Math 308 Solutions Sec. 2.6 Problems(2,5,10,13,20,21,22,23,28) page 1 = 1 ( 1) = 1+1=0, = 1 ( 1) ( 1) = 1+2 1=0and

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1 Math 08 Solutions Sec. 2.6 Problems(2,5,0,,20,2,22,2,28) page In,2 verify that {u, u 2, u }isanorthogonal set. 2.6.(Optional) u =, u 2 =, u = 2 Orthogonal set just means that u T i u j = 0for i j. Sowecompute these quantities. (Since x T y = y T x, (as a T = a for any a), only of the 6 products need to be checked). u T u 2 = [ ] = () =+=0, u T u = [ ] 2 = () () =+2=0and u T 2 u = [ 0] 2=() () () = + 0 = 0. So {u, u 2, u }isorthogonal u = 0, u 2 =, u = Again, orthogonal set just means that u T i u j = 0for i j. Sowecompute these quantities. 0 u T u 2 = [ 0 ] = () = + = 0, u T u = [ 0 ] = = = 0and u T 2 u = [ 0] =() = = 0. So set is orthogonal. In 5,8 we must find a, b, c so that {u, u 2, u }isanorthogonal set. 2 a 2.6.5u =, u 2 = 2, u = b. 4 c 2 ) Check that u T u 2 = 0: [ ] 2= ( 4) = = ) Now weneed u T u = 0and u T 2 u = 0. But this can be written in matrix form as a b = c 0 Thus we do row ops to solve: 0 ===> 0 R + /6R R 2 2R. Thus ===> a = b we need 6c = 0 ===> c = 0. Thus u = b for any fixed, but arbitrary, b gives all possibili-

2 Math 08 Solutions Sec. 2.6 Problems(2,5,0,,20,2,22,2,28) page 2 ties for which {u, u 2, u }isorthogonal (Optional) 2 u =, u 2 = a b, u =. c First u T u 2 = 0says [2 ] a = 2 a + + ( 2) = 2a + 2 = 0. Thus a =, and we use this b 0 = [2 ] = 2b + c value in u 2. Now we need u T u = 0and u T 2 u = 0, or c. Bring the s to b 0 = [ ] = b + c c the other side of the equalities, to see we must solve 2 b =. Now row equivalences solve c this: 2 ===> 0 0. Hence b = 0 and c =. (To actually write down the row ops orthogonal set, plug back into u to get u =, u 2 =, u =.) 2.6.9(Optional) Express v = in terms of the orthogonal basis B = u =, u 2 =, u = 2 from Exercise. First note that none of the u i are zero, so the given set B is an independent set by Theorem. The set is an orthogonal basis for R by the corollary to that theorem, since dim(r )=, and there are independent vectors in B. Now to find the a i with v = a u + a 2 u 2 + a u we exploit the orthogonality to avoid performing row ops on the augmented matrix for this system. Multiply both sides of this equality by each u T i in turn and use u T i u j = 0, i j to get: a u T u = u T v or a [ ] = [ ] ===> a = 2 a = 2/. a 2 u T 2 u 2 = u T 2 v or a 2 [ 0] = [0] ===> 2a 2 = a 2 = /2. a u T u = u T v or a [ 2] 2 = [ 2] ===> 6a = a = /6. Thus v = = Express v = in terms of the orthogonal basis B (from Exersize 9). 2

