Chapter 4. Vector Space Examples. 4.1 Diffusion Welding and Heat States

Chapter 4 Vector Space Examples 4.1 Diffusion Welding and Heat States In this section, we begin a deeper look into the mathematics for diffusion welding application discussed in Chapter 1. Recall that diffusion welding can be used to adjoin several smaller rods into a single longer rod, leaving the final rod just after welding with varying temperature along the rod but with the ends having the same temperature. Recall that we measure the temperature along the rod and obtain a heat signature like the one seen in Figure 1.4 of Chapter 1. Recall, also, that the heat signature shows the temperature difference from that at the ends of the rod. Thus, the initial signature (along with any subsequent signature will show values of 0 at the ends. The heat signature along the rod can be described by a function f 0 : [0, L] R, where L is the length of the rod and f 0 (0 = f 0 (L = 0. The quantity f 0 (x is the temperature difference on the rod at a position x in the interval [0, L]. Because we are collecting and storing heat measurements along the rod, it makes sense that we are only able to collect finitely many such measurements. Thus, we discretize the heat signature f by sampling at only m locations along the bar. If we space the m sampling locations equally, then for Δx = L, we can choose the sampling locations to be m+1 Δx, 2Δx,..., mδx. Since the heat measurement is zero (and fixed at the endpoints we do not need to sample there! The set of discrete heat measurements at a given time is called a heat state, as opposed to a heat signature, which, as discussed earlier, is defined at every point along the rod. We can 57

58 CHAPTER 4. VECTOR SPACE EXAMPLES record the (discretized heat state as the vector u = [u 0, u 1, u 2,..., u m, u m+1 ] = [0, f(δx, f(2δx,..., f(mδx, 0]. Here, if u j = f(x for some x [0, L] then u j+1 = f(x + Δx and u j 1 = f(x Δx. Figure 4.1 shows a (continuous heat signature as a solid blue curve and the corresponding measured heat state indicated by the regularly sampled points marked as circles. Figure 4.1: An example 1D heat signature, f, is shown as a blue line. A heat state is the discrete collection of m + 2 regularly sampled temperatures, {u 0, u 1,, u m+2 }, shown as red circles. Both heat signature and heat state have zero temperature at the end points x = 0 and x = L. As the heat diffuses through the rod, the new heat signatures will also be described by functions f t : [0, L] R, where t is the time measured since the welding was completed. The discretized heat states corresponding to these signatures will form vectors as well. We define scalar multiplication and vector addition of heat states componentwise (in the same way we define the operations on vectors in R m+2. Denote the set of all heat states with m + 2 entries (assumed to have zero temperature at the endpoints by H m (R. H m (R is a vector space (see Exercise 1. 4.2 Function Spaces We ve seen that the set of discretized heat states of the preceding example forms a vector space. These discretized heat states can be viewed as realvalued functions on the set of points that are locations along the rod. In

4.2. FUNCTION SPACES 59 fact, function spaces such as H m (R are very common and useful constructs for solving many physical problems. The following are some such function spaces. Example 4.2.1. Let F = {f : R R}, the set of all functions whose domain is R and whose range is a subset of R. F is a vector space with scalars taken from R. We define addition and scalar multiplication (on functions pointwise. That is, given two functions f and g and a real scalar α, we define the sum f + g by (f + g(x := f(x + g(x and the scalar product αf by (αf(x := α (f(x. Now, for f, g, h F and α, β R, we verify the 10 properties of Definition 3.2.1: f : R R and g : R R. Based on the definition of addition, f + g : R R. So F is closed over addition. Similarly, F is closed under scalar multiplication. Addition is commutative: (f + g(x = f(x + g(x = g(x + f(x = (g + f(x. So, f + g = g + f. Addition is associative: ((f + g + h(x =(f + g(x + h(x = (f(x + g(x + h(x = f(x + (g(x + h(x So (f + g + h = f + (g + h. Scalar multiplication is associative: =f(x + (g + h(x = (f + (g + h(x. (α (β f(x = (α (βf(x = (αβf(x = ((αβ f(x. So α (β f = (αβ f. Scalar multiplication distributes over vector addition: (α (f + g(x = α (f + g(x = α (f(x + g(x So α(f + g = αf + αg. = α f(x + α g(x = (α f + α g(x.

