Vector spaces The vector spaces Rn, properties of vectors, and generalizing Now that you have a grounding in working with vectors in concrete form, we go abstract again. Here’s the plan: (1) The set of all vectors of a given length n has operations of addition and scalar multiplication on it. (2) Vector addition and scalar multiplication have certain properties (which we’ve proven already) (3) We say that the set of vectors of length n with these operations form a vector space Rn (4) So let’s take any mathematical object (like a m × n matrix or even a function ) and define operations that we’ll call “addition” and “scalar multiplication” on that object. If the set of those objects satisfies all the same properties that vectors do ... we’ll call them ... I know ... we’ll call them vectors, and say that they form a vector space. Functions can be vectors! Matrices can be vectors! So, we know A vector in Rn is a mathematical object that we visualize as an n dimensional directed line segment: The vector <v1,v2, ..., vn > describes a ray with tail at the origin < 0, 0, ...0 > and head located at <v1,v2, ..., vn >. Formally, the vector space Rn consists of a set of ordered n-tuples V ,andafieldK (the real numbers). The elements of V are the vectors,andtheelementsofK are the scalars. In English, that simply means that R3 is the set of all 2 three component vectors <v1,v2,v3 >, R is the set of all two component vectors <v1,v2 >, and so on. Operations are defined on these objects: • Vector addition: Let u =<u1,u2, ..., un > and v =<v1,v2, ..., vn > be elements of V .Thenw = u + v is in V as well and is defined by w =<w1,w2, ..., wn >:=<u1 + v1,u2 + v2, ..., un + vn > • Scalar multiplication: Let u =<u1,u2, ..., un >∈ V and k ∈ K. Then Then w = ku is in V as well and is defined by w =<w1,w2, ..., wn >:=<ku1,ku2, ..., kun > And, we’ve done all this - worked with adding vectors and multiplying scalars. One thing we’ve noticed is that there’s a compatibility thing- we can add three component vectors to other three component vectors [e.g. < 1, 2, 3 > + < −1, 4, 7 >=< 0, 6, 10 >], two component vectors to two component vectors [e.g < 1, −6 > + < −3, 1 >=< −2, 5 >], but we can’t add a three component vector to a two component vector [< 1, 2, 3 > + < −1, 5 > makes no sense]. The vector spaces are self contained in this way - there’s no overlap between R3 and R2. This feature is part of what is known as closure. In fact, there are ten properties we can prove about vectors in a vector space, and have in fact proven many of them already. The elements of V and K with these operations exhibit the following properties: Closure axioms: • u + v ∈ V . [The sum of two vectors is still a vector (of length n). This is true by the definition of vector addition.] We say Rn is closed under addition. • ku ∈ V . [The scalar multiple of a vector is stil of vector (of length n). This is true by the definition of scalar multiplication.] We say Rn is closed under addition. Addition axioms: • Commutativity of vector addition and scalar multiplication: u + v = v + u and kv = vk • Associativity of vector addition: u +(v + w)=(u + v)+w • An additive identity, called the zero vector: u + 0 = u • Each element u of V has an additive inverse, denoted −u: u + −u = 0 Scalar multiplication axioms • Scalar multiplication distributes over vector addition: k(u + v)=ku + kv • A vector distributes over a sum of scalars: (k + m)u = ku + mu • Scalar-vector associativity: k(mu)=(km)u • There is a multiplicative identity, denoted 1, for the operation of scalar multiplication: 1u = u Vector spaces are also known as linear spaces. And if anyone asks to to “show a set with some operations is a vector space”, it means ... you must work one by one through that list of ten properties, proving ALL of them for that