Question

given an n m matrix a, the row space of a, denoted row(a), is the span of the vectors that make up the rows of a. (a) show that row(a) is a subspace of r m.

Answer 1

Let A be an m by n matrix. The space spanned by the rows of A is called the row space of A, denoted RS(A); it is a subspace of R n . The space spanned by the columns of A is called the column space of A, denoted CS(A); it is a subspace of R m .

The collection { r 1, r 2, …, r m } consisting of the rows of A may not form a basis for RS(A), because the collection may not be linearly independent. However, a maximal linearly independent subset of { r 1, r 2, …, r m } does give a basis for the row space. Since the maximum number of linearly independent rows of A is equal to the rank of A,

Similarly, if c 1, c 2, …, c n denote the columns of A, then a maximal linearly independent subset of { c 1, c 2, …, c n } gives a basis for the column space of A. But the maximum number of linearly independent columns is also equal to the rank of the matrix, so

Therefore, although RS(A) is a subspace of R n and CS(A) is a subspace of R m , equations (*) and (**) imply that

even if m ≠ n.

Example 1: Determine the dimension of, and a basis for, the row space of the matrix

A sequence of elementary row operations reduces this matrix to the echelon matrix

The rank of B is 3, so dim RS(B) = 3. A basis for RS(B) consists of the nonzero rows in the reduced matrix:

Another basis for RS(B), one consisting of some of the original rows of B, is

Note that since the row space is a 3‐dimensional subspace of R 3, it must be all of R 3.

Criteria for membership in the column space. If A is an m x n matrix and x is an n‐vector, written as a column matrix, then the product A x is equal to a linear combination of the columns of A:

By definition, a vector b in R m is in the column space of A if it can be written as a linear combination of the columns of A. That is, b ∈ CS(A) precisely when there exist scalars x 1, x 2, …, x n such that

Combining (*) and (**), then, leads to the following conclusion: