8.3 Vector Spaces and Subspaces Performance Criterion: 8. (d) Determine whether a subset of Rn is a subspace. If so, prove it; if not, give an appropriate counterexample. Vector Spaces The term “space” in math simply means a set of objects with some additional special properties. There are metric spaces, function space, topological spaces, Banach spaces, and more. The vectors that we have been dealing with make up the vector spaces called R2, R3 and, for larger values, Rn. In general, a vector space is simply a collection of objects called vectors (and a set of scalars) that satisfy certain properties. Definition 8.3.1: Vector Space A vector space is a set V of objects called vectors and a set of scalars (usually the real numbers R), with the operations of vector addition and scalar multiplication, for which the following properties hold for all u, v, w in V and scalars c and d. 1. u + v is in V 2. u + v = v + u 3. (u + v)+ w = u + (v + w) 4. There exists a vector 0 in V such that u + 0 = u. This vector is called the zero vector. 5. For every u in V there exists a vector −u in V such that u + (−u) = 0. 6. cu is in V . 7. c(u + v)= cu + cv 8. (c + d)u = cu + du 9. c(du) = (cd)u 10. 1u = u Note that items 1 and 6 of the above definition say that the vector space V is closed under addition and scalar multiplication. When working with vector spaces, we will be very interested in certain subsets of those vector spaces that are the span of a set of vectors. As you proceed, recall Example 8.2(b), where we showed that the span of a set of vectors is closed under addition and scalar multiplication. Subspaces of Vector Spaces As you should know by now, the two main operations with vectors are multiplication by scalars and addition of vectors. (Note that these two combined give us linear combinations, the foundation of almost everything we’ve done.) A given vector space can have all sorts of subsets; consider the following subsets of R2. • The set S1 consisting of the first quadrant and the nonnegative parts of the two axes, or all vectors of the x1 form such that x1 ≥ 0 and x2 ≥ 0. x2 110 3 • The set S2 consisting of the line containing the vector . Algebraically this is all vectors of the form 2 3 t where t ranges over all real numbers. 2 • The set S3 consisting of the first and third quadrants and both axes. This can be described as the set of all x1 vectors with x1x2 ≥ 0. x2 Our current concern is whether these subsets of R2 are closed under addition and scalar multiplication. With a bit of thought you should see that S1 is closed under addition, but not scalar multiplication when the scalar is negative: S1 S1 u + v u w v cw c< 0 In some sense we can solve the problem of not being closed under scalar multiplication by including the third quadrant as well to get S3, but then the set isn’t closed under addition: S3 S3 u w S3 cw u + v S3 c< 0 v Finally, the set S2 is closed under both addition and scalar multiplication. That is a bit messy to show with a 3 diagram, but consider the following. S2 is the span of the single vector , and we showed in the last section 2 that the span of any set of vectors is closed under addition and scalar multiplication. It turns out that when working with vector spaces the only subsets of any real interest are the ones that are closed under both addition and scalar multiplication. We give such subsets a name: Definition 8.3.2: Subspace of Rn A subset S of Rn is called a subspace of Rn if for every scalar c and any vectors u and v in S, cu and u + v are also in S. That is, S is closed under scalar multiplication and addition. You will be asked whether certain subsets of R2, R3 or Rn are subspaces, and it is your job to back your answers up with some reasoning. This is done as follows: • When a subset IS a subspace a general proof is required. That is, we must show that the set is closed under scalar multiplication and addition, for ALL scalars and ALL vectors. We may have to do this outright, but if it is clear that the set of vectors is the span of some set of vectors, then we know from the argument presented in Example 8.2(b) that the set is closed under addition and scalar multiplication, so it is a subspace. • When a subset IS NOT a subspace, we demonstrate that fact with a SPECIFIC example. Such an example is called a counterexample. Notice that all we need to do to show that a subset is not a subspace is to show 111 that that it is not closed under scalar multiplication OR vector addition. If either is the case, then the set in question is not a subspace. Even if both are the case, we need only show one. The following examples illustrate these things. x1 ⋄ Example 8.3(a): Show that the set S1 consisting of all vectors of the form such that x1 ≥ 0 and x2 2 x2 ≥ 0 is not a subspace of R . As mentioned before, this set is not closed under multiplication by negative scalars, so we just need to give a specific 3 3 −6 example of this. Let u = and c = −2. Clearly u is in S1 and cu = (−2) = , which is 5 5 −10 2 not in S1. Therefore S1 is not closed under scalar multiplication so it is not a subspace of R . ♠ 3 ⋄ Example 8.3(b): Show that the set S2 consisting of all vectors of the form t , where t ranges over 2 all real numbers, is a subspace of R2. 3 3 Let c be any scalar and let u = s and v = t . Then u and v are both in S2 and 2 2 3 3 3 3 3 cu = c s = (cs) and u + v = s + t = (s + t) . 2 2 2 2 2 2 Because cs and s + t are scalars, we can see that both cu and u + v are in S2, so S2 is a subspace of R . ♠ This last example demonstrates the general method for showing that a set of vectors is closed under addition and scalar multiplication. That said, the given subspace could have been shown to be s subspace by simply observing 3 that it is the span of the set consisting of the single vector . From Theorem 8.2.2 we know that the span of 2 any set of vectors is a subspace, so the set described in the above example is a subspace of R2. ⋄ Example 8.3(c): Determine whether the subset S of R3 consisting of all vectors of the form x = 2 4 5 + t −1 is a subspace. If it is, prove it. If it is not, provide a counterexample. −1 3 We recognize this as a line in R3 passing through the point (2, 5, −1), and it is not hard to show that the line does not pass through the origin. Remember l that what we mean by the line is really all position vectors (so with tails at the origin) whose tips are on the line. Considering a similar situation in R2, we see that u is such a vector for the line l shown. It should be clear that if we u multiply u by any scalar other than one, the resulting vector’s tip will not lie on the line. Thus we would guess that the set S, even though it is in R3, is probably not closed under scalar multiplication. 2 4 6 Now let’s prove that it isn’t. To do this we first let u = 5 +1 −1 = 4 , which is in S. Let −1 3 2 12 c =2, so 2u = 8 . We need to show that this vector is not in S. If it were, there would have to be a scalar 4 12 2 4 2 10 4 t such that 8 = 5 + t −1 . Subtracting 5 we get 3 = t −1 . We can see 4 −1 3 −1 5 3 that the value of tthat would be needed to give the correct second component would be −3, but this would 112 12 result in a third component of −9, which is not correct. Thus there is no such t and the vector 8 is not 4 3 in S. Thus S is not closed under scalar multiplication, so it is not a subspace of R . ♠ We should compare the results of Examples 8.3(b) and 8.3(c). Note that both are lies in their respective Rn’s, but the line in 8.3(b) passes through the origin, and the one in 8.3(c) does not. It is no coincidence that the set in 8.3(b) is a subspace and the set in 8.3(c) is not. If a set S is a subspace, being closed under scalar multiplication means that zero times any vector in the subspace must also be in the subspace. But zero times a vector is the zero vector 0.