INTRODUCTION

In a general m × n matrix,

,

the rows

u₁ = (a₁₁ a₁₂ . . . a_1n), u₂ = (a₂₁ a₂₂ . . . a_2n), . . . , u_m = (a_m1 a_m2 . . . a_mn)

and columns

are called the row vectors of A and the column vectors of A, respectively.

A Definition

As vectors, the set u₁, u₂, . . . , u_m is either linearly independent or linearly dependent. We have the following definition.

EXAMPLE 1 Rank of a 3 × 4 Matrix

Row Space

In the terminology of the preceding chapter, the row vectors u₁, u₂, u₃ of the matrix (1) are a set of vectors in the vector space R⁴. Since R_A = Span(u₁, u₂, u₃) (the set of all linear combinations of the vectors u₁, u₂, u₃) is a subspace of R⁴, we are justified in calling R_A the row space of the matrix A. Now the set of vectors u₁, u₂ is linearly independent and also spans R_A; in other words, the set u₁, u₂ is a basis for R_A. The dimension (the number of vectors in a basis) of the row space R_A is 2, which is rank(A).

Rank by Row Reduction

Example 1 notwithstanding, it is generally not easy to determine the rank of a matrix by inspection. Although there are several mechanical ways of finding rank(A), we examine one way that uses the elementary row operations introduced in the preceding section. Specifically, the rank of A can be found by row reducing A to a row-echelon matrix B. To understand this, first recall that an m × n matrix B is row equivalent to an m × n matrix A if the rows of B were obtained from the rows of A by applying elementary row operations. If we simply interchange two rows in A to obtain B, then the row space R_A of A and the row space R_B of B are equal because the row vectors of A and B are the same. When the row vectors of B are linear combinations of the rows of A, it follows that the row vectors of B are in the row space R_A, and so R_B is a subset of R_A (written R_B ⊆ R_A). Conversely, A is row equivalent to B, since we can obtain A by applying row operations to B. Hence the rows of A are linear combinations of the rows of B, and so it follows that R_A is a subset of R_B (R_A ⊆ R_B). From R_B ⊆ R_A and R_A ⊆ R_B, we conclude that R_A = R_B. Finally, if we row-reduce A into a row-echelon matrix B, then the nonzero rows of B are linearly independent. (Why?) The nonzero rows of B form a basis for the row space R_A, and so we have the result that rank(A) = dimension of R_A.

We summarize these conclusions in the next theorem.

EXAMPLE 2 Rank by Row Reduction—Example 1 Revisited

We row-reduce a matrix A to a row-echelon form B in exactly the same manner as we row-reduced the augmented matrix of a system of linear equations to an echelon form when we solved the system using Gaussian elimination. Using the matrix (1) in Example 1, elementary row operations give

Since the last matrix is in row-echelon form, and since the last matrix has two nonzero rows, we conclude from (iii) of Theorem 8.3.1 that rank(A) = 2. ≡

EXAMPLE 3 Linear Independence/Dependence

Determine whether the set of vectors u₁ = 〈2, 1, 1〉, u₂ = 〈0, 3, 0〉, u₃ = 〈3, 1, 2〉, in R³ is linearly dependent or linearly independent.

SOLUTION

It should be clear from the discussion above that if we form a matrix A with the given vectors as rows, and if we row-reduce A to a row-echelon form B with rank 3, then the set of vectors is linearly independent. If rank(A) < 3, then the set of vectors is linearly dependent. In this case, it is easy to carry the row reduction all the way to a reduced row-echelon form:

Thus rank(A) = 3 and the set of vectors u₁, u₂, u₃ is linearly independent. ≡

As mentioned previously, the vectors in the row-echelon form of a matrix A can serve as a basis for the row space of the matrix A. In Example 3, we see that a basis for the row space of A is the standard basis 〈1, 0, 0〉, 〈0, 1, 0〉, 〈0, 0, 1〉 of R³.