3 Math 08 Solutions Sec. 2.6 Problems(2,5,0,,20,2,22,2,28) page Instead of writing out the equations, as in solution to 9, with explicit values for v and the u i,you can use the symbolic results expressed in the left sides of the equations in 9. (See formulas (5a) and (5b) in this section). This means that a i u T i u i = u T i v or a i = ut i v. Thus a u T = = i u i =, a 2 = = 2 () =, a () 2 = = 0 = 0. Thus () () v = = Use Gram-Schmidt to generate an orthogonal set from the given linearly independent set w = [0 0 0] T, w 2 = [ 2 ] T,w = [ 0 ] T. The G-S process starts with u = w = [0 0 0] T.Then proceeds to produce u 2 from w 2 by finding a with u 2 = w 2 + au and u T u 2 = 0tohave orthogonality. Substituting the first equality into the second shows 0 = u T w 2 + au T u = 2 + a( 2 ), so a = 2/ = 2and u 2 = w 2 2u = [ 0 ] T. (Note the formula for a = ut w 2 u T u from Theorem 4 could be used directly.) Finally, u = w + cu + du 2 with c, d chosen so u T u = 0and u T 2 u = 0. These last two equations, using the fact that u T u 2 = 0, have 2 = = 2 2. Thus, solutions c = ut w u T u u = [ 2 0 ]T. = = and d = ut 2w u T 2 u Find abasis for R(A) and N(A) where A = 2 obtain orthogonal bases. Use reduction to REF to get basis for R(A) and N(A): Then use Gram-Schmidt to 4 ===> 0 9 ===> 0 9 R R 2 0 A R 2 + R /5R ===> 0 R 2R R + 7new R Thus {A, A 2 }, the columns corresponding to the leading s, is a basis for {w = [ 2] T,w 2 =[ 2 ] T } R(A), i.e x 2 x 4 x 5 x x 4 2 x 5 From the REF of A, N(A) = {x: Ax = θ } = { x = x : x, x 4, x 5 arbitrary}. So, separating out the x, x 4, x 5 parts of x in N(A), gives v =, v x 4 x = 0, v = 0 as a basis for N(A). Now toget orthogonal bases for R(A) and N(A) use Gram-Schmidt as instructed. Thus for R(A) an orthogonal basis comes from w, w 2 using the formula from Theorem 4: u = w = and 2

4 Math 08 Solutions Sec. 2.6 Problems(2,5,0,,20,2,22,2,28) page 4 9/6 u 2 = w 2 ut w 2 u u T u = 2 = /6 form an orthogonal basis for R(A). 2 +() / For anorthogonal bases for N(A) westart with v, v 2, v and use the same orthogonalization formu- 2 lae: z = v =, z 2 = v 2 zt v 2 z z T z = 2 z = v zt v z z T z zt 2 v z z T 2 z 2 = 9 2 () 2 +() () =,and / 5/ = 2/ Argue that any set of four or more non-zero vectors in R cannot be an orthogonal set. Any set of nonzero orthogonal vectors is independent by Theorem. Since dim(r )=,any set with 4 or more vectors is dependent by results in Section 2.5. Thus any set with with four or more vectors cannot be an orthogonal set Let S = {u, u 2, u }beanorthogonal set of non-zero vectors, and set A = [u u 2 u ]. Show A is nonsingular and A T A = D is diagonal. Since any set of non-zero orthogonal vectors in R is independent (and hence forms a basis as u T dim(r )=), the square matrix A is non-singular. Now form A T = u T 2. Then just compute the product (symbolically): A T A = u T u T u T u T u u T u 2 u T u u T u [u u T u 2 u ] = u T 2 u u T 2 u 2 u T 2 u = 0 u T 2 u u T u u T u 2 u T 2 0 = D, u 0 0 u T u diagonal. Of course we used orthogonality to introduce all those zeros Let W be a p-dimensional subspace of R n. If v is a vector in W such that v T w = 0for every w in W, show that v = θ.(hint: Consider w = v.) : If v = w in W, then v T w = 0becomes w 2 = 0. Hence w = θ as only θ has length zero Let B = {u, u 2,...,u p }beanorthonormal basis for a subspace W. Let v be an vector in W, where v = a u + a 2 u a p u p.show that v 2 = a 2 + a a p 2. This is just the pythagorean theorem and is is proved by just multiplying out the terms in the product v 2 = v T v,or v 2 = (a u + a 2 u a p u p ) T (a u +a 2 u a p u p ) =(a u T +a 2 u 2 T +...+a p u p T )(a u + a 2 u a p u p ) =((a )a u T u + a a 2 u T u a a p u T u p + mult by a 2 (u 2 ) T :((a 2 )a u 2 T u +a 2 a 2 u 2 T u a 2 a p u 2 T u p + : : mult by a p (u p ) T :((a p )a u p T u +a p a 2 u p T u a p a p u p T u p. Now the orthonormality of u i says u i T u j = 0, i j, soeach line after the last = has only one non-zero term, so

5 Math 08 Solutions Sec. 2.6 Problems(2,5,0,,20,2,22,2,28) page 5 v 2 = a 2 u 2 + a 2 2 u a p 2 u p 2 = a 2 + a a p 2,since u i = bythe normality part of orthonormality.