60 CHAPTER 4. VECTOR SPACE EXAMPLES Scalar multiplication distributes over scalar addition: ((α + β f(x = (α + β f(x = α f(x + β f(x So, (α + β f = α f + β f. = (α f + β f(x. F contains the zero vector. Consider the function z(x = 0 for every x R. We see that (z+f(x = z(x+f(x = 0+f(x = f(x = f(x+0 = f(x+z(x = (f+z(x. That is, z + f = f + z = f. Thus, the function z is the zero vector in F. Every vector f has in additive inverse f in F. Observe, (f +( f(x = f(x + ( f(x = f(x f(x = 0 = z(x, where z is defined above. So, f + ( f = z. The scalar identity is the real number 1. Notice: (1 f(x = 1 f(x = f(x. So, 1 f = f. In the vector space F, vectors are functions. Example vectors in F include sin(x, x 2 + 3x 5, and e x + 2. Not all functions are vectors in F (see Exercise 5. Example 4.2.2. Let P n (R be the set of all polynomials of degree less than or equal to n with coefficients from R. That is, P n (R = {a 0 + a 1 x + + a n x n a k R, k = 0, 1,, n}. Let f(x = a 0 + a 1 x + + a n x n and g(x = b 0 + b 1 x + + b n x n be polynomials in P n (R and α R. Define addition and scalar multiplication component-wise: (f + g(x = (a 0 + b 0 + (a 1 + b 1 x + + (a n + b n x n, (αf(x = (αa 0 + (αa 1 x + + (αa n x n. With these definitions, P n (R is a vector space. Vectors in this space are polynomials. Example 4.2.3. A function f : R R is called an even function if f( x = f(x for all x R. The set of even functions is a vector space over R with the definitions of vector addition and scalar multiplication given in Example 4.2.1.

4.3. MATRIX SPACES 61 4.3 Matrix Spaces A matrix is an array of real numbers arranged in a rectangular grid, for example, let ( 1 2 3 A =. 5 7 9 The matrix A has 2 rows (horizontal and 3 columns (vertical, so we say it is a 2 3 matrix. In general, a matrix B with m rows and n columns is called an m n matrix. We say the dimensions of the matrix are m by n. Any two matrices of the same dimension are added together by adding their entries (individual numbers component-wise. A matrix is multiplied by a scalar by multiplying all of its entries by that scalar (that is, multiplication of a matrix by a scalar is also component-wise as in Example 3.2.3. Example 4.3.1. Let Then A = ( 1 2 3 5 7 9 ( 1 0 1, B = 2 1 0 A + B = ( 2 2 4 3 8 9 ( 1 2, and C = 3 5 but since A M 2 3 and C M 2 2, the definition of matrix addition does not work. That is, A + C is undefined. Using the definition of scalar multiplication, we get 3 A = ( 3 1 3 2 3 3 3 5 3 7 3 9 =, ( 3 6 9 15 21 27 With this understanding of operations on matrices, we can now discuss (M m n, +, as a vector space over R... Theorem 4.3.1. Let m and n be in N, the natural numbers. The set M m n of all m n matrices with real entries, together with the operations of addition and scalar multiplication is a vector space over R.

