = 0 x2 – 2x3 – 12 x4 – 16 x5+ 5 x6 = 0 ß It follows that the general solution of the system is
x1 = 4 r + 28 s + 37 t – 13 u, x2 = 2 r + 12 s + 16 t – 5u, x3 = r, x4 = s, x5 = t, x6 = u or − x1 4 28 37 13 − x2 2 12 16 5 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
24 Example
¢ What is the maximum possible rank of an matrix A that is not square? ¢ Solution: The row space of A is at most n-dimensional and the column space is at most m-dimensional. Since the rank of A is the common dimension of its row and column space, it follows that the rank is at most the smaller of m and n.
25 Theorem 4.8.2
¢ Theorem 4.8.2 (Dimension Theorem for Matrices)
ß If A is a matrix with n columns, then rank( A) + nullity( A) = n. ¢ Proof: ¢ 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
26 Theorem 4.8.2
¢ 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
27 Example
−1204 5 − 3 3− 720 1 4 A = 2− 524 6 1 4− 92 − 4 − 47
¢ This matrix has 6 columns, so rank( A) + nullity( A) = 6 ¢ In previous example, we know rank( A) = 4 and nullity( A) = 2
28 Theorem 4.8.3
¢ Theorem 4.8.3
ß If A is an m×n 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.
29 Example
¢ Find the number of parameters in the general solution of Ax = 0 if A is a 5 ×7 matrix of rank 3. ¢ Solution:
ß nullity( A) = n – rank( A) = 7 – 3 = 4
ß Thus, there are four parameters.
30 Theorem 4.8.4 (Equivalent Statements) × n → n ¢ If A is an n n matrix, and if TA : R R is multiplication by A, then the following are equivalent: ß A is invertible. ß Ax = 0 has only the trivial solution.
ß The reduced row-echelon form of A is In. ß A is expressible as a product of elementary matrices. ß Ax = b is consistent for every n×1 matrix b. ß Ax = b has exactly one solution for every n×1 matrix b. ß det( A)≠0. ß The column vectors of A are linearly independent. ß The row vectors of A are linearly independent. n ß The column vectors of A span R . n ß The row vectors of A span R . n ß The column vectors of A form a basis for R . n ß The row vectors of A form a basis for R . ß A has rank n. ß A has nullity 0.
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