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VECTOR SPACES: FIRST EXAMPLES

PIETER HOFSTRA

1. Definition So far in the course, we have been studying the spaces Rn and their subspaces, bases, and so on. In , it is common practice to ask what features of the objects you study are essential for the theory to work, and which are coincidental. We then separate the essential from the inessential, and leap into abstraction: we the essential features into a definition, so that we can develop the theory on a more general level. In the case of linear , it turns out that the essential features that make things work for Rn are the fact that you can add vectors and that you can do (and that they behave properly). For example, the notions of , linear independence, and only refer to and . This leads to the following definition: Definition 1.1 (Abstract Vector ). A consists of three things: • A V (to whose elements we sometimes refer as vectors). • An addition operation V × V → V , written, for v, w ∈ V , as (v, w) 7→ v + w. • A scalar multiplication operation R × V → V , written, for v ∈ V, c ∈ R, as (c, v) 7→ cv. These operations are required to satisfy the following axioms: (1) v + w = w + v for all v, w ∈ V (commutativity of + ) (2)( v + w) + z = v + (w + z) for all v, w, z ∈ V (associativity of + ) (3) There is an element 0 ∈ V such that 0 + v = v = v + 0 for all v ∈ V (4) For all v ∈ V there is a unique w with v + w = 0 (5)( cd)v = c(dv) for all c, d ∈ R and v ∈ V (6)1 v = v for all v ∈ V (7)( c + d)v = cv + dv for all c, d ∈ R and v ∈ V (distributivity 1) (8) c(v + w) = cv + cw for all c ∈ R and v, w ∈ V (distributivity 2)

2. Examples Of course, the motivating example of a vector space is Rn, with the usual addition of vectors and scalar multiplication. However, there are many other examples. Several of those will be closely related to Rn, but especially the last one ( spaces) is quite different in nature. Example 2.1. Let V be C, the set of complex . When discussing the algebra of complex numbers, we noted that we could define addition of complex numbers by (a + bi) + (c + di) = (a + c) + (b + d)i. Also, given a complex z = a + bi and a scalar () c, we can multiply them to get cz = c(a + bi) = (ca) + (cb)i.

With these notions of addition and scalar multiplication, it is easily verified that C forms a vector space. When we define complex numbers, we make use of the real numbers: given z = a + bi, the two parts a and b are just real numbers. That is, a z = a + bi is completely determined by the two real numbers a and b, and conversely, any two real numbers a and b uniquely determine a complex number a + bi. We might say that C is like R2, but with a different notation: in R2 we write elements as (a, b), while in C we write them as a + bi. Technically, however, they are different sets; in mathematics,

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such a situation is made precise by saying that the two sets, while different, are in one-one correspondence with each other. Schematically: C / R2

a + bi / (a, b) But more is true: addition of complex numbers (a + bi) + (c + di) agrees with that of addition of vectors in R2. On the one hand we have (a + bi) + (c + di) = (a + c) + (b + d)i, while on the other hand we have (a, b) + (c, d) = (a + c, b + d). This tells us that not only can we “go back and forth” between the complex numbers and R2, but this passage respects the addition: first adding two complex numbers and then turning it into a vector in R2 gives the same result as first turning both complex numbers into vectors in R2 and then adding them there. Diagrammatically: (a + bi) , (b + di) / (a, b) , (c, d) _ _ add individually add   (a + c) + (b + d)i / (a + c, b + d)

The same holds for scalar multiplication: given z = a + bi and a scalar c, first turning z into (a, b) ∈ R2 and then forming c(a, b) = (ca, cb) gives the same result as first doing scalar multiplication in C and then turning the result into a vector in R2:

c , (a + bi) / c , (a, b) _ _ multiply multiply   (ca) + (cb)i / (ca, cb) We may summarize all this informally by saying that not only is C just R2 in disguise as a set, but C is also R2 in disguise as a vector space! √ Example 2.2. A is an expression like p(x) = 2x2 − x + 3, or q(x) = x4 − 5, or r(x) = 1 3 2 x − x + 5. Formally: Definition 2.3. A polynomial (of degree n) is an expression of the form n n−1 anx + an−1x + ··· + a1x + a0