62 CHAPTER 4. VECTOR SPACE EXAMPLES Proof. The proof proceeds along the same lines as the alternate proof for Example 3.2.3. Notice that because the matrix addition and scalar multiplication are defined component-wise, the size of the matrix is not changed. Thus, M m n is closed under addition and scalar multiplication. Now, since the entries of every matrix in M m n are all real, and addition and scalar are defined component-wise, then M m n inherits the vector space properties from R. Notation: If we want to refer to an entry (or one of the individual numbers of a matrix, we can refer to its position. For example, we write A = (a i,j to mean that we will use lower case a with subscripts designating the position to denote the entries. In this case, the number in the first row and second column of the matrix A above is denoted by a 1,2 = 2. More generally, the value in the i-th row and j-th column is a i,j. With this notation, we can formally define component-wise addition and scalar multiplication. Definition 4.3.1. Let A = (a i,j and B = (b i,j be m n matrices. We define matrix addition by A + B = C = (c i,j, where c i,j = a i,j + b i,j. For fun, try using this notation to describe the product ca where c R and A M m n. Caution: The spaces M m n and M n m are not the same! The matrix 1 2 3 M 1 = 0 4 10 7 7 8 1 0 0 is a 4 3 matrix, whereas M 2 = 1 0 7 1 2 4 7 0 3 10 8 0 is a 3 4 matrix. As we saw above, these two matrices cannot even be added together!

4.3. MATRIX SPACES 63 As with R, M m n has other operations with which it is associated. In fact, we are able to multiply matrices, but unlike matrix addition the matrix product is not defined component-wise. Definition 4.3.2. Given an n m matrix A = (a i,j and an m l matrix B = (b i,j, we define the matrix product C = AB where c i,j = m k=1 a i,kb k,j. We will call this operation matrix multiplication. Notice that there is no between the matrices, rather they are written in juxtaposition to show a difference between the notation of a scalar product and the notation of a matrix product. Notice that the definition requires that the number of columns of A is the same as the number of rows of B for the product AB. Example 4.3.2. Let P = 1 2 3 4 5 6, Q = 2 1 1 1 2 1, and R = ( 2 0 1 2 Since both P and Q are 3 2 matrices, we see that the number of columns of P is not the same as the number of rows of Q. Thus, P Q is not defined. But, since P has 2 columns and R has 2 rows, P R is defined. Let s compute it. We can compute each entry as in Definition 4.3.2. Position Computation (i, j p i,1 r 1,j + p i,2 r 2,j (1, 1 1 2 + 2 1 (1, 2 1 0 + 2 ( 2 (2, 1 3 2 + 4 1 (2, 2 3 0 + 4 ( 2 (3, 1 5 2 + 6 1 (3, 2 5 0 + 6 ( 2 Typically, when writing this out, we write it as P R = 1 2 3 4 5 6 ( 2 0 1 2 = 1 2 + 2 1 1 0 + 2 ( 2 3 2 + 4 1 3 0 + 4 ( 2 5 2 + 6 1 5 0 + 6 ( 2. = 4 4 10 8 16 12.

64 CHAPTER 4. VECTOR SPACE EXAMPLES In the above example, the result of the matrix product was a matrix of the same size as P. Let s do another example to show that this is not always the case. Example 4.3.3. Let M 1 = ( 1 0 2 2 1 1 and M 2 = 2 1 1 0 1 1 1 1 2 1 1 2. Notice that M 1 is 2 3 and M 2 is 3 4. So, we can multiply them and the resulting matrix will be 2 4. Indeed, we can find the (i, j entry of M 1 M 2 by multiplying the ith row of M 1 by the jth column of M 2. We do this for each entry as follows Position Computation (i, j a i,1 b 1,j + a i,2 b 2,j + a i,3 b 3,j (1, 1 1 2 + 0 1 + 2 2 (1, 2 1 1 + 0 ( 1 + 2 1 (1, 3 1 ( 1 + 0 1 + 2 1 (1, 4 1 0 + 0 1 + 2 2 (2, 1 2 2 + ( 1 1 + 1 2 (2, 2 2 1 + ( 1 ( 1 + 1 1 (2, 3 2 ( 1 + ( 1 1 + 1 1 (2, 4 2 0 + ( 1 1 + 1 2 Thus, AB = ( 1 0 2 2 1 1 2 1 1 0 1 1 1 1 2 1 1 2 = ( 6 3 1 4 5 4 2 1 Notice that in Example 4.3.2, RP is not defined so RP P R. Similarly, in Example 4.3.2, BA AB. But, one might wonder whether matrix multiply is commutative when the matrix product is defined both ways. That is, we might wonder if AB = BA when both AB and BA are defined. Note that in order for AB and BA to be defined, both must be square matrices (the number of rows and columns are the same. Let us explore this in the next example..

