160 CHAPTER 9. SUBSPACES SPANS AND BASES

9.1 SubspacesA subspace is a set of vectors with the property that linear combinations of these vectorsremain in the set. Geometrically, subspaces are like lines and planes which contain theorigin. More precisely, the following definition is the right way to think of this.

Definition 9.1.1 Let V be a nonempty collection of vectors in Fn. Then V is called asubspace if whenever α,β are scalars and u,v are vectors in V, the linear combinationαu+βv is also in V .

There is no substitute for the above definition or equivalent algebraic definition! How-ever, it is sometimes helpful to look at pictures at least initially. The following are foursubsets of R2. The first is the shaded area between two lines which intersect at the origin,the second is a line through the origin, the third is the union of two lines through the origin,and the last is the region between two rays from the origin. Note that in the last, multipli-cation of a vector in the set by a nonnegative scalar results in a vector in the set as does thesum of two vectors in the set. However, multiplication by a negative scalar does not take avector in the set to another in the set.

not subspace not subspacesubspace not subspace

Observe how the above definition indicates that the claims posted on the picture arevalid. Now here are the two main examples of subspaces.

Theorem 9.1.2 Let A be an m×n matrix. Then Im(A) is a subspace of Fm. Also let

ker(A)≡ N (A)≡ {x ∈ Fn such that Ax= 0}

Then ker(A) is a subspace of Fn.

Proof: Suppose Axi is in Im(A) and a,b are scalars. Does it follow that aAx1 +bAx2is in Im(A)? The answer is yes because

aAx1 +bAx2 = A(ax1 +bx2) ∈ Im(A)

this because of the above properties of matrix multiplication. Note that A0 = 0 so 0 ∈Im(A) and so Im(A) ̸= /0.

Now suppose x,y are both in N (A) and a,b are scalars. Does it follow that ax+by ∈N (A)? The answer is yes because

A(ax+by) = aAx+bAy = a0+b0 = 0.

Thus the condition is satisfied. Of course N (A) ̸= /0 because A0 = 0. ■Subspaces are exactly those subsets of Fn which are themselves vector spaces. Recall

that a vector space is something which satisfies the vector space axioms on Page 55.