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# 4solution - ECEN 455 Assignment 4 Problems 1(LA 8.1.2 Let V...

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Unformatted text preview: ECEN 455: Assignment 4 Problems: 1. (LA: 8.1.2) Let V be a vector space over F . Show that the sum of two inner products on V is an inner product on V . Is the difference of two inner products an inner product? Show that a positive multiple of an inner product is an inner product. Let (·|·) a and (·|·) b be two inner products on V . Furthermore, for v ,u ∈ V , define ( u | v ) = ( u | v ) a + ( u | v ) b . Then, the following properties hold. (a) For any u ,v ,w ∈ V , ( u + v | w ) = ( u + v | w ) a + ( u + v | w ) b = ( u | w ) a + ( v | w ) a + ( u | w ) b + ( v | w ) b = ( u | w ) + ( v | w ) . (b) For any v ,w ∈ V and s ∈ F , ( sv | w ) = ( sv | w ) a + ( sv | w ) b = s ( v | w ) a + s ( v | w ) b = s ( v | w ) . (c) For any v ,w ∈ V , ( v | w ) = ( v | w ) a + ( v | w ) b = ( w | v ) a + ( w | v ) b = ( w | v ) . (d) If v negationslash = 0 then ( v | v ) a > and ( v | v ) b > , which implies that ( v | v ) = ( v | v ) a + ( v | v ) b > . That is, the sum of two inner products on V is itself an inner product on V . The difference of two inner products is not necessarily an inner product. Suppose that (·|·) a is an inner product on V . Then (·|·) a − (·|·) a = 0 is not an inner product, whereas 2 (·|·) a − (·|·) a = (·|·) a is obviously an inner product. A positive multiple of an inner product is also an inner product. Let (·|·) be an inner product on V and let c be a positive number. Then, for all u ,v ,w ∈ V and s ∈ F , we have (a) c ( u + v | w ) = c ( u | w ) + c ( v | w ) (b) c ( sv | w ) = cs ( v | w ) = sc ( v | w ) (c) c ( v | w ) = c ( v | w ) = c ( v | w ) (d) c ( v | v ) > if v negationslash = 0 . 2. (LA: 8.1.9) Let V be a real or complex vector space with an inner product. Show that the quadratic form determined by the inner product satisfies the parallelogram law bardbl α + β bardbl 2 + bardbl α − β bardbl 2 = 2 bardbl α bardbl 2 + 2 bardbl β bardbl 2 . The parallelogram law can be shown as follows, bardbl α + β bardbl 2 + bardbl α − β bardbl 2 = ( α + β | α + β ) − ( α − β | α − β ) = ( α | α ) + ( α | β ) + ( β | α ) + ( β | β ) + ( α | β ) − ( α | β ) − ( β | α ) + ( β | β ) = 2 ( α | α ) + 2 ( β | β ) = 2 bardbl α bardbl 2 + 2 bardbl β bardbl 2 ....
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## This note was uploaded on 11/13/2011 for the course ECEN 455 taught by Professor Staff during the Spring '08 term at Texas A&M.

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4solution - ECEN 455 Assignment 4 Problems 1(LA 8.1.2 Let V...

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