L E C T U R E 1 2 : R N A N D V E C T O R S PA C E S. Monday, October 3

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1 L E C T U R E 1 2 : R N A N D V E C T O R S PA C E S Monday, October 3 In the last few lectures we considered various properties of matrices in order to better understand the coefficient matrices of our linear systems. In this lecture we re going to consider shift gears and consider the solutions of our linear systems. Suppose we have a linear system with m equations and n variables. Then we can express this system as Ax b, where A is the coefficient matrix of the linear system and b is the constant vector. The solution to this system, if it exists, is the vector» fi x x 1 x 2 ffi ffi ffi. fl where each of the x i are real numbers. That is, x is a n-tuple of real numbers. Since writing column vectors as above takes up a lot of space, we often them as follows:» fi x 1 x 2 ffi ffi ffi. fl px 1, x 2,..., x n q T x 1 x 2 x n, x n where the superscript of T denotes the transpose. x n Definition 0.1. We denote the set of all n-tuples of real numbers by R n. In set builder notation R n tpx 1, x 2,..., x n q : x i P Ru In the next section we want to study the set R n, in particular the algebraic structure of this set, which we call a vector space. 1

2 1 vector spaces To motivate the definition of a vector space, let s consider a set whose elements are particularly easy to visualize, namely R 2 tpx, yq : x, y P Ru. An element px, yq in the set R 2 can be thought of as the arrow from the origin p0, 0q to the given point px, yq, and it is this perspective in which these elements are often considered. These arrows were typically used to represent physical notations like force or velocity, because they give a direction (whichever way the arrow is pointing) and a magnitude (the length of the arrow). But their utility doesn t end there. For example, if I were to apply a force to an object, let s say pushing a chair to the north side of a room, and at the same time you were to push the chair to the west side of the room, then the chair will move northwest. The resulting arrow can be obtained by adding the two arrows giving our respective contributions. How do we add elements of R 2? Componentwise, that is, if v 1 px 1, y 1 q and v 2 px 2, y 2 q are two elements of R 2, then v 1 ` v 2 px 1 ` x 2, y 1 ` y 2 q. v 1 ` v 2 v 1 v 2 One can show that this vector addition is commutative and associative. In addition to adding two elements from R 2 together, we can also scale these elements. If a P R and v px, yq P R 2, then av pax, ayq P R 2. Depending on the value of a, the resulting arrow av could be longer or shorter than the original vector v. 2

3 2 v v 1 2 v So to summarize, the set R 2 can be thought to consist of sets of arrows that we can add together and scale by multiplying by a constant. We want to generalize this basic structure to arbitrary sets. 2 definition of an abstract vector space Definition 2.1. A vector space V over R is a set with two operations: pvector additionq A function that assigns to each v, w P V an element v ` w P V. pscalar multiplicationq A function that assigns to each pair a P R and v P V an element av P V. such that the following conditions are satisfied: pcommutativityq v ` w w ` v for all v, w P V passociativityq pv ` wq ` u v ` pw ` uq for all v, w, u P V, apbvq pabqv for all v P V and a, b P R. padditive identityq There exists an element 0 P V, such that v ` 0 v for all v P V. padditive inverseq For every v P V, there exists an element w P V such that v ` w 0. pmultiplicative identityq 3

4 1v v for all v P V. pdistributive propertiesq apv ` wq av ` aw for all v, w P V and a P R, pa ` bqv av ` bv for all v P V and a, b P R. An element of V is called a vector. An element of R is called a scalar. Now, that was a long and abstract definition, so let s look at some concrete examples to help make sense of it. We start with our motivating example. Example 1. Recall that R n tpx 1,..., x n q : x i P Ru. If we define vector addition and scalar multiplication by px 1,..., x n q ` py 1,..., y n q px 1 ` y 1,..., x n ` y n q apx 1,..., x n q pax 1,..., ax n q, one can check (and you should!) that R n is a vector space over the field R. Example 2. Denote the set of all polynomials with coefficients in R by PpRq. That is, every element of the set PpRq is of the form a n x n ` a 1 x ` a 0 for some n ě 0, where a i P R for all i. If we define vector addition and scalar multiplication is the natural way, i.e. pp ` qqpxq ppxq ` qpxq papqpxq appxq then PpRq is a vector space over R. 4

5 Example 3. Let M 2 prq denote the set of 2 ˆ 2 matrices with entries from R. Then M 2 prq is a vector space over R with respect to vector addition and scalar multiplication given by ˆa c b d ˆe ` g ˆa α c f h b d ˆa ` e b ` f c ` g d ` h ˆαa αb. αc αd Let s look at a couple of examples that aren t vector spaces. Example 4. Let s once again consider the set R n. Suppose we define vector addition and scalar multiplication by pa 1,..., a n q ` pb 1,..., b n q pa 1 b 1,..., a n b n q apa 1,..., a n q paa 1,..., aa n q. Is F n with these operations still a vector space? No, because, for example, addition is not commutative. Example 5. Let V R. Define addition and scalar multiplication as follows u ` v uv av av. Is V a vector space over R? No. Note that in this case, the additive identity would be 1 since v ` 1 v 1 v for all v P V. However, this means that there is no real number v that can be the additive inverse of 0 since 0 ` v 0 v 0 1 for all v. These operations would also fail to satisfy the distributive properties. 5