where x is a and where the a0, . . . , an are real numbers, called the coefficients of the polynomial. n n Given two p(x) = anx + ··· + a1x + a0 and q(x) = bnx + ··· + b1x + b0, we have p(x) = q(x) precisely when ai = bi for all i = 0, . . . , n. The set of all polynomials of degree n is denoted Pn[x], and the set of all polynomials (of arbitrary degree) P[x]. We now explain how to add polynomials and how to define scalar multiplication (although you already n n know this from calculus!). Given two polynomials p(x) = anx + ··· + a1x + a0 and q(x) = bnx + ··· + b1x + b0, we define p(x) + q(x) = (an + bn)x + ··· + (a1 + b1)x + (a0 + b0). That is, we add polynomials by adding their coefficients. n For scalar multiplication, consider Given two polynomials p(x) = anx + ··· + a1x + a0 and c ∈ R, and define n cp(x) = (can)x + ··· + (ca1)x + (ca0). With these definitions, Pn[x] becomes a vector space. Moreover, using the fact that we can always regard a polynomial of degree n as one of higher degree by adding coefficients of zero, we also get that P[x] is a vector space as well. n Just as with complex numbers, a polynomial p(x) = anx + ··· a1x + a0 is completely determined by n+1 its coefficients (an, . . . , a0), which form an element of R . Moreover, addition and scalar multiplication n+1 are defined coefficient-wise; this means that Pn[x] is just R in disguise! VECTOR SPACES: FIRST EXAMPLES 3

Example 2.4. Matrices. We know from algebra that matrices of the same size can be added entry-wise. Also, we can multiply any matrix by a scalar. This means that when we fix m, n, the set Mm,n of all mxn matrices is a vector space. Since an mxn matrix can be regarded as n column vectors in Rm (or, of course, as m row vectors in n mn R ), we find that Mm,n is R in disguise. For example, we can translate a 3x2 matrix to a vector in R6 as follows: a b c 7−→ (a, b, c, d, e, f) ∈ 6 d e f R Example 2.5. Our last example is different from the previous ones, in the sense that it is not a disguised form of Rn. Consider F[R], the set of all functions from the reals to the reals. So an element of F[R] is a function f : R → R. (Remember, a function has to give a unique output f(x) for every input x, so something like f(x) = 1/x is not an element of F[R], since it is not defined for x = 0.) From calculus, we know that it is possible to add functions (pointwise, as it is sometimes called): given two functions f, g : R → R, define a new function f + g : R → R by (f + g)(x) = f(x) + g(x) Note that we’re defining this new function f +g by specifying what it’s output is given an arbitrary input value x: to know a function h is to know h(x), for all possible values of x. Note also that this uses the + symbol twice: on the left hand side, it means addition of functions (which is what we’re trying to define) and on the right hand side it’s ordinary addition of real numbers (something we assume we understand already). For scalar multiplication, consider f : R → R and a real number c ∈ R. Define a new function cf : R → R by (cf)(x) = c · f(x) That is, on input x, the function cf returns the output value c times f(x). With these definitions of addition and scalar multiplication, F[R] is a vector space. Note that the “” in this vector space is the constant function z(x) = 0.

3. A bit of linear algebra We now proceed to examine the concepts of linear combinations and linear (in)dependence in the examples given above. Recall first the definition of linear combination: it makes sense in any vector space.

Definition 3.1 (Linear Combination). Let V be a vector space, and let v1, . . . , vk be elements of V .A linear combination of v1, . . . , vk is a vector v of the form

v = a1v1 + ... + akvk

for some real numbers a1, . . . , ak. Here are some examples:

2 2 Example 3.2. We work in P2[x]. The polynomial p(x) = 4x −4x is a linear combination of q(x) = x +2 and r(x) = 2x2 + x + 3 because p(x) = 12q(x) − 4r(x). Example 3.3. Is p(x) = 4x2 − 4x a linear combination of x2, x2 − x + 1 and 2x − 2? The answer is no. To see this, suppose we had 4x2 − 4x = a(x2) + b(x2 − x + 1) + c(2x − 2) for some numbers a, b, c. Collecting powers of x in the right hand side gives 4x2 − 4x = (a + b)x2 + (2c − b)x + (b − 2c). For this equation to hold, we need a + b = 4; (2c − b) = −4; b − 2c = 0. 4 PIETER HOFSTRA

This is a system of linear in the unknowns a, b, c, which is readily seen to be inconsistent. Hence it is not possible to find numbers a, b, c with the desired property, and therefore p(x) is not a linear combination of the given polynomials. Exercise 1. Is x3 − 3 a linear combination of 2x3 + 1, x2 − x + 1, x + 2, x3 + 1 and x2 − 1? √ 1 Example 3.4. This example is more a warning than anything else: while 5x, 4 x, 2x and 0 are all linear combinations (scalar multiples) of x, p(x) = x2 is not. Yes, x2 = x · x, but this involves multiplying x with itself, which is not a linear operation.

