Math 110 Fall 05 Lectures notes 35 Nov 21 Monday We begin Chapter 6 starting with a summary of the main points of the chapter The main new concept is orthogonality of vectors in a vector space The most familiar example of this is when two vectors x and y in R 2 form a right angle 90 degrees or pi 2 radians Write x x1 x2 rx cos tx rx sin tx where rx length x and tx angle between x and horizontal axis y y1 y2 ry cos ty ry sin ty where ry length y and ty angle between y and horizontal axis then we note x1 y1 x2 y2 rx ry cos tx cos ty sin tx sin ty rx ry cos tx ty or cos angle between x and y cos tx ty x1 y1 x2 y2 rx ry Thus x and y orthogonal tx ty pi 2 radians cos tx ty 0 x1 y1 x2 y2 0 The quantity x1 y1 x2 y2 is an example of what we will define as an inner product of vectors x and y which can be used to measure angles between vectors in any vector space Since vector spaces are more general than vectors in R 2 or R n we see that we will need a more general general definition of inner product Now suppose we have two vectors in R n x x 1 x n and y y 1 y n What is the angle between them We recall a result from analytic geometry the Law of Cosines if a triangle has sides of length a b and c and t is the angle opposite side c then c 2 a 2 b 2 2 a b cos t ASK WAIT Why is this true Hint Draw picture with perpendicular to side b We apply this to a triangle with a length x b length y c length x y picture By the Pythagorean Theorem a 2 length x 2 sum i 1 to n x i 2 b 2 length y 2 sum i 1 to n y i 2 c 2 length x y 2 sum i 1 to n y i x i 2 sum i 1 to n y i 2 2 x i y i x i 2 b 2 2 sum i 1 to n x i y i a 2 1 a 2 b 2 2 a b sum i 1 to n x i y i a b Comparing to the Law of Cosines we see we must have cos t sum i 1 to n x i y i a b sum i 1 to n x i y i length x length y Def 1 If x and y are vectors in R n then x y sum i 1 to n x i y i x t y is called the dot product of x and y Def 2 if x y 0 then x and y are called orthogonal Lemma 1 length x sqrt x x Proof this follows from the definitions Def 3 norm x x sqrt x x If x 1 x is called a unit vector Thm 1 If t angle between x in R n and y in R n then cos t x y sqrt x x y y Proof done above using Law of Cosines Def 4 A real matrix Q q 1 q n all of whose columns are 1 unit vectors q i q i 1 for all i 2 pairwise orthogonal q i q j 0 for i neq j is called an orthogonal matrix Ex I is orthogonal So is Q cos t sin t sin t cos t Lemma 3 Q is orthogonal if and only if Q t Q I So if Q is square Q t Q 1 Proof Q t Q ij sum k 1 to n Q t ik Q kj sum k 1 to n Q ki Q kj q i q j Thus saying that Q t Q is the identity is equivalent to saying q i q i 1 and q i q j 0 for i neq j Orthogonal matrices are very convenient because they are easy to invert just by transposition and inverting transposing and multiplying orthogonal matrices gives you other orthogonal matrices Ma113 students may note that this means the orthogonal matrices form a group Multiplying by them also does not change lengths of vectors Q x x 2 or angles between vectors Q x Q y x y Here is a summary the most important uses of orthogonal matrices that we will cover in Chapter 6 we state these results without proof and come back to the proofs later Just as previous chapter had matrix factorizations that summed up many of the important properties chapter 6 has such a factorization too called the QR factorization Thm 2 Let A A can 1 Q 2 R be be is is a real m by n matrix with linearly independent columns Then decomposed as A Q R where an m by n orthogonal matrix an n by n invertible upper triangular matrix We will give a procedure for computing a QR decomposition which will prove Thm 2 called the Gram Schmidt Orthogonalization process which is a bit like Gaussian elimination In the meantime we give an example of what it is good for fitting curves to data Ex Suppose we have a collection of points x 1 y 1 x m y m in the plane and we want to draw a straight line that somehow best approximates these points This means we would like to pick constants a and b so that the line y a x b passes as closely as possible to the points x 1 y 1 x m y m picture To do this we will use the same procedure that Gauss invented centuries ago called least squares choose a and b to minimize the sum of squares of the errors in approximating each point x i y i where the error is measured by the vertical distance from the line to the point picture e i a x i b y i e i 2 a x i b y i 2 sum i 1 to m e i 2 sum i 1 to m a x i b y i 2 e e Thus the vector e of errors is a function of a and b and we want to minimize e as a function of a and b To do so write A x 1 1 z a y y 1 x 2 1 b y 2 x m 1 y m 3 and note that e e 1 x 1 a 1 b y 1 A z y e 2 x 2 a 1 b y 2 e m x m a 1 b y m Thus we may state our problem as given the matrix A and vector y to choose the vector z a b to minimize the length of the vector e A z y Thm 3 Suppose A is …