6 3 subspaces Definition 3.1. Let V be a vector space over R. A subset U Ă V is a subspace of V if U is a vector space with respect to the addition and scalar multiplication defined on V. Example 6. Let n ě 1 be an integer. Recall that the set R n tpx 1,..., x n q : x i P Fu is a vector space with addition and scalar multiplication given by px 1,..., x n q ` py 1,..., y n q px 1 ` y 1,..., x n ` y n q Consider the subset apx 1,..., x n q pax 1,..., ax n q. U tpx 1,..., x n 1, 0q : x i P Ru Ă R n. One can verify that U is vector space over R under the above operations of addition and scalar multiplication, that is, U is a subspace of R n. Example 7. Recall that the set of polynomials with coefficients in R, which we denote PpRq, is a vector space over R with addition and scalar multiplication given by Consider the subset pp ` qqpxq ppxq ` qpxq papqpxq appxq. U tppxq P PpFq : pp1q 0u Ă PpRq. One can verify that U is vector space over R under the above operations of addition and scalar multiplication, that is, U is a subspace of PpRq. We ll also prove this later. 6

7 Example 8. Suppose we have a homogeneous linear system win m equations and n variables. Then we can express this system as Ax 0, where 0 here denotes the zero vector. The solutions to this system form a subspace of R n It turns out that verifying a subset of a vector space is a subspace is easier than verifying that a set is a vector space. The reason for this is that many of the conditions that a set must satisfied in order to be a vector space, are automatically satisfied by virtue of the fact that we are considering a subset of a vector space. In fact, we only need to check three things! Theorem 3.1. Let V be a vector space, and let U be a subset of V. Then U is a subspace of V if and only if the following three conditions hold: (i) U is non-empty, (ii) x ` y P U for all x, y P U, (iii) ax P U for all x P U and a P R. In other words, in order for a subset U of a vector space V to be a subspace, it must be non-empty, closed under addition, and closed under scalar multiplication. Proof of Theorem 3.1. The statement of this theorem is an if and only if statement, therefore we have to show two things. First, we have to show that if U is a subspace of V, then it satisfies conditions (i)-(iii) (this is the only if part). Second, we must also show that if U satisfies conditions (i)-(iii), then it is a subspace of V (this is the if part). Let s begin with the only if part. Suppose U is a subspace of V. Then U is a vector space, which implies it has an additive identity. Therefore, U is non-empty. Furthermore, the fact that U is a vector space implies that it is closed under addition and scalar multiplication, that is, it satisfies (ii) and (iii). Next, we consider the if part. 7

8 Suppose U satisfies conditions (i)-(iii). Since U is a subset of V, we know that addition is associative and commutative, 1v v for all v P V, addition and scalar multiplication satisfy the distributive properties. Therefore, all we need to show is that U has an additive identity, and every v P U has an additive inverse. Let v P U. By (iii) we know that both 0v and 1v are elements of U, and we know that 0v 0 and 1v v. Hence, U has an additive identity and every element of U has an additive inverse. When you want to show that a subset U of a vector space V is non-empty, it is often easiest to show that U contains the additive identity of V. In fact, we could replace the first condition of the above theorem with 0 P U, where 0 is the additive identity of V. Let s use the above theorem to show that the subsets we considered in Examples 6 and 8 are subspaces. Example 9. Recall the subset U tpx 1,..., x n 1, 0q : x i P Fu Ă R n. To show that U is a subspace of R n, we first want to show that U is non-empty. To do so, we note that p0,..., 0q P U. Next, we want to show that U is closed under addition. px 1,..., x n 1, 0q, py 1,..., y n 1, 0q P U, then we have Let px 1,..., x n 1, 0q ` py 1,..., y n 1, 0q px 1 ` y 1,..., x n 1 ` y n 1, 0q, which is an element of U. Hence, U is closed under addition. Finally, we want to show that U is closed under scalar multiplication. Let px 1,..., x n 1, 0q P U and a P R. Then, apx 1,..., x n 1, 0q pax 1,..., ax n 1, 0q, which is an element of U. Hence, U is closed under scalar multiplication. Since U is non-empty, closed under addition, and scalar multiplication, it is a subspace of V. 8

9 Example 10. Recall the subset U tppxq P PpRq : pp1q 0u Ă PpRq. To show that U is a subspace of V we first want to show that it is non-empty. Since the zero polynomial 0 P U, we know that this is the case. Next, we want to show that U is closed under addition. Let ppxq, qpxq P U. We want to show that pp ` qqpxq ppxq ` qpxq is an element of U. We have pp ` qqp1q pp1q ` qp1q 0 which means pp ` qqpxq P U. Hence, U is closed under addition. Finally, we want to show that U is closed under scalar multiplication. Let a P R and ppxq P U. We want to show that papqpxq appxq is an element of U. We have papqp1q app1q a0 0, which implies papqpxq P U. Hence, U is closed under scalar multiplication. Since U is non-empty, closed under addition, and scalar multiplication, it is a subspace of V. Example 11. Suppose we have a homogeneous linear system Ax 0 with n-variables. Let U denote the set of solutions to this system. Then U Ă R n. To show that U is a subspace of R n we first want to show that it is non-empty. Since the zero vector 0 P U, we know that this is the case. Next, we want to show that U is closed under addition. Let x 1, x 2 P U. We want to show that x 1 ` x 2 is also a solution. By the distributive property of matrix multiplication, we have Apx 1 ` x 2 q Ax 1 ` Ax 2 0 ` 0 0, which means x 1 ` x 2 P U. Hence, U is closed under addition. Finally, we want to show that U is closed under scalar multiplication. Let a P R and let x P U. We want to show that ax is also a solution. We have Apaxq apaxq a0 0, which implies ax P U. Hence, U is closed under scalar multiplication. Since U is non-empty, closed under addition, and scalar multiplication, it is a subspace of R n. 9