and u and v are orthogonal if and only if u v = 0. u v = x1x2 + y1y2 + z1z2. 1. In R 3 the dot product is defined by

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1 Linear Algebra [] 4.2 The Dot Product and Projections. In R 3 the dot product is defined by u v = u v cos θ. 2. For u = (x, y, z) and v = (x2, y2, z2), we have u v = xx2 + yy2 + zz2. 3. cos θ = u v u v, and u and v are orthogonal if and only if u v =.

2 Linear Algebra [2] 4.3 Planes A nonzero vector n is called a normal to a plane if it is orthogonal to every vector in the plane. A point P is on the plane with normal n through the point P if and only if n ( PP ) =.

3 Linear Algebra [3] and If n = (a, b, c), P(x, y, z) and P (x, y, z), then PP = OP OP = (x x, y y, z z) (a, b, c) (x x, y y, z z) =. Hence, the plane through P(x, y, z) with normal n = (a, b, c) is given by a(x x) + b(y y) + c(z z) =. Eg. An equation of the plane through P(,, 3) with normal n = (3,, 2) is 3(x ) (y + ) + 2(z 3) =. This simplifies to 3x y + 2z =.

4 Linear Algebra [4] 7. Orthogonality in R n Given X = x. and Y = xn X and Y is defined by y. in R n, the dot product of yn X Y = X T Y = xy + x2y2 + + xnyn. The length X of X is defined by X = X X = x 2 + x x2 n.

5 Linear Algebra [5] Eg. If X = and Y = 2 2, then 2 X Y = [ 2 ] 2 = =, 2 and X 2 = = 7 and X = 7.

6 Linear Algebra [6] Thm.. X Y = Y X 2. X (Y + Z) = X Y + X Z, (X + Y ) Z = X Z + Y Z. 3. (kx) Y = X (ky ) = k(x Y ), k R. 4. X, and X = X = O. 5. kx = k X, k R.

7 Linear Algebra [7] Eg. X + Y 2 = (X + Y ) (X + Y ) = X X + X Y + Y X + Y Y = X 2 + Y 2 + 2(X Y ). X X has length. Indeed, X X = X X =.

8 Linear Algebra [8] Def.. Two vectors X and Y are orthogonal if X Y =. 2. A set {X, X2, Xm} of nonzero vectors in R n is called an orthogonal set if Xi Xj = for i j. 3. An orthogonal set {X, X2,, Xm} is orthonormal if Xi = for all i. Rmk. If {X, X2,, Xm} is orthogonal, then { } X X, X2 X2,, Xm Xm is orthonormal.

9 Linear Algebra [9] Eg. X = form an orthogonal set and X 2 =, 3 4 2, X2 = and X3 = X2 3 2 = and X = 2 2 form an orthonormal set.

10 Linear Algebra [] Thm. Every orthogonal set of vectors in R n is linearly independent. Proof. Let {X, X2,, Xm} be orthogonal. Consider rx + r2x2 + + rmxm = O. = Xi O = Xi (rx + r2x2 + + rmxm) = r(xi X) + r2(xi X2) + + rm(xi Xm) = ri Xi 2. Hence ri = for each i.

11 Linear Algebra [] Thm. If {X, X2,, Xn} is an orthogonal basis of R n, then X = X X X 2 X + X X 2 X2 2 X X X n Xn 2 X n for every X in R n. Proof. If X = rx + r2x2 + + rnxn, then X Xi = ri(xi Xi) = ri Xi 2. Therefore, ri = X X i Xi 2.

12 Linear Algebra [2] Eg. X = 2, X2 = orthogonal basis of R 3. X = 3 and X3 = 3 = rx + r2x2 + r3x3. 4 form an 4 r = X X X 2 = 2, r 2 = X X 2 X2 2 = 3, r 3 = X X 3 X3 2 =. Hence, X = 2 X + 3 X 2.

13 Linear Algebra [3] Projections Def. If U is a subspace of R n, we define the orthogonal complement U of U by U = {X R n X Y = for all Y U}. Observe that if U = span{x,, Xm}, then U = {X R n X Xi = for all i}.

14 Linear Algebra [4] Eg. Find U if U = span, 2 x y Solution. Let X = U. z w 2 3 in R 4. Then X = and X = yield x y + 2z =, x 2z + 3w =.

15 Linear Algebra [5] [ ] [ ], x y = s z w 2 4 +t 3 3. U = span 2 4, 3 3.

16 Linear Algebra [6] Def. Let {X, X2,, Xm} be an orthogonal basis of a subspace U of R n. Given X in R n, we define proj U (X) = X X X 2 X + X X 2 X2 2 X X X m Xm 2 X m and call it the orthogonal projection of X on U.

17 Linear Algebra [7] Thm. If U is a subspace of R n and X R n, write P = proj U (X). Then. P U and X P U. 2. X P X Y for all Y U. 3. dim U + dim U = n. Proof.. Clearly, P U. (X P ) Xi = X Xi P Xi = X Xi X X i Xi 2 X i Xi = Thus X P U.

18 Linear Algebra [8] 2. Write X Y = (X P ) + (P Y ). Then P Y is in U and X P U. X Y 2 = (X P ) + (P Y ) 2 = X P 2 + P Y 2 + 2(X P ) (P Y ) = X P 2 + P Y 2 X P Let {X,, Xm} and {Y,, Yk} be orthogonal basis of U and U, respectively. Then {X,, Xm, Y,, Yn} is orthogonal, so linearly independent. If X R n, then X = P + (X P ). Thus {X,, Xm, Y,, Yn} spans R n.

19 Linear Algebra [9] Eg. Let U = span,. If X = 3 2, find the vector in U closest to X and express X as the sum of a vector in U and a vector in U. Solution. Note that, is orthogonal.

20 Linear Algebra [2] P = proj U (X) = X X X 2 X + X X 2 X2 2 X 2 = 4 3 X + 3 X 2 = = 3 X = P + (X P ) =

21 Linear Algebra [2] Gram-Schmidt Orthogonalization Algorithm Question : Given a basis B = {Y,, Ym} of U, how can we obtain an orthogonal basis from B? Answer : Construct X,, Xm in U as follows. X = Y, X2 = Y2 Y 2 X X 2 X, X3 = Y3 Y 3 X X 2 X Y 3 X2 X2 2 X 2, Xm = Ym Y m X X 2 X Y m X2 X2 2 X 2 Y m Xm Xm 2 X m. Then {X,, Xm} is an orthogonal basis of U.

22 Linear Algebra [22] Proof. Let U = span{x}, U2 = span{x, X2},, Um = span{x,, Xm }. {X} is orthogonal. X2 = Y2 proj U (Y 2), X2 U {X, X2} is orthogonal. X3 = Y3 proj (Y U 2 3), X3 U 2 {X, X2, X3} is orthogonal. Continue the process. Xm = Ym proj (Y U m m), Xm U m {X,, Xm} is orthogonal.

23 Linear Algebra [23] Eg. Let U be the subspace of R 4 with basis {Y, Y2, Y3}, where Y =, Y2 = Find an orthogonal basis., Y3 =. Solution. X = Y, X2 = Y2 Y 2 X X 2 X = ( 2 3 ) = ,

24 Linear Algebra [24] X3 = Y3 Y 3 X X 2 X Y 3 X2 X2 2 X 2 Thus = ( 3 ), 2, ( 2 5 ) 2 = 3 is an orthogonal basis

25 Linear Algebra [25] How can we get an orthonormal basis? 3, 5 2,