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column space of ; these are

Transposing again and adjusting the notation appropriately yields the basis vectors
and
for the row space of A.
We know from Theorem 5.5.5 that elementary row operations do not alter relationships of linear independence and linear
dependence among the column vectors; however, Formulas 6 and 7 imply an even deeper result. Because these formulas actually
have the same scalar coefficients , , … , , it follows that elementary row operations do not alter the formulas (linear
combinations) that relate linearly dependent column vectors. We omit the formal proof.

EXAMPLE 9 Basis and Linear Combinations

(a) Find a subset of the vectors

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)

We begin by constructing a matrix that has , , …, as its column vectors:

(8)

The first part of our problem can be solved by finding a basis for the column space of this matrix. Reducing the matrix to reduced
row-echelon form and denoting the column vectors of the resulting matrix by , , , , and yields