4.4. MISCELLANEOUS VECTOR SPACES 65 Example 4.3.4. Let A = ( 1 2 3 1 Let us compute both AB and BA. and B = ( 2 0 2 3. ( ( 1 2 2 0 AB = 3 1 2 3 ( 2 6 = 8 3 ( = 1 2 + 2 ( 2 1 0 + 2 3 3 2 + ( 1 ( 2 3 0 + ( 1 3 ( ( 2 0 1 2 BA = 2 3 3 1 ( 2 4 = 7 7 ( 2 1 + 0 3 2 2 + 0 ( 1 = 2 1 + 3 3 2 2 + 3 ( 1 Notice, even in the simple case of 2 2 matrices, matrix multiply is not commutative. Note: Because matrix multiply is not generally defined between two matrices in M m n, we can see that M m n is not, in general, closed under matrix multiply. Since the matrix product ( is not commutative, we can also see that (M n n,, is not a vector space over R. Although matrix products are not ideal as a vector space operation, we will find them very useful in later chapters. 4.4 Miscellaneous Vector Spaces The next few examples are related to linear equations, which will be fundamental objects of study in this course. We start with the definition of a linear equation. Definition 4.4.1. A linear equation is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where b, a 1,, a n F are called coefficients and x 1, x 2,, x n are called variables.

66 CHAPTER 4. VECTOR SPACE EXAMPLES For problems of two variables we often use variable names x and y and the equation looks like αx + βy = b. For problems of three variables we often use variable names x, y and z. Note that the scalars a 1, a 2,, a n and b are not necessarily nonzero. Some examples of linear equations are x=0, 3x + 2y 5z = 7, and x 1 + 3x 2 x 3 = 0. A solution to a linear equation with n variables is a point in R n whose coordinates satisfy the equation. More precisely, Definition 4.4.2. Let a 1 x 1 + a 2 x 2 + + a n x n = b be a linear equation in n variables, x 1, x 2,, x n. Then (v 1, v 2,, v n R n is a solution to the linear equation if a 1 v 1 + a 2 v 2 + + a n v n = b. Notice that in Definition 4.4.2, we substitute the components of a solution into the corresponding variable. When we do this, the resulting equation holds. Example 4.4.1. The vector (2, 3, 1 R n is a solution to the linear equation because 3x + 2y 5z = 7 3(2 + 2(3 5(1 = 7. What does all of this have to do with vector spaces? We can consider the set of all solutions to a linear equation, and we can ask whether this set satisfies the properties of a vector space. Example 4.4.2. Let V R 3 be the set of all solutions to the equation x 1 + 3x 2 x 3 = 0. That is, x 1 V = x 2 R 3 x 1 + 3x 2 x 3 = 0 x 3. The set V, together with the operations + and inherited from R 3 forms a vector space.

