# Vectors. Arrows that have the same direction and length represent the same vector.

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1 Part 1. 1 Part 1. Vectors A vector in the plane is a quantity that has both a magnitude and a direction. We can represent it by an arrow. Arrows that have the same direction and length represent the same vector. A special case is the zero vector which has zero magnitude and no particular direction (we just draw it as a dot), We ll usually denote a vector by a lower case letter. If u is a vector, u denotes the magnitude of u. A scalar is a quantity that has magnitude but no direction, i.e., just a

2 Part 1. 3 Part 1. number. If v is a vector, v denotes the vector that has the same length but the opposite direction. Thus, v = v. If v is a vector an s is a scalar we define the vector sv as follows. If s = 0, sv is the zero vector. If s > 0, sv has the same direction as v and sv = s v. If s < 0, then sv = s ( v), i.e., it points in the direction opposite to v and sv = s v. If u and v are vectors, as in the picture, we define u + v by the parallelogram law, as in the following diagram.

3 Part 1. 5 Part 1. The diagram shows that vector addition is commutative, i.e., u + v = v + u. Subtraction u v is defined as u + ( v), but the diagram is worth drawing

4 Part 1. 7 Part 1. A choice of a coordinate system sets up a 1-1 correspondence between points in the plane and pairs of real numbers. It also gives a 1-1 correspondence between vectors and pairs of real numbers. How do scalar multiplication and vector addition look in coordinates? Consider the case shown in the diagram We just write u = (u 1, u 2 ). The magnitude of the vector is given by u = u u2 2. The set of ordered pairs of real numbers is denoted R 2. so s > 0. Let the coordinates of the tip of sv be (a, b). From similar triangles we have v 1 v = a sv = a s v. from which we conclude that a = sv 1.

5 Part 1. 9 Part 1. 1 Similarly, we have and so v 2 v = b s v b = sv 2. Thus, the components of sv are (sv 1, sv 2 ). In other words, s(v 1, v 2 ) = (sv 1, sv 2 ). For vector addition, consider the diagram The vector operations satisfy the following rules 1. (u + v) + w = u + (v + w) (Associative Law) 2. u + v = v = v + u. (Commutative Law) v = v. (Additive identity) 4. v + ( v) = 0. (Additive inverse) 5. s(tv) = (st)v. (Associative Law) 6. (s + t)v = sv + tv. (Distributive Law) 7. s(u + v) = su + sv. (Distributive Law) 8. 1v = v. We conclude (u 1, u 2 ) + (v 1, v 2 ) = (u 1 + v 1, u 2 + v 2 ).

6 Part Part 1. 1 We can define the angle between two vectors. Recall the law of cosines: The dot product of two vectors is defined by u v = u v cos(θ). c 2 = a 2 + b 2 2ab cos(θ). Apply this to the diagram (This is zero if u or v is zero.) Observe u u = u 2. We get u v 2 = u 2 + v 2 2 u v cos(θ), which we recognize as u v 2 = u 2 + v 2 2u v.

7 Part Part 1. 1 Thus, we get 2u v = u 2 + v 2 u v 2 = u u v1 2 + v2 2 [(u 1 v 1 ) 2 + (u 2 v 2 ) 2 ] We can use this to compute angles between vectors in terms of the components cos(θ) = u v u v. = u u v1 2 + v2 2 (u 2 1 2u 1 v 1 v1) 2 (u 2 2 2u 2 v 2 + v2) 2 = u u v1 2 + v2 2 u u 1 v 1 v1 2 u u 2 v 2 v2 2 = 2u 1 v 2 + 2u 2 v 2. Thus, we have the formula u v = u 1 v 1 + u 2 v 2. Properties of the dot product. 1. u v = v u. 2. (su) v = s(u v). 3. (u + v) w = u w + v w.

8 Part Part 1. 1 Matrices An m n matrix is a rectangular array of numbers with m rows and n columns. For example, a 11 a 12 a a 1n a 21 a 22 a a 2n A = a 31 a 32 a a 3n a m1 a m2 a m3... a mn In this notation, a ij denotes the entry in row i and column j. To multiply a matrix by a scalar, multiply each entry by the scalar Matrices of the same size can be added by adding corresponding entries, like vectors. An m n matrix A and an n p matrix B can be multiplied to give an m p matrix AB. For the product of an 1 n matrix (row vector) and a n 1 matrix (column vector) the product should be 1 1, i.e., just a number. The definition is b 1 ] b 2 [a 1 a 2 a 3... a n b 3 =.. b n a 1 b 1 + a 2 b 2 + a 3 b a n b n. If A is m n and B is n p, C = AB is m p and is given by c ij = Row i (A) Col j (B).

