Math 276, Spring 2007 Additional Notes on Vectors

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1 Math 276, Spring 2007 Additional Notes on Vectors 1.1. Real Vectors. 1. Scalar Products If x = (x 1,..., x n ) is a vector in R n then the length of x is x = x x2 n. We sometimes use the notation x instead of x. If x = (x 1,..., x n ) and y = (y 1,..., y n ) are two vectors in R n, the dot product or scalar product is defined to be x y = x 1 y 1 + x 2 y x n y n. In particular, x x = x 2. Note that the dot product of two vectors gives us a scalar, and satisfies the following properties. Let x, y, u be vectors (in R n ) and let α be a (real) scalar. Then: (i) (ii) (iii) (iv) (v) x y = y x (α x) y = α ( x y) x (α y) = α ( x y) ( x + u) y = x y + u y x ( y + u) = x y + x u (vi) x x 0 for any vector x (vii) x x = 0 x = 0. Theorem 1 (Cauchy-Schwarz Inequality). For any vectors x and y in R n, x y x y. (1) Proof. If the vector y is the zero vector, then both sides of the inequality in equation (1) are equal to zero, and so the inequality is true (with = instead of ). Thus in proving (1) we can assume that y 0, and hence y = 0. Let t R. Then using properties (i) through (vii) we have 0 x + t y 2 = ( x + t y) ( x + t y) = x x + t( y x) + t( x y) + t 2 ( y y) = x 2 + 2t( x y) + t 2 y 2. Think of x 2, ( x y), and y 2 as constants, and consider the quadratic polynomial in t given by P (t) = x 2 + 2t( x y) + t 2 y 2. The above inequalities show that 0 P (t). We can find where P attains its minimum value by solving P (t) = 0. This gives 2( x y) + 2t y 2 = 0, or equivalently, t = ( x y) y 2.

2 2 If we plug this value into the inequality we get 0 x 2 ( x y) 2 y 2 0 x 2 + 2t( x y) + t 2 y 2, ( x y) + ( x y)2 y 4 y 2 = x 2 ( x y)2 y 2. It follows that ( x y) 2 x 2 y 2, which is equivalent to the statement of the theorem Complex Vectors. The situation for complex vectors in C n is quite similar, except that one needs to use the complex conjugates of certain complex numbers. Recall that if z = x+iy is a complex number, its complex conjugate is z = x iy, that z z = x 2 + y 2, and that z = x 2 + y 2. If z = (z 1,..., z n ) is a vector in C n then the length of z is z = z z n 2. If z = (z 1,..., z n ) and w = (w 1,..., w n ) are two vectors in C n, the dot product or scalar product is defined to be In particular, x w = z 1 w 1 + z 2 w z n w n. z 2 = z z. Note again that the dot product of two vectors gives us a (complex) scalar, and satisfies the following properties. Let z, w, u be vectors (in C n ) and let λ be a (complex) scalar. Then: (i) (ii) (iii) (iv) (v) z w = w z (λ z) w = λ ( z w) z (λ w) = λ ( z w) ( z + u) w = z w + u w z ( w + u) = z w + z u (vi) z z 0 for any vector z (vii) z z = 0 z = 0. Theorem 2 (Cauchy-Schwarz-Bunyakovski Inequality). For any vectors z and w in R n, z w z w. (2) Proof. If the vector w is the zero vector, then both sides of the inequality in equation (2) are equal to zero, and so the inequality is true (with = instead of ). Thus in proving (2) we can assume that w 0, and hence w = 0.

3 3 Let λ C. Then using properties (i) through (vii) we have 0 z + λ w 2 = ( z + λ w) ( z + λ w) = z z + λ( w z) + λ( z w) + λ 2 ( w w) = z 2 + λ( z w) + λ( z w) + λ 2 w 2. In analogy with the proof in the real case, we plug the value into the inequality We get ( z w) λ = w 2 0 z 2 + λ( z w) + λ( z w) + λ 2 w 2. 0 z 2 2 = z 2 z w 2 z w 2 + w 2 w 2 z w 2 w 2. It follows that z w 2 z 2 w 2, which is equivalent to the statement of the theorem Another example. Consider the space of all complex-valued continuous functions on the interval [0, 1]. For two such functions f and g, define f, g = 1 0 f(x) g(x) dx. Note that given two functions, the product f, g gives us a complex scalar. It is not hard to check that this product satisfies the following conditions: (i) f, g = g, f (ii) λf, g = λ f, g (iii) f, λg = λ f, g (iv) f + h, g = f, g + h, g (v) f, g + h = f, g + g, h (vi) f, f 0 for any function f (vii) f, f = 0 f = 0. Since these are the same properties as we used for complex vectors, we obtain the following version of the C-S-B theorem: 1 0 ( 1 f(x)g(x) dx 0 f(x) 2 ) 1 2 ( 1 0 g(x) 2 ) 1 2.

