3 Two-Dimensional Linear Algebra

3 Two-Dimensional Linear Algebra 3.1 Basic Rules Vectors are the way we represent points in 2, 3 and 4, 5, 6, ..., n, ...-dimensional space. There even exist spaces of vectors which are said to be infinite- dimensional! A vector can be considered as an object which has a length and a direction, or it can be thought of as a point in space, with coordinates representing that point. In terms of coordinates, we shall use the notation µ ¶ x x = y for a column vector. This vector denotes the point in the xy-plane which is x units in the horizontal direction and y units in the vertical direction from the origin. We shall write xT = (x, y) for the row vector which corresponds to x, and the superscript T is read transpose. This is the operation which flips between two equivalent represen- tations of a vector, but since they represent the same point, we can sometimes be quite cavalier and not worry too much about whether we have a vector or its transpose. To avoid the problem of running out of letters when using vectors we shall also write vectors in the form (x1, x2) or (x1, x2, x3, .., xn). If we write R for real, one-dimensional numbers, then we shall also write R2 to denote the two-dimensional space of all vectors of the form (x, y). We shall use this notation too, and R3 then means three-dimensional space. By analogy, n-dimensional space is the set of all n-tuples {(x1, x2, ..., xn): x1, x2, ..., xn ∈ R} , where R is the one-dimensional number line. This n-dimensional space is written Rn. 1 3.2 Vector operations Vectors have several important operations associated with them. Firstly, 1. there is a vector addition x + y = (x1, x2) + (y1, y2) = (x1 + y1, x2 + y2), 2. and there is scalar multiplication tx = t(x1, x2) = (tx1, tx2), where t ∈ R is said to be a scalar. Let us note that the terms scalar and real number are synonyms. Rule 2, scalar multiplication, implies that there is always a so-called zero vector. Just set t = 0, 0 = (0, 0) and note that this satisfies x + 0 = x for every vector x. Example 3.1 To perform vector addition, just add componentwise: viz, (0, 1) + (0, −1) = (0, 1 − 1) = (0, 0), and (2, 1) + (3, 3) = (2 + 3, 1 + 3) = (5, 4). Let us note that every vector (x, y) ∈ R2 can be written in the form µ ¶ µ ¶ µ ¶ x 1 0 = x + y . y 0 1 The two vectors that we use to form this decomposition are called basis vectors and are the building blocks of two-dimensional space and we shall use i and j to denote them. In three-dimensional space, R3, we also have a set of basis vectors and if (x, y, z) ∈ R3 then 0 1 0 1 0 1 0 1 x 1 0 0 @ y A = x @ 0 A + y @ 1 A + z @ 0 A , z 0 0 1 2 Figure 1: The parallelogram law of vector addition a a+b b 0 and we shall write this x = xi + yj + zk. Throughout, bold font (x, y, z) is used to denote vectors and Greek mi- nuscules and roman fonts (λ, µ, s, t) are used to denote scalars. You are encouraged to use underlined characters, x, y, z, to denote vectors in order to distinguish them from scalars. Sometimes, you will also see vectors written using an arrow −→x , and you may use this notation too. The operation of vector addition must be well-defined, that is, it must be clear how to perform it. To explain what this means, try and perform the addition 2 + (99, 31) ! This makes no sense, and this is because scalars and vectors are distinct objects. Of course, we can perform the addition (2, 0) + (99, 31), but (2, 0) and the number 2 are not the same type of object: one is a vector and one is a scalar. 3.3 Scalar Products and Angles T 2 T 2 If x = (x1, x2) ∈ R and y = (y1, y2) ∈ R , then we define their inner, dot or scalar product x • y by x • y = x1y1 + x2y2. 3 Example 3.2 For instance, if we recall the basis vectors i = (1, 0) and j = (0, 1), then x • i = x1 and x • j = x2. Also, i • j = 0. i • i = 1, j • j = 1, We shall also write the scalar product in the form µ ¶ T y1 x • y = x y = (x1, x2) = x1y1 + x2y2. y2 Example 3.3 µ ¶ µ ¶ µ ¶ 2 9 9 • = (2, 3) = 18 + 3 = 21. 