4.4. MISCELLANEOUS VECTOR SPACES 67 Proof. Note that Properties (P3-(P7 and (P10 of Definition 3.2.1 are automatically satisfied because the operations + and on R 3 satisfy these properties. Hence we need only check properties (P1, (P2, (P8, and (P9. (P1 Let u = (u 1, u 2, u 3 and v = (v 1, v 2, v 3 be vectors in V. Then we have u + v = (u 1 + v 1, u 2 + v 2, u 3 + v 3. To determine if u + v is in V, we compute (u 1 + v 1 + 3(u 2 + v 2 (u 3 + v 3. Since u and v solve the equation, u 1 + 3u 2 u 3 = 0 and v 1 + 3v 2 v 3 = 0, and so (u 1 + v 1 + 3(u 2 + v 2 (u 3 + v 3 = (u 1 + 3u 2 u 3 + (v 1 + 3v 2 v 3 = 0. Since u and v are arbitrary vectors in V it follows that V is closed under addition. (P2 Let u = (u 1, u 2, u 3 be in V and α R. Then αu = (αu 1, αu 2, αu 3, and so αu 1 + 3αu 2 αu 3 = α(u 1 + 3u 2 u 3 = α0 = 0. Hence V is closed under scalar multiplication. (P8 The vector 0 in R 3 is contained in V because 0 + 3 0 0 = 0. (P9 Let u = (u 1, u 2, u 3 be in V. Then the additive inverse of u is 1 u = ( u 1, u 2, u 3, which by property (P 2 is in V. Hence V contains additive inverses. We can also visualize this space on the coordinate axes. In this case the space V is a plane in R 3 that passes through the points (0, 0, 0, (1, 0, 1 and (0, 1, 3. Here is a very similar equation, however, whose solution set does not form a vector space. Example 4.4.3. Let V R 3 be the set of all solutions to the equation x 1 +3x 2 x 3 = 5. The solution set to this equation is a plane in R 3 that passes through the points (0, 0, 5, (1, 0, 4 and (0, 1, 2, in other words, it is the translation of the plane in Example 4.4.2 by 5 units down (in the negative x 3 direction. The set V is not a vector space. The proof is Exercise 9. In fact, we can generalize the preceding examples by considering the set of linear equations that are homogeneous.

68 CHAPTER 4. VECTOR SPACE EXAMPLES x 3 x 2 x 1 Figure 4.2: The vector space (V, +, of solutions to the equation x 1 + 3x 2 x 3 = 0 is a plane in R 3. Definition 4.4.3. We say that a linear equation a 1 x 1 + a 2 x 2 + + a n x n = b is homogeneous if b = 0. Example 4.4.4. Let {a 1, a 2,..., a n } R and consider the set V R n of all solutions to the homogeneous linear equation a 1 x 1 + a 2 x 2 + + a n x n = 0. The set V, together with the operations + and inherited from R n forms a vector space. The proof is Exercise 10. Question: Is the set of all solutions to an inhomogeneous linear equation (a linear equation that is not homogeneous a vector space? Example 4.4.5. A sequence of real numbers is a function s : N R. That is, s(n = a n for n = 1, 2, where a n R. A sequence is denoted {a n }. Let S(R be the set of all sequences. Let {a n } and {b n } be sequences in S(R and α in R. Define vector addition and scalar multiplication: {a n } + {b n } = {a n + b n }, and α {a n } = {αa n } S(R is a vector space with these (elementwise operations. Example 4.4.6. (Eventually Zero Sequences Let S fin (R be the set of all sequences that have a finite number of nonzero terms. Then S fin (R is a vector space with operations as defined in Example 4.4.5. Example 4.4.7. Let k R and denote by J k the set of all bar graphs with k bins. Here, we consider a bar graph to be a function from the set {1,..., k}

4.4. MISCELLANEOUS VECTOR SPACES 69 to R, and we visualize such an object in the familiar graphical way, as shown in Figure 4.3. Define addition and scalar multiplication on such bar graphs as follows. Let J 1 and J 2 be two bar graphs in J k and let α be a scalar in R. 1 2 3 4 5 6 7 8 Figure 4.3: A graphical representation of a vector in J 8 We define the 0 bar graph to be the bar graph where each of the k bars has height 0. J 1 + J 2 is defined to be the bar graph obtained by summing the height of the bars in corresponding bins of J 1 and J 2. α J 1 is defined to be the bar graph obtained by multiplying each bar height of J 1 by α. With these definitions, check that J k is a vector space. Note that the space of bar graphs is actually a discrete function space with the added understanding that the domain of the functions has a geometric representation (the bins for the bar graphs are all lined up in order from 1 to k. Example 4.4.8. Consider the image set of 7-bar LCD characters, D(Z 2, where Z 2 is the field that includes only the scalar values 0 and 1. Figure 4.4 shows ten example characters along with the image geometry.