4 Geometric definition. 2. The cross product Let A and B be two vectors in R 3. We define the cross-product of A and B by setting A B = A B sin(θ) n where θ is the angle between the vectors A and B, and n is a unit vector (a vector of length 1) which is perpendicular to both A and B, and is ermined by the right hand rule: if the fingers of your right hand point in the direction of A, and if these fingers curl in the direction of B, then n points in the direction of your thumb. The cross product is useful in finding a vector perpendicular to two given vectors. As we will see in class, we have the following facts about the cross product. (1) If two sides of a triangle are given by A and B, then the area of the triangle is 1 2 A B. (2) if A, vecb, and C are three vectors in R 3, then the volume of the parallelopiped with these vectors as edges is given by A B C Determinants. We want to define the erminant of an n n array of numbers. There are many ways of doing this, and we will see some of these next semester. At this point, we will use a very unintuitive definition by induction. When n = 1, we put a = a. When n = 2, we define a b c d = ad bc, or equivalently a 1 a 2 b 2 b 2 = a 1b 2 a 2 b 1. Note that this can be written a 1 a 2 b 2 b 2 = a 1 b 2 a 2 b 1. Next consider a 3 3 array of numbers a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 Given any entry in the top row, if we cross out the column containing the entry and the top row, we are left with a 2 2 array of numbers. Thus a 1 corresponds to the array b 2 b 3 c 2 c 3, a 2 corresponds to the array b 1 b 3 c 1 c 3, and a 3 corresponds to the array b 1 b 2 c 1 c 2. We then define a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = +a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2.

5 5 Note that as we go across the first row, the signs alternate. We have +a 1, then a 2, then +a 3. Of course, we can now use the definition of a 2 2 erminant to complete the calculation. We have a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = +a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2 = +a 1 (b 2 c 3 b 3 c 2 ) a 2 (b 1 c 3 b 3 c 1 ) + a 3 (b 1 c 2 b 2 c 1 ) = a 1 b 2 c 3 a 1 b 3 c 2 a 2 b 1 c 3 + a 2 b 3 c 1 + a 3 b 1 c 2 a 3 b 2 c 1. To define a 4 4 erminant, we follow the same pattern: a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 b 2 b 3 b 4 c 1 c 2 c 3 c 4 = a 1 c 2 c 3 c 4 d 1 d 2 d 3 d 4 d 2 d 3 d 4 a b 2 b 3 b 4 2 c 2 c 3 c 4 d 2 d 3 d 4 b 2 b 3 b 4 + a 3 c 2 c 3 c 4 d 2 d 3 d 4 a b 2 b 3 b 4 4 c 2 c 3 c 4 d 2 d 3 d 4 This can now be evaluated using the definition of a 3 3 matrix Unit vectors in R 3. We introduce special notation for the three special unit vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1) in R 3. We set (1, 0, 0) = î (0, 1, 0) = ĵ (0, 0, 1) = ˆk Note that with this definition we can write the vector A = (a 1, a 2, a 3 ) = a 1 (1, 0, 0) + a 2 (0, 1, 0) + a 3 (0, 0, 1) = a 1 î + a 2 ĵ + a 3ˆk. These three vectors are mutually perpendicular, so we have î î = ĵ ĵ = ˆk ˆk = 1 î ĵ = î ˆk = ĵ ˆk = 0. Also, we have the following formulas for the cross products: î ĵ = ˆk ĵ î = ˆk î î = 0 ĵ ˆk = î ˆk ĵ = î ĵ ĵ = 0 ˆk î = ĵ î ˆk = ĵ ˆk ˆk = 0.