Rank and Linear Systems

The concept of rank can be related back to solvability of linear systems of algebraic equations. Suppose AX = B is a linear system and that (A|B) denotes the augmented matrix of the system. In Example 7 of Section 8.2, we saw that the system

was inconsistent. The inconsistency of the system is seen in the fact that, after row reduction of the augmented matrix (A|B),

(2)

the last row of the reduced row-echelon form is nonzero. Of course, this reduction shows that rank(A|B) = 3. But note, too, that the result in (2) indicates that rank(A) = 2 because

We have illustrated a special case of the next theorem.

In Example 6 of Section 8.2, we saw that the system

(3)

was consistent and had an infinite number of solutions. We solved for two of the variables, x₁ and x₂, in terms of the remaining variable x₃, which we relabeled as a parameter t. The number of parameters in a solution of a system is related to the rank of the coefficient matrix A.

For the system (3), we can see from the row reduction

that rank(A) = rank(A|B) = 2, and so the system is consistent by Theorem 8.3.2. With n = 3, we see from Theorem 8.3.3 the number of parameters in the solution is 3 − 2 = 1.

FIGURE 8.3.1 outlines the connection between the concept of rank of a matrix and the solution of a linear system.

REMARKS

We have not mentioned the connection between the columns of a matrix A and the rank of A. It turns out that the maximum number of independent columns that a matrix A can have must equal the maximum number of independent rows. In the terminology of vector spaces, the row space R_A of matrix A has the same dimension as its column space C_A. For example, if we take the transpose of the matrix in (1) and reduce it to a row-echelon form

we see that the maximum number of rows of A^T is 2, and so the maximum number of linearly independent columns of A is 2.

8.3 Exercises Answers to selected odd-numbered problems begin on page ANS-18.

In Problems 1–10, use (iii) of Theorem 8.3.1 to find the rank of the given matrix.

In Problems 11–14, determine whether the given set of vectors is linearly dependent or linearly independent.

11. u₁ = 〈1, 2, 3〉, u₂ = 〈1, 0, 1〉, u₃ = 〈1, −1, 5〉
12. u₁ = 〈2, 6, 3〉, u₂ = 〈1, −1, 4〉, u₃ = 〈3, 2, 1〉, u₄ = 〈2, 5, 4〉
13. u₁ = 〈1, −1, 3, −1〉, u₂ = 〈1, −1, 4, 2〉, u₃ = 〈1, −1, 5, 7〉
14. u₁ = 〈2, 1, 1, 5〉, u₂ = 〈2, 2, 1, 1〉, u₃ = 〈3, −1, 6, 1〉, u₄ = 〈1, 1, 1, −1〉
15. Suppose the system AX = B is consistent and A is a 5 × 8 matrix and rank(A) = 3. How many parameters does the solution of the system have?
16. Let A be a nonzero 4 × 6 matrix.
    1. What is the maximum rank that A can have?
    2. If rank(A|B) = 2, then for what value(s) of rank(A) is the system AX = B, B ≠ 0, inconsistent? Consistent?
    3. If rank(A) = 3, then how many parameters does the solution of the system AX = 0 have?
17. Let v₁, v₂, and v₃ be the first, second, and third column vectors, respectively, of the matrix

    

    What can we conclude about rank(A) from the observation 2v₁ + 3v₂ − v₃ = 0? [Hint: Read the Remarks at the end of this section.]

Discussion Problems

18. Suppose the system AX = B is consistent and A is a 6 × 3 matrix. Suppose the maximum number of linearly independent rows in A is 3. Discuss: Is the solution of the system unique?
19. Suppose we wish to determine whether the set of column vectors

    

    

    is linearly dependent or linearly independent. By Definition 7.6.3, if

    c₁v₁ + c₂v₂ + c₃v₃ + c₄v₄ + c₅v₅ = 0 (4)

    only for c₁ = 0, c₂ = 0, c₃ = 0, c₄ = 0, c₅ = 0, then the set of vectors is linearly independent; otherwise the set is linearly dependent. But (4) is equivalent to the linear system

    

    Without doing any further work, explain why we can now conclude that the set of vectors is linearly dependent.

Computer Lab Assignment

20. A CAS can be used to row-reduce a matrix to a row-echelon form. Use a CAS to determine the ranks of the augmented matrix (A|B) and the coefficient matrix A for

    

    Is the system consistent or inconsistent? If consistent, solve the system.