Note that each time we try to answer a question about Pn[x], we end up solving a linear system. This is not a coincidence, and can be made more explicit by “translating” the question into one about Rn+1. This is an important idea: it allows us to use all the old techniques and results. For example, consider again the problem in Ex. 3.3. We solved this by reasoning directly about polynomials, then setting up a system, then solving that system and then drawing a conclusion from that. But what’s much nicer and quicker is to translate the problem into one about R3. The key is that that a polynomial ax2 + bx + c can be regarded as a vector (a, b, c) in R3. So we translate all the polynomials involved, and we get the following question about vectors in R3: Example 3.5 (Translated version of Ex. 3.3). Is the vector (4, −4, 0) a linear combination of (1, 0, 0), (1, −1, 1) and (0, 2, −2)? Here, we can use any method we like, for example row reduction: 1 1 0 | 4  1 1 0 | 4 0 −1 2 | −4 ∼ 0 1 −2 | 4 0 1 −2 | 0 0 0 0 | 4 leading to the immediate conclusion that the system is inconsistent. Schematically, here’s what’s happening:

Problem about P3[x] 99K Solution to the original problem translate ↓ ↑ translate back Problem about R3 −→ Solution to problem inR3 Exercise 2. Try the same approach for Ex. 3.2.

For our next example, we work in M2,2. 1 2 2 0 0 1 0 0 Example 3.6. The matrix is a linear combination of , and , because 3 4 0 2 1 1 1 1 1 2 1 2 0 0 1 0 0 = + 2 + 3 4 2 0 2 1 1 1 1 1 2 2 0 0 1 0 0 However, is not a linear combination of , and . 3 4 0 2 1 1 0 1 Exercise 3. Verify the last claim by considering the equation 1 2 2 0 0 1 0 0 = a + b + c 3 4 0 2 1 1 0 1 and then solving for a, b, c. Example 3.7. Is the set 1 2 2 0 0 1 0 0 , , , 3 4 0 2 1 1 1 1 linearly independent? Solution. No, from Ex. 3.6 we know that the first is a linear combination of the other three.

4 We’ll now indicate how to solve questions like these using the idea that M2,2 is R in disguise. VECTOR SPACES: FIRST EXAMPLES 5

1 2 2 0 0 1 Example 3.8 ( of Ex. 3.6). Is the matrix is a linear combination of , 3 4 0 2 1 1 0 0 and ? 1 1 Translate to R4: is the vector (1, 2, 3, 4) a linear combination of (2, 0, 0, 2), (0, 1, 1, 1) and (0, 0, 1, 1)?

   1  2 0 0 | 1 1 0 0 | 2 0 1 0 | 2 0 1 0 | 2   ∼   0 1 1 | 3 0 0 1 | 1 2 1 1 | 4 0 0 0 | 0 Thus the answer to the translated problem is Yes, and hence the answer to the original problem is Yes, too.

Finally, some examples in F[R]. Example 3.9. f(x) = 2x+1 + x2 is a linear combination of g(x) = 2x − 1 and h(x) = x2 + 2, because 2g(x) + h(x) = 2(2x − 1) + x2 + 2 = 2 · 2x − 2 + x2 + 2 = 2 · 2x + x2 = 2x+1 + x2. Example 3.10. g(x) = sin2(x) is a linear combination of h(x) = cos2(x) and k(x) = 1 because of the Pythagorean sin2(x) + cos2(x) = 1, from which we get g(x) = k(x) − h(x)

1 2 2 Exercise 4. Use the same idea to show that m(x) = 2 cos (x) is a linear combination of 2 sin (x) − 3 and 3 sin2(x) + 5. Example 3.11. Recall the double angle formula sin(2x) = 2 sin(x) cos(x). One might be tempted to jump to the conclusion that sin(2x) is therefore a linear combination of sin(x) and cos(x), but that would be incorrect. Indeed, it is impossible to find a, b ∈ R with 2 sin(x) cos(x) = a sin(x) + b cos(x). To see why, keep in mind that this equality would have to hold for all x ∈ R. So if it were true, it should be true for x = 0, giving 2 sin(0) cos(0) = a sin(0) + b cos(0) which, because of sin(0) = 0, becomes 0 = b cos(0) = b. This forces b = 0, reducing the original equation to 2 sin(x) cos(x) = a sin(x). However, this should in particular be true of x = π/2, giving π π π 2 sin( ) cos( ) = a sin( ). 2 2 2 Because of cos(π/2) = 0, this becomes 0 = a sin(π/2) = a, forcing a = 0 as well. Contradiction, and therefore such a, b cannot exist.

4. Subspaces The notion of subspace of Rn only referred to linear independence and to spanning. Therefore it makes sense for general vector spaces: Definition 4.1. Let V be a vector space, and let U ⊆ V be a subset of V . Then U is a subspace of V provided the following three conditions hold. (1)0 ∈ U (2) v, w ∈ U implies v + w ∈ U (3) v ∈ U, c ∈ R implies cv ∈ U. Thus, a subspace of V is a collection of elements of V containing the zero vector, and is closed under addition and scalar multiplication. All three conditions have to hold; as soon as one of them fails, you no longer have a subspace. 6 PIETER HOFSTRA

2 Example 4.2. Consider the subset U = {ax | a ∈ R} of P2[x]. Show that U is a subspace, give a basis and find its . Solution 1. We verify the three axioms. Throughout, keep in mind that to say that a polynomial p(x) is in U, is exactly to say that is of the form p(x) = ax2 + 0x + 0 for some a.