6 The formula for cross products. Theorem 3. Let A = (a 1, a 2, a 3 ) and B = (b 1, b 2, b 3 ) be any two vectors in R 3. Then A B î ĵ ˆk = a 1 a 2 a 3 b 1 b 2 b 3 = î a 2 a 3 b 2 b 3 ĵ a 1 a 3 b 1 b 3 + ˆk a 1 a 2 b 1 b 2 = (a 2 b 3 a 3 b 2 )î (a 1 b 3 a 3 b 1 )ĵ + (a 1 b 2 a 2 b 1 )ˆk ( ) = (a 2 b 3 a 3 b 2 ), (a 1 b 3 a 3 b 1 ), +(a 1 b 2 a 2 b 1 ) Discussion. To see where this formula comes from, let us try to find a vector N = (x, y, z) which is perpendicular to both vectors A = (a 1, a 2, a 3 ) and B = (b 1, b 2, b 3 ). We must have xa 1 + ya 2 + za 3 = N A = 0, xb 1 + yb 2 + zb 3 = N B = 0. Let us temporarily think of z as fixed, and try to solve for x and y. We then have two equations in two unknowns xa 1 + ya 2 = a 3 z xb 1 + yb 2 = b 3 z, If we multiply the first equation by b 2 and the second by a 2 and subtract, we get x(a 1 b 2 a 2 b 1 ) = (a 2 b 3 a 3 b 2 )z, or x = a 2b 3 a 3 b 2 a 1 b 2 a 2 b 1 z. If we multiply the first equation by b 1 and the second by a 1 and subtract, we get y(a 2 b 1 a 1 b 2 ) = (a 3 b 1 a 1 b 3 )z, or y = a 1b 3 a 3 b 1 a 1 b 2 a 2 b 1 z. Thus if we let z = a 1 b 2 a 2 b 1, we see that the vector N = ( + (a 2 b 3 a 3 b 2 ), (a 1 b 3 a 3 b 1 ), (a 1 b 2 a 2 b 1 ) ) is a vector perpendicular to both A and B. We can compute the square of the length of N and the square of A B. We have N 2 = (a 2 b 3 a 3 b 2 ) 2 + (a 1 b 3 a 3 b 1 ) 2 + (a 1 b 2 a 2 b 1 ) 2 = a 2 2b a 2 3b a 2 1b a 2 3b a 2 1b a 2 2b 2 1 2a 2 a 3 b 2 b 3 2a 1 a 3 b 1 b 3 2a 1 a 2 b 1 b 2, ( A B) 2 = (a 1 b 2 + a 2 b 2 + a 3 b 3 ) 2 = a 2 1b a 2 2b a 2 3b a 1 a 2 b 1 b 2 + 2a 1 a 3 b 1 b 3 + 2a 2 a 3 b 2 b 3.

7 7 It follows that N 2 + ( A B) 2 = a 2 2b a 2 3b a 2 1b a 2 3b a 2 1b a 2 2b a 2 1b a 2 2b a 2 3b 2 3 Thus Thus = (a a a 2 2)(b b b 2 3) = A 2 B 2 N 2 = A 2 B 2 ( A B) 2 = A 2 B 2 A 2 B 2 cos 2 (θ) = A 2 B 2( 1 cos 2 (θ) ) = A 2 B 2 sin 2 (θ) N = A B sin(θ) where θ is the angle between A and B. Since N is perpendicular to both A and B, it follows that N = A B sin(θ) n where n is perpendicular to both A and B. We can check that n points in the direction given by the right hand rule by seeing what happens if A = î = (1, 0, 0) and B = ĵ = (0, 1, 0). Then î ĵ ˆk = î ĵ ˆk = 0î 0ĵ + 1ˆk = ˆk.

8 8 Review for Final Exam Thursday, May 17 2:45 PM 4:45 PM, in 6102 Social Science The final exam will be cumulative, and will have questions on material covered by the three earlier exams as well as on the most recent material on vectors. You should consult the earlier review sheets for the earlier topics. For the material on vectors, you should be sure to read the assignment in Chapters 12 and 13 of the text. You should also know the following from our discussion of vectors. (1) The definitions (both geometric and algebraic) of the dot product. (2) The statement and proof of the Cauchy-Schwarz inequality. (3) The statement and proof of the triangle inequality. (4) The definitions (both geometric and algebraic) of the cross product. (5) How to compute 2 2 and 3 3 erminants. (6) What it means that two vectors are perpendicular. (7) How to find equations of planes in R 3. (8) How to find equations of lines in R 3. (9) How to find the projection of a vector A in the direction of another vector B. (10) How to find the area of a triangle, two of whose sides are A and B.

Worksheet 1.3: Introduction to the Dot and Cross Products

Boise State Math 275 (Ultman Worksheet 1.3: Introduction to the Dot and Cross Products From the Toolbox (what you need from previous classes Trigonometry: Sine and cosine functions. Vectors: Know what