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4.3: Subspaces

( \newcommand{\kernel}{\mathrm{null}\,}\)

As mentioned in the last section, there are countless examples of vector spaces. One particularly important source of new vector spaces comes from looking at subsets of a set that is already known to be a vector space.

Definition 4.3.1. Let V be a vector space over F, and let U be a subset of V . Then we call U a subspace of V if U is a vector space over F under the same operations that make V into a vector space over F.

To check that a subset U of V is a subspace, it suffices to check only a few of the conditions of a vector space.

Lemma 4.3.2. Let UV be a subset of a vector space V over F. Then U is a subspace of V if and only if the following three conditions hold.

  1. additive identity: 0U;
  2. closure under addition: u,vUu+vU;
  3. closure under scalar multiplication: aF, uUauU.

Proof. Condition 1 implies that the additive identity exists. Condition 2 implies that vector addition is well-defined and, Condition 3 ensures that scalar multiplication is well-defined. All other conditions for a vector space are inherited from V since addition and scalar multiplication for elements in U are the same when viewed as elements in either U or V .

Remark 4.3.3. Note that if we require UV to be a nonempty subset of V, then condition 1 of Lemma 4.3.2 already follows from condition 3 since 0u=0 for uU.

Example 4.3.4. In every vector space V , the subsets 0 and V are easily verified to be subspaces. We call these the trivial subspaces of V .

Example 4.3.5. (x1,0)|x1R is a subspace of R2 .

Example 4.3.6. U=(x1,x2,x3)F3|x1+2x2=0 is a subspace of F3 . To see this, we need to check the three conditions of Lemma 4.3.2.

The zero vector (0,0,0)F3 is in U since it satisfies the condition x1+2x2=0. To show that U is closed under addition, take two vectors v=(v1,v2,v3) and u=(u1,u2,u3). Then, by the definition of U, we have v1+2v2=0 andu1+2u2=0. Adding these two equations, it is not hard to see that the vector

v+u=(v1+u1,v2+u2,v3+u3) satises (v1+u1)+2(v2+u2)=0.

Hence v+uU. Similarly, to show closure under scalar multiplication, take u=(u1,u2,u3)U and aF. Then au=(au1,au2,au3) satisfies the
equation au1+2au2=a(u1+2u2)=0, and so auU.

Example 4.3.7. U=pF[z]|p(3)=0 is a subspace of F[z]. Again, to check this, we need to verify the three conditions of Lemma 4.3.2.

Certainly the zero polynomial p(z)=0zn+0zn1++0z+0 is in U since p(z) evaluated at 3 is 0. If f(z), g(z) \in U\), then f(3)=g(3)=0 so that (f+g)(3)=f(3)+g(3)=0+0=0. Hence f+gU, which proves closure under addition. Similarly, (af)(3)=af(3)=a0=0 for any aF, which proves closure under scalar multiplication.

Example 4.3.8. As in Example 4.1.6, let DR be a subset of R, and let C(D) denote the set of all smooth (a.k.a. continuously differentiable) functions with domain D and codomain R. Then, under the same operations of pointwise addition and scalar multiplication, one can show that C(D) is a subspace of C(D).

figure 4.3.1 - Copy.jpg

Figure 4.3.1: The intersection UU of two subspaces is a subspace.

Example 4.3.9. The subspaces of R2 consist of 0, all lines through the origin, and R2 itself. The subspaces of R3 are {0}, all lines through the origin, all planes through the origin, and R3. In fact, these exhaust all subspaces of R2 and R3 , respectively. To prove this, we will need further tools such as the notion of bases and dimensions to be discussed soon. In particular, this shows that lines and planes that do not pass through the origin are not subspaces (which is not so hard to show!).

Note that if U and U are subspaces of V , then their intersection UU is also a subspace (see Proof-writing Exercise 2 and Figure 4.3.1). However, the union of two subspaces is not necessarily a subspace. Think, for example, of the union of two lines in R2 , as in Figure 4.4.1 in the next chapter.


This page titled 4.3: Subspaces is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.

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