70 CHAPTER 4. VECTOR SPACE EXAMPLES Figure 4.4: The ten digits of a standard 7-bar LCD display. For these images, white corresponds to the value zero and green corresponds to the value one. With element-wise definitions of addition and scalar multiplication as defined for the field Z 2, D(Z 2 is a vector space. Here are two examples of vector addition in D(Z 2 : + = + + = One can show that D(Z 2 is a vector space. (Exercise 14 4.5 Is My Set a Vector Space? We now have a variety of tests which can be used to determine if a given set is, or is not, a vector space.

4.6. EXERCISES 71 A set V (with given operations of vector addition and scalar multiplication is a vector space if it satisfies each of the ten properties of Definition 3.2.1. One must show that these properties hold for arbitrary elements of V and arbitrary scalars in R. In order to determine if a set is not a vector space, we have a few possible tests, most of which are direct consequence of one of the theorems of the previous chapter. A set V (with given operations of vector addition and scalar multiplication is not a vector space if any one of the following statements is true. 1. For some element(s in V and/or scalar(s in R, any one of the ten properties of Definition 3.2.1 is not true. 2. For some elements x, y, and z in V with x y, we have x + z = y + z. 3. The zero element of V is not unique. 4. Any element of V has a non-unique additive inverse. 5. If for some element x in V, 0 x 0. That is, the zero scalar multiplied by some element of V does not equal the zero element of V. 4.6 Exercises 1. Show that H m (R, the set of all heat states with m + 2 real entries, is a vector space. 2. Plot a possible heat state u for a rod with m = 12. In the same graph, plot a second heat state that corresponds to 2u. Describe the similarities and differences between u and 2u. 3. Plot a possible heat state u for a rod with m = 12. In the same graph, plot the heat state that corresponds to u + v, where v is another heat

72 CHAPTER 4. VECTOR SPACE EXAMPLES state that is not a scalar multiple of u. Decribe the similarities and differences between u and u + v. 4. Let P 2 = {ax 2 +bx+c a, b, c R}. Show that P 2 is a vector space with scalars taken from R and addition and scalar multiplication defined in the standard way for polynomials. 5. Determine whether or not the following functions are vectors in F as defined in Example 4.2.1. f(x = tan(x, g(x = x 5 5, h(x = ln(x. 6. Show that the set of even functions (see Example 4.2.3 is a vector space. 7. Let C([a, b] be the set of continuous functions on the interval [a, b]. Is C([a, b] a vector space with the definitions of vector addition and scalar multiplication given in Example 4.2.1? 8. Show that the set P n (R of Example 4.2.2 is a vector space. 9. Let V R n denote the set of all solutions to the linear equation x 1 + 3x 2 x 3 = 5. (a Use the algebraic definition of a vector space to show that V is not a vector space. (b Give a geometric argument for the fact that V is not a vector space. 10. Let {a 1, a 2,..., a n } R and consider the set V R n of all solutions to the homogeneous linear equation a 1 x 1 + a 2 x 2 + + a n x n = 0. Show that the set V, together with the operations + and inherited from R n forms a vector space. 11. Show that the set S of Example 4.4.5 is a vector space. 12. Show that the set S fin of Example 4.4.6 is a vector space. 13. Many of the sets above are subsets of larger vector spaces. Which of the 10 vector space properties do not necessarily carry down into the smaller subset? 14. Prove that D(Z 2 from Example 7.0.10 is a vector space.