Introduction to Abstract Algebra and Vector Spaces

Home , Associative property, Magnitude (mathematics), Vector space, Zero element

A vector space is an algebraic structure that generalizes the geometrical concept of vector. The reader has hopefully been introduced to vectors in a previous math or physics course. We now review some of the basic properties of geometrical vectors that we desire to capture in the algebraic deﬁnition of a vector space.

A geometric vector is an entity that has both a direction and magnitude (see Figure ). If we place a vector in the Cartesian plane with its tail at the origin, then the vector’s direction and magnitude is completely determined by the location of its head in the plane, which can be represented with a point (x, y). This is considered one of the great achievements of the philosopher and mathematician Rene Descartes, who showed how a geometrical object, such as a vector, can be algebraically represented, in this case using a pair of numbers (x, y).

Using the Cartesian representation of a vector, the following properties of vectors become more apparent.

Addition and Subtraction Two vectors u = (x1, y1) and v = (x2, y2) can be added to form the new vector u+v = (x1 +x2, y1 +y2), and subtracted to form the new vector u−v = (x1 −x2, y1 −y2).

Figure 1: A geometrical vector

1 Notice that vector addition is associative: u + (v + w) = (u + v) + w since the performed component-wise additions are all associative. It is also commutative: u + v = v + u.

Zero Vector The vector 0 = (0, 0) represents the zero vector which has zero magnitude, and hence no direction.

Additive Inverse Associated with each vector v = (x, y) is its additive inverse −v = (−x, −y), which has the same magnitude as v, but has the opposite direction. Moreover, v + (−v) = 0.

Scaling Given real number r, every vector v = (x, y) can be scaled by r to form the new vector r · v = r(x, y) = (rx, ry). If r > 0, then r · v has the same direction as v, but its magnitude has been multiplied by a factor of r. If r = 0, then r · v = 0, where as r < 0 multiplies the magnitude of v by |r|, and r · v has the same direction as −v.

Magnitude The magnitude of a vector v = (x, y) can be computed using the Pythagorean Theorem as |v| = px2 + y2.

Angle Between Vectors If Θ(v) denotes the counterclockwise angle that v makes with the x-axis, then the angle between two vectors u and v is equal to |Θ(u) − Θ(v)|.

Example 1. Demonstrate the above vector properties with vectors in the Cartesian plane.

2 Abelian Groups

We now move from geometry to algebra. The simplest kind of algebraic structure is a set A, along with an operation that is closed under A, and that obeys a set of properties. For example, let A = I be the set of integers, and consider the addition operation +. This operation is closed under A since the addition of two integers results in another integer. Moreover, addition satisﬁes the following properties. Given arbitrary integers x, y, and z, then

Associative x + (y + z) = (x + y) + z

Commutative x + y = y + x

Zero Element x + 0 = 0 + x = x

Additive Inverse there exists an integer −x for which x + (−x) = 0

We can write this algebraic structure as a pair: (I, +), where the ﬁrst component is the set, and the second is the operation that is deﬁned and closed over the set.

It turns out that (I, +) is not the only algebraic structure that satisﬁes the above four properties. For example (Q, +) and (R, +) also satisfy these properties, where Q is the set of rational numbers (i.e. fractions of the form p/q where p and q 6= 0 are integers), and R is the set of real numbers. In general, any algebraic structure that satisﬁes the above four properties is called an abelian group, while the above four properties referred to as the abelian-group axioms. Thus, when a mathematician says “such and such is an abelian group”, she means that “such and such” is referring to a set, along with an addition operation closed over the set that is associative, commutative, allows for additive inverses, and for which the set has a zero element with respect to the addition.

Once a class, such as abelian groups, of algebraic structures has been deﬁned, it is the job the algebraist to study the logical implications of an algebraic structure that belongs in the class. For example, if an algebraic structure is an abelian group, then what other properties, if any, must it satisfy? Suppose an algebraist proves that all abelian groups have property P . Then she has not only proved that (I, +), (Q, +), and (R, +) have property P , but has also proved it for any other algebraic structure that meets the criteria for being an abelian group.

3 Example 2. Let R2 denote the set of ordered pairs (x, y), where x and y are real numbers. R2 is 2 referred to as the Cartesian plane. Verify that (R , +) is an abelian group, where, for u = (x1, y1) and v = (x2, y2), u + v = (x1 + x2, y1 + y2).

Example 3. Let (R−0, ·) be an algebraic structure, where R−0 is the set of real numbers, excluding 0, and · is the multiplication operation for real numbers. Verify that (R − 0, ·) is an abelian group.