9 Part Part 1. 1 Here are the two cases that will be most important to us. First a 2 2 matrix A times a two entry column vector u Au = a 11 a 12 u 1 = a 11u 1 + a 12 u 2, a 21 a 22 u 2 a 21 u 1 + a 22 u 2 yielding a two entry column vector. Second, the product of two 2 2 matrices A and B AB = a 11 a 12 b 11 b 12 = a 21 a 22 b 21 b 22 a 11b 11 + a 12 b 21 a 11 b 12 + a 12 b 22. a 21 b 11 + a 22 b 21 a 21 b 12 + a 22 b 22 Matrix multiplication is not commutative! = = Exercise: Let A and[ B be 2 ] 2 matrices. Write B = b 1 b 2 where b 1 and b 2 are the columns of B considered as column [ vectors. Show AB = Ab 1 Ab 2 ]. Similarly, write A = a 1 a 2 where a 1 and a 2 are the rows of A considered as row vectors. Show that AB = a 1B. a 2 B

10 Part Part 1. 2 Rules of Matrix Algebra. Matrix addition and scalar multiplication obey the same rules as addition and scalar multiplication of vectors. In addition, we have the following rules (The matrices are assumed to have compatible sizes) 1. A(BC) = (AB)C. (Multiplication is Associative) 2. A(B + C) = AB + AC. (Left Distributive Law) 3. (A + B)C = AC + BC. (Right Distributive Law). 4. (sa)b = s(ab) = A(sB) The 2 2 identity matrix is I = A (2 2) matrix A is invertible if there is a (2 2) matrix B so that AB = BA = I. It s easy to show there is at most one matrix B with this property. If B exists, we write B = A 1. Note that if A 1 exists, (A 1 ) 1 = A. The determinant of the 2 2 matrix A = a c b d is defined by det(a) = ad bc. If A and B have compatible sizes, IA = A and BI = B.

11 Part Part 1. 2 Theorem Let A be a 2 2 matrix. Then, the following conditions are equivalent. 1. A is invertible. 2. The only vector v such that Av = 0 is v = det(a) 0. To show that (1) = (2), suppose (1) holds and that Av = 0. Then we have A 1 Av = A 1 0 = 0. Then 0 = A 1 Av = Iv = v, so v = 0. Thus, (2) holds. To show that (2) = (3), it will suffice to show that (3) = (2). So assume det(a) = ad bc = 0. If A = 0 then Av = 0 for all v so (2) fails. So, suppose A 0. Notice that a c d ad bc = = 0 = 0 b d b bd bd 0 a c c ac + ac = = 0. b d a bc + ad If A 0, one of these vectors is nonzero, so (2) fails. Finally, to show that (3) = (1), assume that det(a) 0. Just check that the following formula works A 1 1 = d c. det(a) b a Exercise: Define the trace of A, denoted tr(a), by tr a c = a + d, b d (i.e., the sum of the diagonal entries).

12 Part Part Show that tr(ab) = tr(ba). (Use brute force.) 2. Use the first part to show tr(p 1 AP ) = tr(a). Exercise: det(a) = 0 if and only if one of the columns of A is a multiple of the other. Linear Transformations Let e 1 and e 2 be the vectors e 1 = 1, e 2 = Note that u = u 1 = u u 2 0 = u 1 e 2 + u 2 e u 2 If A is a 2 2 matrix then Ae 1 = Col 1 (A) and Ae 2 = Col 2 (A). A transformation T : R 2 R 2 is linear if T (αu + βv) = αt (u) + βt (v) for all vectors u and v and scalars α and β. An example is the transformation T (v) = Av for a 2 2 matrix A.

13 Part Theorem If T : R 2 R 2 is linear, there is a unique matrix A such that T (v) = Av. In fact, [ A = T (e 1 ) T (e 2 ) ]. To see this, suppose T is linear and v is a vector. Then, T (v) = T v 1 v 2 = T (v 1 e 1 + v 2 e 2 ) = v 1 T (e 1 ) + v 2 T (e 2 ) [ ] = T (e 1 ) T (e 2 ) v 1. v 2

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