2152assign8solutions

MATH 2152 Winter 2022 Assignment on Section 8 Solutions The following questions are to be turned in to be graded. 1. (2 marks) Let V be an inner product space. Suppose u is orthogonal to both v and w . Prove that for any scalars c and d , u is orthogonal to c v + d w . Solution : Since u is orthogonal to both v and w , we have h u , v i = 0 and h u , w i = 0. Thus h u , c v + d w i = h u , c v i + h u , d w i = c h u , v i + d h u , w i = c (0) + d (0) = 0 . Thus u is orthogonal to c v + d w . 2. (3 marks) Let u and v be vectors in an inner product space V . Prove that || u + v || 2 + || u - v || 2 = 2 ( || u || 2 + || v || 2 ) . This is called the parallelogram equality , due to the picture of the equality with the Euclidean inner product on R 2 , given below. Here, the equality states that the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of all sides. Note that your proof should work for any inner product (not just the Euclidean inner product). Solution : We have || u + v || 2 = h u + v , u + v i = h u , u + v i + h v , u + v i = h u , u i + h u , v i + h v , u i + h v , v i . = || u || 2 + h u , v i + h v , u i + || v || 2 . Also, || u - v || 2 = h u - v , u - v i = h u , u - v i - h v , u - v i

= h u , u i - h u , v i - h v , u i + h v , v i = || u || 2 - h u , v i - h v , u i + || v || 2 . Adding these two equations yields || u + v || 2 + || u - v || 2 = 2 || u || 2 + 2 || v || 2 . 3. (2 marks) Let V be an inner product space over R . Suppose u , v ∈ V have the same norm. Prove that u + v and u - v are orthogonal. Solution : We have h u + v , u - v i = h u , u - v i + h v , u - v i = h u , u i - h u , v i + h v , u i - h v , v i = || u || 2 - h u , v i + h u , v i - || v || 2 = 0 (we used the fact that || u || = || v || , and that since the field of scalars is R , inner product space axiom P5 yields h u , v i = h v , u i ). 4. (2 marks) Let x 1 , ..., x n ∈ R . Prove that ( x 1 + x 2 + ... + x n ) 2 ≤ n ( x 2 1 + x 2 2 + ... + x 2 n ) (hint: use Cauchy-Schwarz). Deduce that the square of an average of real numbers is less than or equal to the average of the squares. Solution : We use (the square of) the Cauchy-Schwarz inequality in R n with u = (1 , 1 , ..., 1) and v = ( x 1 , x 2 , ..., x n ). This yields ( x 1 + x 2 + ... + x n ) 2 = h u , v i 2 ≤ || u || 2 || v || 2 = n ( x 2 1 + x 2 2 + ... + x 2 n ) . Now, starting with the square of the average of x 1 , x 2 , ..., x n , we have x 1 + x 2 + ... + x n n 2 = ( x 1 + x 2 + ... + x n ) 2 n 2 ≤ n ( x 2 1 + x 2 2 + ... + x 2 n ) n 2 = x 2 1 + x 2 2 + ... + x 2 n n , which is the average of the squares. 5. (2 marks) Let a, b, c > 0 be real numbers. Prove that abc ( a + b + c ) ≤ a 3 b + b 3 c + c 3 a (hint: use Cauchy-Schwarz on a √ c , b √ a , c √ b and ( √ c, √ a, √ b )).

Solution : We use (the square of) the Cauchy-Schwarz inequality in R 3 with u = a √ c , b √ a , c √ b and v = √ c, √ a, √ b . Since a, b, c > 0, the square roots are real, and we are not dividing by 0. This yields ( a + b + c ) 2 = h u , v i 2 ≤ || u || 2 || v || 2 = a 2 c + b 2 a + c 2 b ( c + a + b ) . Dividing both sides by a + b + c , and then multiplying both sides by abc (both of which are positive) yields abc ( a + b + c ) ≤ a 3 b + b 3 c + c 3 a. 6. (3 marks) Consider a triangle with sides of length a , b , and c . Let d be the length of the line segment from the midpoint of the side of length c to the opposite vertex. Apollonius' Identity is a 2 + b 2 = 1 2 c 2 + 2 d 2 . Prove this identity. Hint: let u be the side of length b and v be half the side of length c . Then express a , b , c , and d in terms of norms of linear combinations of u and v . Use these to show that a 2 + b 2 - 1 2 c 2 - 2 d 2 = 0. Note : this is the picture on the front of the Axler text. Solution : We let u be the side of length b (orienting the vector so that it is heading towards the upper right) and v be half the side of length c (orienting the vector so that it is heading to the right). Then the side with length c is 2 v , the side with length d is u + v , and the side with length a is u + 2 v . Thus a 2 + b 2 - 1 2 c 2 - 2 d 2 = || u + 2 v || 2 + || u || 2 - 1 2 || 2 v || 2 - 2 || u + v || 2 = h u + 2 v , u + 2 v i + h u , u i - 1 2 h 2 v , 2 v i - 2 h u + v , u + v i