set with those operations. Here are the proofs of all the properties for the vector space R3: Example: 3 Consider R - the set of all vectors in the form v =<v1,v2,v3 > along with real number scalars. Let vector addition be defined in the usual way: w = u + v is defined by adding components <w1,w2,w3 >=<u1 + v1,u2 + v2,u3 + v3 > And scalar multiplication w = kv by <w1,w2,w3 >=<ku1,ku2,ku3 > Prove R3 is a vector space by showing all the properties hold. Solution: So, you just go down the line: 1. Is R3 closed under addition? I.e. if you add two three component vectors, do you get another three component vector? YES, by definition. 2. Is R3 closed under scalar multiplication? I.e. if you multiply a scalar times a three component vectors, do you get another three component vector? YES, by definition. 3. Is vector addition commutative? YES. u + v =<u1 + v1,u2 + v2,u3 + v3 >=<v1 + u1,v2 + u2,v3 + u3 >= v + u 4. Is vector addition associative? YES. u+(v+w)=<u1+(v1+w1),u2+(v2+w2),u3+(v3+w3) >=< (u1+v1)+w1, (u2+v2)+w2, (u3+v3)+w3 >=(u+v)+w 5. Is there a zero vector that acts as an additive identity? YES, the vector < 0, 0, 0 >. u + 0 =<u1,u2,u3 > + < 0, 0, 0 >=<u1 +0,u2 +0,u3 +0>=<u1,u2,u3 >= u 6. Is there a vector that acts as a negative (additive inverse)? YES, the vector −u for any u. u +(−u)=<u1,u2,u3 > + < −u1, −u2, −u3 >=<u1 +(−u1),u2 +(−u2),u3 +(−u3) >=< 0, 0, 0 >= 0 7. Does a scalar distribute over the sum of two vectors? YES. k(u + v)=k<u1 + v1,u2 + v2,u3 + v3 > = <k(u1 + v1),k(u2 + v2),k(u3 + v3 >) = <ku1 + kv1,ku2 + kv2,ku3 + kv3 > = <ku1,ku2,ku3 > + <kv1,kv2,kv3 > = k<u1,u2,u3 > +k<v1,v2,v3 >= ku + kv 8. Does a vector distribute over the sum of two scalars? YES. (k + c)u =(k + c) <u1,u2,u3 > = < (k + c)u1, (k + c)u2, (k + c)u3 > = < (ku1 + cu1,ku2 + cu2,ku3 + cu3 > = <ku1,ku2,ku3 > + <cu1,cu2,cu3 > = k<u1,u2,u3 > +c<u1,u2,u3 >= ku + cu 9. Associativity with scalar multiplication? YES. (kc)u =(kc) <u1,u2,u3 > = < (kc)u1, (kc)u2, (kc)u3 > = <k(cu1),k(cu2),k(cu3) > = k<cu1,cu2,cu3 >= k(cu) 10. Scalar muliplicative identity? YES, the scalar k =1 1u =1<u1,u2,u3 >=< 1u1, 1u2, 1u3 >=<u1,u2,u3 >= u Notice that all of the proceeding are just typical element proofs - we break it down to the component level and work with the fact that the individual components are scalars. This is exactly what we’ve been doing with matrices, and, of course, vectors are matrices - they’re arrays which are n × 1or1× n. Where we’re going with this is not an exhaustive proof of all these things for other items - what we want to get out of this is that a vector space is something that satisfies all these properties, and move toward the idea of a subspace. For example, consider the set of all vectors of the form <u1,u2, 0 > - vectors where the third component is zero [e.g. < 1, 5, 0 >, < −6, 10, 0 >]. These vectors are a subset of R3, but are they a subspace of R3? In other words, can you prove the ten properties about them? But wait - you don’t need to prove all ten of the properties again! Vectors are vectors - all the adding and scalar multiuplying works just as before. The only thing that can possibly change is CLOSURE, and that’s the only one I need to check. If I add two of these vectors, do I stay in the space? (Does the result still have a zero as the third component?) YES: <u1,u2, 0 > + <v1,v2, 0 >=<u1 + v1,u2 + v2, 0+0>=<u1 + v1,u2 + v2, 0 > And scalar multiply? YES: k<u1,u2, 0 >=<ku1,ku2,k(0) >=<ku1,ku2, 0 > 3 That’s all we need - the set of vectors in the form <u1,u2, 0 > is closed, and a subspace of R . Compare that to vectors in the form <u1,u2, 1 >.Ifweaddtwoofthesevectors, <u1,u2, 1 > + <v1,v2, 1 >=<u1 + v1,u2 + v2, 1+1>=<u1 + v1,u2 + v2, 2 > we’ve gone out of the set - the third component of the result is 2, not 1. This set is not closed, and not a subspace of R3..