(1) We must show that the zero vector is in U. The zero vector in P2[x] is the polynomial 0, or, more formally, 0x2 + 0x + 0. Hence this polynomial is in U, as required. (2) Consider p(x), q(x) ∈ U. This means that p(x) = ax2 for some number a, and q(x) = bx2 for some number b. Now to show that p + q is in U, note that p(x) + q(x) = ax2 + bx2 = (a + b)x2. Thus p + q is also a multiple of x2, and hence is in U. (3) Consider p(x) ∈ U and c ∈ R. Then we have p(x) = ax2 for some number a. Now cp(x) = (ca)x2, and therefore cp(x) is again a multiple of x2 and hence in U. For a basis, observe that ax2 is a scalar multiple of x2; thus {x2} is a basis and dim(U) = 1. Solution 2. We can write U = Span{x2}. Since the span of any set of vectors is a subspace, we’re done; it is clear that {x2} is linearly independent (having only one vector in it) and hence is a basis, resulting in dim(U) = 1.

Solution 3. We can translate this into a problem about subspaces of R3. To do so, recall that an arbitrary polynomial ax2 + bx + c is translated as the vector (a, b, c). Thus a polynomial ax2 is translated as (a, 0, 0). We thus get: is the subset U 0 = {(a, 0, 0)|a ∈ R} of R3 a subspace? We can now use any method we like to answer this question, e.g. we can regard it as the span of {(1, 0, 0)}.

T Example 4.3. Is U = {A | A = A } a subspace of M2,2? If so, find a basis and find its dimension. Solution. First note that U is the set of all symmetric 2x2 matrices. We could easily verify the three axioms for a subspace here, but we’ll instead translate the problem to R4. Observe that for a matrix a b A = , we have A = AT precisely when b = c. This means that the translated subset is: c d

0 4 U = {(a, b, c, d) ∈ R | b = c} which can also be written as U 0 = {(a, b, c, d) ∈ R4 | b − c = 0}. Therefore it is the of a homogeneous system, hence is a subspace. For a basis, we solve this system (keep in mind, this system has four variables a, b, c, d and one equation b − c = 0, so has as matrix 0 1 −1 0 ) . We get three parameters, say a = t, c = s, d = r, and one leading variable b, which becomes b = c = s. Therefore the general solution is (a, b, c, d) = t(1, 0, 0, 0) + s(0, 1, 1, 0) + r(0, 0, 0, 1). This means that we have three basic solutions forming the basis for U 0, namely {(1, 0, 0, 0), (0, 1, 1, 0), (0, 0, 0, 1)}. Consequently, dim(U) = 3. Translating this back to M2,2, we see that a basis for U is 1 0 0 1 0 0 , , 0 0 1 0 0 1 and that dim(U) = 3.

Example 4.4. Which of the following sets are subspaces of F[R]? • A = {f | f(1) = 0} • B = {f | f(x) ≥ 0 for all x} • C = {f | f(4) − 2f(1) = f(−2)} • D = {f | f(x) = 2f(x + 1) for all x} • E = {f | f(0)f(1) = 1} VECTOR SPACES: FIRST EXAMPLES 7

Solution. As a rough guideline, we use the idea that subsets defined by homogeneous linear equations are subspaces, but subsets defined by nonlinear equations aren’t. For example, A is defined by the equation f(1) = 0, which is linear; similarly, C and D are defined by linear equations, but E is not, because (i) the equation uses a of two values of f and (ii) the right hand side is 1, making it non-homogeneous. We’ll prove that B and E are not subspaces first. The set B contains the zero vector, is closed under addition but not under scalar multiplication. Consider for example f(x) = ex. This function is positive, hence is in B. However, −3ex (a scalar multiple of f) is negative, hence not in B. The set E doesn’t contain the zero vector: recalling that the zero vector is z(x) = 0 for all x, we get z(0)z(1) = 0 6= 1. We now show that D is a subspace. First of all, the zero vector satisfies z(x) = 2z(x + 1) because both sides are 0. Next, consider f, g ∈ D. That is, f(x) = 2f(x + 1) and g(x) = 2g(x + 1) for all x. Then to show that f + g ∈ D, consider (f + g)(x) = f(x) + g(x) = 2f(x + 1) + 2g(x + 1) = 2(f + g)(x + 1) so that f + g ∈ D. Finally, if f ∈ D and c ∈ R we get (cf)(x) = cf(x) = c(2f(x + 1)) = 2cf(x + 1) so that cf ∈ D. The items A and C are left as an exercise.