4 Example 4. Let (F(R,R), +) be an algebraic structure, where F(R,R) is the set of functions f : R → R having the real numbers as its domain and co-domain. and, given functions f, g ∈ F(R,R), f +g is the function for which (f +g)(x) = f(x)+g(x), for all real numbers x. Prove that (F(R,R), +) is an abelian group.

Example 5. Let (M2,2, ·) be an algebraic structure, where M2,2 is the set of 2 × 2 matrices having real-valued entries, and · represents matrix multiplication. Show that (M2,2, ·) is not an abelian group. However, show that (M2,2, +) is an abelian group, where + is matrix addition.

5 Vector Spaces

An algebraic structure can have more than one operation. For example, the set of integers has both addition and multiplication operations which can be brought together to form the algebraic structure (I, +, ·). For this structure, the addition operation satisﬁes all the abelian-group axioms. Moreover, the multiplication operation is associative, commutative, and has 1 as its multiplicative identity. Finally, addition and multiplication, together, satisfy the distributive property

x · (y + z) = x · y + x · z.

It is these properties that depend on both operations that make the algebraic structure interesting to study.

In this section we study a class of algebraic structures called vector space. Like integer arithmetic, a vector space consists of a set V and two operations that are closed over V. The ﬁrst operation is addition, and it satisﬁes all the abelian-group axioms. In other words, if we ignore the second operation, then the algebraic structure (V, +) is an abelian group. Thus, every vector space is an abelian group.

The second vector space operation is called the scaling operation, and is usually given the multiplication symbol ·. This operation is described as follows. left input The left input is a real number r. right input The right input is a vector v ∈ V. output The output is denoted r · v, or simply rv, and is a vector in V.

The left input r has the eﬀect of scaling the right input v. For example, if r = 2, then 2v is a vector that has the same “direction” as v, but has doubled in “magnitude” (here we place “direction” and “magnitude” in quotes, because we do not yet know what these terms mean in an abstract vector space). For this reason r is referred to as a scalar.

With the terminology now in place, a vector space is an algebraic structure (V, +, ·), where (V, +) is an abelian group, and · is a scaling operation that satisﬁes the following properties. In what follows, r and s are arbitrary scalars, while u and v are arbitrary vectors. Moreover, e.g., r · v is expressed as rv.

Unit Scalar 1v = v

Associative r(sv) = (rs)v

Distributive over V r(u + v) = ru + rv

Distributive over R (r + s)v = rv + sv

6 These properties will henceforth be referred to as the scalar axioms. The scalar axioms together with the abelian-group axioms form what are called the vector-space axioms. Note that an element of a vector space is referred to as a vector.

Example 6. We show that (R, +, ·), is a vector space, where R is the set of real numbers, and + and · are the usual addition and multiplication operations over R. Certainly, (R, +) is an abelian group since real-number addition is associative and commutative, 0 is the additive identity, and every real has an additive inverse. Moreover, · is a scaling operation, since its left input is a real, while its right input and output are both real (in this case the scalar set R is equal to the vector space set R). All that remains to show is that the scaling operation satisﬁes the scalar axioms. But notice that, since the scaling operation is just multiplication of two real numbers, these axioms are just the axioms of real-number multiplication, meaning that i) 1 is a multiplicative identy for R, and that ii) real-number multiplication is associative, commutative, and distributive.

Example 7. Let (R2, +, ·) be an algebraic structure, where (R2, +) is the abelian group from Example 2, and · is the scaling operation deﬁned as follows. Given r ∈ R and (a, b) ∈ R2, r · (a, b) = r(a, b) = (ra, rb). We show that (R2, +, ·) is a vector space.

From Example 2 we know that (R2, +) is an abelian group. It remains to verify the scaling axioms. In doing so, let r and s be arbitrary scalars, and u = (a, b), v = (c, d) be arbitrary vectors.

Unit Scalar 1v = 1(a, b) = (1a, 1b) = (a, b) = v. Notice that, in (1a, 1b), 1a refers to real-number multiplication, and not the scaling operation. The same is true for 1b.

Associative r(sv) = r(s(a, b)) = r(sa, sb) = (r(sa), r(sb)) = ((rs)a, (rs)b) = (rs)(a, b) = (rs)v, where the fourth equality is from the associative property of real-number multiplication.

Distributive over V

r(u + v) = r((a, b) + (c, d)) = r(a + c, b + d) = (r(a + c), r(b + d)) = (ra + rc, rb + rd) =

(ra, rb) + (rc, rd) = r(a, b) + r(c, d) = ru + rv, where the fourth equality is from the distributive property of real-number multiplication.

