v = v 1 2 +v 2 2. Two successive applications of this idea give the length of the vector v R 3 :

Transcription

1 Length, Angle and the Inner Product The length (or norm) of a vector v R 2 (viewed as connecting the origin to a point (v 1,v 2 )) is easily determined by the Pythagorean Theorem and is denoted v : v = v 1 2 +v 2 2. Two successive applications of this idea give the length of the vector v R 3 : v = v 1 2 +v 2 2 +v 3 2. By analogy then, we define the length of a vector v R n to be v = v 1 2 +v 2 2 +L+v n 2. Notice that for any scalar c, we have cv = c v (why?); in particular, if v is not the zero vector (so that v 0), we can take our constant multiplier to be c = 1/ v, whence

2 v v = 1 v = 1. v That is, the vector u = v is a unit vector, one v whose length equals 1. The process of scalar multiplying v by the reciprocal of its length to produce a unit vector is called normalizing v. Given two nonzero vectors u and v (in either R 2 or R 3 ), if we picture these two vectors as positioned at the origin, then the triangle having these vectors as two sides has u v (or v u) as the third side. It makes sense, therefore, to define the distance between the vectors u and v to be the length of the vector u v. The Law of Cosines, which generalizes the Pythagorean Theorem, says that any triangle with side lengths a, b and c satisfies c 2 = a 2 +b 2 2ab cosθ where θ is the angle between sides a and b (i.e., the angle opposite side c). (To prove this, introduce h, the height of the altitude on side c, and apply the Pythagorean Theorem to the two right triangles made by h and a or by h and b.)

3 This relation allows us to determine the angle θ made between the two vectors u and v, since we obtain u v 2 = u 2 + v 2 2 u v cosθ, 2 u v cosθ = u 2 + v 2 u v 2 = (u 1 2 +u 2 2 ) +(v 1 2 +v 2 2 ) ([u 1 v 1 ] 2 + [u 2 v 2 ] 2 ) = 2(u 1 v 1 +u 2 v 2 ) (where in R 3 we need an extra component with subscript 3 in each term above). Thus, or u v cosθ = u 1 v 1 +u 2 v 2 cosθ = u 1v 1 +u 2 v 2 u v. Again, this gives us the means to define by analogy the angle between any pair of nonzero vectors u and v in R n to be the value of θ (between 0 and π) that satisfies cosθ = u 1v 1 +u 2 v 2 +L+u n v n u v.

4 Notice that the formulas we have developed for the length of v and the angle between u and v, v = v 1 2 +v 2 2 +L+v n 2 cosθ = u 1v 1 +u 2 v 2 +L+u n v n u v can be simplified by making use of the fact that [ ] u T v = u 1 u 2 L u n v 1 v 2 M v n = u 1 v 1 +u 2 v 2 +L+u n v n. This expression we turn into a scalar product of the vectors u and v, also called the inner product or dot product, denoting it with a raised dot: u v = u T v = u 1 v 1 +u 2 v 2 +L+u n v n. We can now more simply write the formulas for length and angle as v = v v and cosθ = u v u v.

5 Theorem If u,v,w R n and c is a scalar, then (1) u v = v u [inner product is commutative]; (2) (u + v) w = u w + v w [inner product distributes over vector addition]; (3) (cu) v = c(u v) = u (cv) [inner product is compatible with scalar multiplication]; (4) u u 0, and more specifically, u u = 0 if and only if u = 0. Proof (1)-(3) are clear: u v = u 1 v 1 +u 2 v 2 +L+u n v n = v 1 u 1 +v 2 u 2 +L+v n u n = v u (u + v) w = (u 1 +v 1 )w 1 +(u 2 +v 2 )w 2 +L +(u n +v n )w n = (u 1 w 1 +u 2 w 2 +L+u n w n ) +(v 1 w 1 +v 2 w 2 +L+v n w n ) = u w + v w (cu) v = (cu 1 )v 1 + (cu 2 )v 2 +L+ (cu n )v n = c(u 1 v 1 +u 2 v 2 +L+u n v n ) = c(u v) = u 1 (cv 1 )+u 2 (cv 2 )+L+u n (cv n ) = u (cv)

6 For (4), we simply note that u u = u 1 2 +u 2 2 +L+u n 2 0 where equality can only occur if each of the components of u is 0. // Corollary The inner product is linear in the first factor, i.e., if u 1,u 2,,u m, v R n and c 1,c 2,,c m are scalars, then (c 1 u 1 +c 2 u 2 +L+c m u m ) v = c 1 ( u 1 v) +c 2 (u 2 v) +L+c m (u m v) (By commutativity, it follows that the inner product is linear in the second factor as well.) // In R 2 lines are perpendicular when the angle between them is a right angle. If u and v are angles in R 2, then the angle θ between them is right if and only if 0 = cosθ = u v u v u v = 0.