3 1 1 Using the scalar product, we can define the length of a vector as follows. Definition 3.1 The length of a vector is given by √ |x| = + x • x. This is also sometimes called the norm of x. The vector ˆx = x/|x| is called the unit vector in the x-direction. The function x 7→ |x|, where x ∈ R, is called the modulus or absolute value function and this can be thought of as the length function for one-dimensional vectors! Example 3.4 The vectors i and j are unit vectors as their length equals one. The length of a number t ∈ R is given by √ |t| = + t2. This simply means stripping its sign away, for instance |−5| = 5 and |5| = 5 too. 4 Figure 2: The angle between two vectors. a a . b = |a||b| cos( ) b 0 Let us note that if t is a scalar and x a vector, then we have the result |tx| = |t||x|. Now, the angle θ between two vectors x and y is calculated according to the rule |x||y| cos(θ) = x • y, where we take 0 ≤ θ ≤ π. Example 3.5 The cosine of the angle, θ, between the two vectors (2, 3) and (9, 1) is µ ¶ µ ¶ x • y 1 2 9 21 cos θ = = √ √ • = √ √ . |x||y| 4 + 9 81 + 1 3 1 13 82 Hence θ ≈ 0.8721 radians. Let us now give a mathematical definition of what it means for two vector to lie ‘at right angles’ to one another. Definition 3.2 Two vectors, x and y, are said to be normal or orthogonal if x • y = 0. The following are examples of orthogonal vectors. Example 3.6 The basis vectors i and j are orthogonal, and the pair {i, j} is called an orthogonal basis for this reason. Example 3.7 The following are orthogonal pairs of vectors, {(2, 2), (2, −2)}, {(−1, −1), (−2, 2)}, {(1, 1/2), (−1, 2)}. 5 3.4 The Scalar Product Properties The following properties or rules of the scalar product are easily verified from its definition. Suppose that s, t are scalars and x, y and z are vectors, then 1. x • y = y • x, 2. x • (ty) = tx • y, 3. x • (z + y) = x • y + x • z. For example, using only these properties we can prove, in a strict, mathematical sense, the following theorem. Theorem 3.1 For all vectors x, y ∈ R2, we find |x • y| ≤ |x||y|. Proof 3.1 Let t ∈ R be any scalar, and consider the function of t defined by q(t) = (x + ty) • (x + ty). Using Properties 2 and 3 of the scalar product, we see that q(t) = x • (x + ty) + ty • (x + ty), = x • x + tx • y + ty • x + t2y • y, = |x|2 + 2tx • y + t2|y|2. However, q(t) ≥ 0 since it is the length2 of the vector x + ty, which is necessarily non-negative, and q is a quadratic function of t. Therefore, q has either no real roots or a repeated real root, and its discriminant must therefore be non-positive, so that 4(x • y)2 − 4|x|2|y|2 ≤ 0, and the given inequality follows after taking positive square roots. This theorem gives us the important piece of information that the equation for the angle between two vectors, cos(θ) = x • y/|x||y|, always has a solution θ. If you are unsure as to the reasoning in the above proof, the following information may help you. 6 Remark 3.1 Let us recall that the discriminant of the quadratic at2 + bt + c is the expression b2 − 4ac, and that the solutions to at2 + bt + c = 0 are 1 ³ √ ´ t = −b ± b2 − 4ac . 2a Hence if the discriminant if negative, the given quadratic has no real roots, and if the discriminant is zero then t = −b/2a is a repeated, real root. 3.5 The Equation of a Line 3.5.1 The Parametric Equation A unit vector is one way of representing a direction without having to worry about distances. For instance, if t is a scalar (in R) then as t varies, x `(t) = tˆx = t |x| is a function which represents a line in space along the direction of x which also passes through the origin (0, 0). The length of the vector `(t) is just |t|, that is ¯ ¯ ¯ ¯ ¯ x ¯ ¯ x ¯ |x| |`(t)| = ¯ t ¯ = |t| ¯ ¯ = |t| = |t|. ¯ |x|¯ ¯|x|¯ |x| By shifting this line, which simply means adding another vector to it, we can construct a line between any two points in space. So, let a, b ∈ R2 be any two vectors and define the function L(t) := ta + (1 − t)b, then L(0) = b and L(1) = a. This is the parametric equation of the line which passes through the two points determined by a and b.

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