Distributive over R

(r + s)u = (r + s)(a, b) = ((r + s)a, (r + s)b) = (ra + sa, rb + sb) =

7 (ra, rb) + (sa, sb) = r(a, b) + s(a, b) = ru + su, where the third equality is from the distributive property of real-number multiplication.

n We can generalize the above example and deﬁne R to be the set of n-tuples of the form (a1, . . . , an), where ai ∈ R, for i = 1, . . . , n. Moreover, addition and scalar multiplication is deﬁned analogously to the case n = 2 in Example 7. From a geometrical perspective, we can think of a vector in Rn as representing a point in n-dimensional space.

Example 8. Recall from Example 4 that (F(R,R), +) is an abelian group. Consider the following scalar multiplication deﬁned over F(R,R). Given r ∈ R and f ∈ F(R,R), then rf is the function for which (rf)(x) = rf(x), for every x ∈ R. In words, if x is the input to function rf, then the output is computed as the product of r with f(x). Verify that this scalar multiplication satisﬁes the scalar axioms, and hence (F(R,R), +, ·) is a vector space.

Example 9. Recall from Example 5 that (M2,2, +) is an abelian group. Consider the following scalar multiplication deﬁned over M2,2. Given r ∈ R and A ∈ M2,2, then rA is the matrix whose (i, j) entry equals raij. In other words, the entries of rA are obtained by multiplying the entries of A by r. Verify that this scalar multiplication satisﬁes the scalar axioms, and hence (M2,2, +, ·) is a vector space.

8 As was mentioned earlier, the algebraist is interested in proving additional properties of vector spaces that follow from the abelian-group and scalar axioms. Once a property is proved, it holds true in any specﬁc vector space. Here is an example of some basic results that can be proved from the axioms.

Proposition 1. Let V be a vector space, v a vector in V, and r a real number. Then the following are true.

(a)0 u = 0

(b) r0 = 0

(c)( −1)u = −u

(d) if ru = 0, then either r = 0 or u = 0

We prove (a) and leave the rest as exercises.

0u = (0 + 0)u = 0u + 0u, by the distributive rule over R. But for any vector v that satisﬁes v = v + v, we can add −v to both sides of the equation to get

v + (−v) = 0 = (v + v) + (−v) = v + (v + (−v)) = v + 0 = v, which implies that v = 0. Therefore, since 0u = 0u + 0u, it follows that 0u = 0.

9 Exercises

1. Given geometric vectors u = (1, 2) and v = (−1, 1), compute i) u + v, ii) u − v, iii) −u, iv) 3v, v) |u|, and vi) the angle between u and v. Draw the parallelogram whose sides include u and v, and verify that its diagonals represent u + v and u − v. Hint: for vi) use the law of cosines: a2 = b2 + c2 − 2ab cos θ, where a, b, and c are the lengths of the sides of a triangle, and θ is the angle opposite the side with length a. 2. Give an example that shows why the set of matrices having the form a 1 , 1 b where a and b are real numbers, is not closed under matrix addition. 3. Let F denote the set of real-valued functions f for which f(0) = 1. Explain why F is not closed under function addition.

4. Consider the set Q1 of all geometric vectors that lie in the first quadrant of the Cartesian plane. Verify that this set is closed under vector addition. Why is (Q1, +) not an abelian group? Hint: vector (x, y) ∈ Q1 iff x > 0 and y > 0. 5. Consider the set A of all geometric vectors of the form (x, 0), where x is a real number. (A, +) an abelian group, where + represents normal addition of geometric vectors? If yes, verify each abelian-group axiom. If not, justify your answer. 6. Suppose, we change vector addition in R2 from (a, b) + (c, d) = (a + c, b + d) to (a, b) + (c, d) = (a + c + 1, b + d + 1). Is R2 still an abelian group under this addition? If yes, verify each abelian-group axiom. If not, justify your answer. 7. Given the matrix 1 0 A = , −5 2 provide the matrices 2A, −3A and 0A. 8. Suppose (V, +) is an abelian group. Consider the following scaling operation · over V that is defined by rv = 0, for every r ∈ R, and v ∈ V. Is (V, +, ·) a vector space? Explain. 9. Suppose V consists of the single element “Joe”. Moreover, addition and scaling are defined so that Joe + Joe = Joe and r · Joe = Joe. Prove that (V, +, ·) is a vector space.