7 We use this notion to generalize the concept of perpendicularity to any Euclidean space: we say that vectors u,v R n are orthogonal if u v = 0. Theorem [The Pythagorean Theorem in R n ] Two vectors u,v R n are orthogonal if and only if Proof As u + v 2 = u 2 + v 2. u + v 2 = (u + v) (u + v) = u (u + v)+ v (u + v) = u u + u v + v u + v v = u 2 + v 2 +2u v then u,v R n are orthogonal u v = 0 u + v 2 = u 2 + v 2. //

8 Orthogonal Complements Orthogonality of vectors extends to a property of entire vector spaces. If W is a subspace of R n, then the set of all vectors x in R n which are orthogonal to every vector in W is called the orthogonal complement of W and is denoted W (often read W perp ). This may seem to be a very hard property to check: how can one determine whether a vector x is orthogonal to every vector in the vector space W (especially since vector spaces generally contain infinitely many vectors)? Luckily, this is not a serious problem because of the Theorem A vector x is orthogonal to every vector in the vector space W if and only if it is orthogonal to each vector in a basis for W. Proof Let W have basis B = { b 1, b 2,, b m }. Then it is clear that if x is orthogonal to every vector in W, it must be orthogonal to every one of the b s. Conversely however, if x is orthogonal to every one of the b s, then since every w W has a representation of the form w = c 1 b 1 +L+ c m b m for suitable scalars c 1,,c m, we have

9 w x = (c 1 b 1 +L+c m b m ) x = c 1 ( b 1 x)+l+ c m (b m x ) = 0 whence x is orthogonal to every vector in W. // More important for the theory are the following properties: Theorem If W is a subspace of R n, then so is W. Proof Left as an exercise (#30, p. 383). // One important situation in which orthogonal complements of vector spaces arise naturally is one with which we are already familiar: Theorem Let A be an m n matrix. Then (Row A ) = Nul A and (Col A ) = Nul A T. Proof The vector x R n lies in Nul A if and only if Ax = 0; but this matrix equation is equivalent to the vector equations r 1 x = 0, r 2 x = 0,, r m x = 0

10 where r 1, r 2,, r m are the m row vectors of A, which span Row A. As a subset of these vectors form a basis for Row A, it follows that x lies in Nul A if and only if x lies in (Row A ). Similarly, the vector y R m lies in Nul A T if and only if A T y = 0 y T A = 0 T (by taking the transpose of both sides); this last matrix equation is equivalent to the vector equations y T c 1 = 0, y T c 2 = 0,L, y T c n = 0 where c 1, c 2,K, c n are the n column vectors if A, which span Col A. As a subset of these vectors form a basis for Col A, it follows that y lies in Nul A T if and only if y lies in (Col A ). //

11 Orthogonal and Orthonormal Sets Any finite set of vectors S = { u 1, u 2,,u k } is called an orthogonal set if every pair of vectors in S is orthogonal to each other, i.e., u i u j = 0 whenever i j. The most obvious example of an orthogonal set of vectors in R n is given by the standard basis E = {e 1,e 2,,e n }. Indeed, this leads to the following definition: a basis for a vector space is called an orthogonal basis if it is an orthogonal set of vectors. Orthogonal bases have particularly nice properties. For instance, it is easy to determine the weights needed to express a vector in terms of an orthogonal basis: Theorem If B = {u 1,u 2,, u k } is an orthogonal basis for a subspace W of R n, then the (unique) weights that express a vector w W in terms of the basis vectors in B namely, the c s in the equation w = c 1 u 1 +L+c k u k are given by the formula c i = w u i u i u i (i = 1, 2,, k). Proof Choose any i between 1 and k. Then

12 w u i = (c 1 u 1 +L+ c k u k ) u i = c 1 (u 1 u i ) +L+c k (u k u i ) = c i (u i u i ) from which the formula follows directly (recognizing that u i u i 0 since u i is a basis vector, hence cannot equal 0). //