Section 17.3 Concepts
In this section.
Subsection 17.3.1 Recognizing subspaces
How can we recognize subspaces? To be a subspace, a subcollection of vectors must satisfy all ten vector space axioms on its own. But Axiom A 2, Axiom A 3, and Axioms S 2β5 all concern the algebra of vectors, and don't really take into account where the vectors are considered to βliveβ. Since these algebra axioms are true about all vectors in the larger space, they are automatically true about the vectors in the subcollection. So that leaves Axiom A 1, Axiom A 4, Axiom A 5, and Axiom S 1. For axioms Axiom A 1 and Axiom S 1, we already have addition and scalar multiplication of vectors defined as in the large space. But when we are considering whether the smaller collection is a vector space all on its own, the vectors not in this collection are no longer relevant. So the parts of these axioms that say βthe result is always equal to another in the collection of objectsβ now refer to the subcollection of vectors under consideration. That is, we need to make sure that when vectors in the subcollection are added or scalar multiplied, then the result is again in the subcollection, not somewhere else in the vector space at large. We call this property being closed under the vector operations. Similarly, for Axiom A 5, we already know that every vector in the large space has a negative, hence so does every vector in the subcollection. But we need to check that the subcollection is closed under taking negatives β that the negative of a vector in the subcollection is again in the subcollection. And we know that there is a zero vector in the large space, but the subcollection needs a zero vector too, to satisfy Axiom A 4. The zero vector from the larger space already satisfies the property that v+0=v for all vectors, but again we need the zero vector to be in the subcollection. As we will prove in Proposition 17.5.1, we really only need to check a subcollection for closure under addition and scalar multiplication in order to verify that it is also a vector space.Procedure 17.3.1. Subspace Test.
To test whether a subcollection of vectors in a vector space is a subspace (that is, a vector space on its own), check whether all three of the following conditions are met.
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Nonempty.
The subcollection contains at least one vector.
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Closed under vector addition.
Given vectors w1 and w2 in the subcollection, the sum w1+w2 is also in the subcollection.
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Closed under scalar multiplication.
Given vector w in the subcollection and scalar k, the scaled vector kw is also in the subcollection.
Remark 17.3.2.
- The first condition might seem unnecessary. But in math it is possible to accidentally be considering some collection of objects that in reality contains no objects. For example, consider M1(R), the vector space of all 1Γ1 matrices. We could try to determine whether the subcollection of all 1Γ1 matrices whose square is equal to [β1] is a subspace of of M1(R), but we'd be wasting our time because there are no such matrices.
- The logic of the test works in reverse as well: every subspace satisfies the three statements of the test because it is a vector space all on its own and thus satisfies the ten vector space axioms. (This is the βand only ifβ part of Proposition 17.5.1.) So a subspace is always nonempty because it must contain a zero vector (Axiom A 4), and it is always closed under the vector operations (Axiom A 1 and Axiom S 1).
Subsection 17.3.2 Building subspaces
Suppose we are studying a problem for which certain vectors in a certain vector space are important. We would like to do linear algebra with these certain special vectors, so the fact that they are part of a vector space is essential, but perhaps not all of the vector space in which they live is relevant to the problem. Can we find a smaller subspace which contains these important vectors? Even better, can we determine the smallest subspace which contains these important vectors? As stated above, every subspace must satisfy the three parts of the Subspace Test. So if a subspace contains our special vectors, then it must also contain all scalar multiples of those vectors. And it must also be closed under vector addition, so it must also contain all sums of scalar multiples of the special vectors. Therefore, it must contain every linear combination of the special vectors. (In other words, subspaces are also closed under taking linear combinations.) As we noted in Discovery guide 17.1, and will verify in Subsection 17.5.2 (Proposition 17.5.5), the subcollection of a vector space consisting of all linear combinations of a set of vectors S is always a subspace, called the span of S and written SpanS. So the process of taking all possible linear combinations of a set of vectors can be used to build subspaces. And sometimes, as in Discovery 17.4, the space that span builds ends up being the whole larger space.Subsection 17.3.3 The subspaces of Rn
We saw in Subsection 15.3.2 that a line through the origin in Rn can be described in vector form as x=tp for some vector p that is parallel to the line (and taking βinitialβ point x0=0, since the line passes through the origin). Similarly, we saw that a plane through the origin in R3 can be described in vector form as x=sp1+tp2 for some vectors p1 and p2 that are parallel to the plane but not parallel to each other. With our new, more sophisticated view of vector spaces and subspaces, we can now recognize a line x=tp as the subspace Span{p}, and a plane x=sp1+tp2 as the subspace Span{p1,p2}. In fact, every (nontrivial) subspace of Rn has a geometric interpretation as a line or a plane or some sort of higher-dimensional hyperplane. In particular,-
the subspaces of R2 are precisely
- the zero space {0},
- Span{p} for a nonzero vector p, which builds a line through the origin, and
- Span{p1,p2} for two nonzero, nonparallel vectors p1,p2, which builds the whole plane R2;
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the subspaces of R3 are precisely
- the zero space {0},
- Span{p} for a nonzero vector p, which builds a line through the origin,
- Span{p1,p2} for two nonzero, nonparallel vectors p1,p2, which builds a plane through the origin, and
- Span{p1,p2,p3} for three nonzero, non-coplanar vectors p1,p2,p3, which builds all of space R3;
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the subspaces of R4 are precisely
- the zero space {0},
- Span{p} for a nonzero vector p, which builds a line through the origin,
- Span{p1,p2} for two nonzero, nonparallel vectors p1,p2, which builds a plane through the origin,
- Span{p1,p2,p3} for three nonzero, non-coplanar vectors p1,p2,p3, which builds a hyperplane through the origin, and
- Span{p1,p2,p3,p4} for four nonzero, non-cohyperplanar vectors p1,p2,p3,p4, which builds all of four-dimensional space R4;
- etc.