10. Let Pn denote the set of all polynomials of degree at most n and having real-valued coeﬃcients. 2 For example, 3.4x − 2x + 5 is a polynomial of degree 2, and hence is in P2. Given p(x) = n n anx + ··· + a0 and q(x) = bnx + ··· + b0, deﬁne polynomial addition by

n (p + q)(x) = (an + bn)x + ··· + (a0 + b0), and scalar multiplication by n rp(x) = ranx + ··· + ra0.

Prove that Pn is a vector space under these operations.

10 11. Given a vector space V, prove that (−1)v = −v, for all v ∈ V

12. Given a vector space V, prove that r0 = 0, for any real number r.

13. Given a vector space V, suppose we have rv = 0. Prove that either r = 0 or v = 0.

11 Exercise Solutions √ 1. i) (0, 3), ii) (2, 1), iii) (−1, −2), iv) (−3, 3), v) 5, vi) 50.77◦.

2. Consider 2 1 3 1 5 2 + = . 1 2 1 2 2 4 The matrices being added are in the set, but sum of the two matrices is not a matrix of the set, since its oﬀ-diagonal entries do not equal 1.

3. Given two functions f, g ∈ F. Then f(0) = 1 and g(0) = 1, but (f + g)(0) = f(0) + g(0) = 2, which implies that (f + g) 6∈ F.

4. Addition is closed over Q1 since, if (x1, y1), (x2, y2) ∈ Q1, then x1, x2, y1, y2 > 0, which implies x1 + x2 > 0 and y1 + y2 > 0, and therefore (x1 + x1, y1 + y2) ∈ Q1.(Q1, +) is not an abelian group since there is no zero element.

5. Yes. Associative:

(x, 0) + ((y, 0) + (z, 0)) = (x, 0) + (y + z, 0) = (x + (y + z), 0) = ((x + y) + z, 0) =

((x, 0) + (y, 0)) + (z, 0). Commutative: (x, 0) + (y, 0) = (x + y, 0) = (y + x, 0) = (y, 0) + (x, 0). The zero element is (0, 0) and (−x, 0) is the additive inverse of (x, 0), for all real x.

(x, a) + ((y, b) + (z, c)) = (x, a) + (y + z + 1, b + c + 1) = (x + (y + z + 1) + 1, a + (b + c + 1) + 1) =

((x + y + 1) + z + 1, (a + b + 1) + c + 1) = (x + y + 1, a + b + 1) + (z, c) = ((x, a) + (y, b)) + (z, c). Commutative:

(x, a) + (y, b) = (x + y + 1, a + b + 1) = (y + x + 1, b + a + 1) = (y, b) + (x, a).

The zero element is (−1, −1), since (x, y) + (−1, −1) = (x − 1 + 1, y − 1 + 1) = (x, y). Finally, (−x − 1, −y − 1) is the additive inverse of (x, y).

7. Given the matrix 1 0 A = , −5 2 2 0 −3 0 2A = , − 3A = , −10 4 15 −6 while 0A = 0.

8.( V, +, ·) is not a vector space if there is a nonzero element v ∈ V. This is because 1v = 0 6= v, and the Unit-Scalar axiom is not satisﬁed. On the other hand, if V only consists of one element, the zero element, then (V, +, ·) would be a vector space.

12 9. Since Joe is the only element of V, it would have to equal the zero element. This is conﬁrmed by the fact that v + Joe = v, for all v ∈ V (since v must equal Joe). Also, since

Joe + Joe = Joe,

we see that Joe is its own additive inverse. Finally, notice that the result of any addition or scalar multiplication is always equal to Joe. Therefore, addition must be associative and commutative, while all the scalar axioms hold, since both sides of the equations will always result in Joe.

10. For brevity, only the associative scalar axiom is proved. The others are proved similarly. Let r n and s be scalars, and p(x) = anx + ··· + a0 be an arbitrary polynomial. Then

n n r(sp(x)) = r(sanx + ··· + sa0) = r(san)x + ··· + r(sa0) =

n (rs)anx + ··· + (rs)a0 = (rs)p(x).

11. By Proposition 1a, 0 = 0v = (1 + (−1))v = v + (−1)v, by distributivity over the reals. But this means that (−1)v is the additive inverse of v. In other words, (−1)v = −v.

12. r0 = r(0 + 0) = r0 + r0, by distributivity over vectors. Then subtracting r0 from both sides yields 0 = r0, or r0 = 0.

13. If r = 0, then we are done. So suppose r 6= 0. Then, since rv = 0, we must have (1/r)rv = (1/r)0 = 0 by the previous exercise. But by associativity, (1/r)rv = (1/r · r)v = 1v = v. Therefore, v = 0.