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Theorem 4.1.5 Suppose that υ 1, υ 2 , …,υ n are all mutually orthogonal nonzero vectors in a vector space V with an inner product < , >. Then v 1 , v 2 ,..., v n are linearly independent. Proof : suppose that a 1 v 1 , a 1 v 2 ,..., a n v n = 0. It follows that for each i, 0=¿ υ i, V >¿ ¿ <υ i, α 1 υ 1 + α 2 υ 2 ++α n υ n > ¿ ¿ <υ i, α 1 υ 1 >+ + ¿ υ 1 n υ n >¿ ¿ α 1 < υ i, υ 1 >+ + α 1 <υ i, υ 1 >+…α n <υ i, υ n > ¿ ¿ α 1 0 ++ α 1 <υ i, υ 1 >+ +α n 0 ¿ α 1 < υ i, υ 1 > ¿ Since υ i 0, we find by the positive definite property ¿ υ i, υ 1 0. Thus the equation ¿ α 1 < υ i, υ 1 =0 gives α 1 =0. This shows that υ 1, υ 2 , …,υ n are linearly independent, as desired. // Consider two vectors u,υ∈R 2 . If θ is the angle beween these two vectors, then by elementry trigonometry u cos θ is the length of the projection of the vector u onto the vector ʋ. (See the figure on the next page). Consequently, the projection of u onto ʋ [denoted proj (u)] is the vector u cos θ ( υ υ ) = u cos θ υ υ That υ =( υ.υ ) 1 2 u.v=u ∥∥ υ cos θ, shows proj v ( u )= u ∥∥ v cos θ u ∥∥ υ v ¿ u.v u 2 = u.v v.v v The projection of a vector u onto a vector υ is pictured below

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Theorem 4.1.5 Suppose that are all mutually orthogonal nonzero vectors in a vector space V with an inner product < , >.Then v1, v2,..., vn are linearly independent.Proof : suppose that a1v1, a1v2,..., anvn = 0. It follows that for each i,

Since , we find by the positive definite property Thus the equation gives This shows that are linearly independent, as desired. //Consider two vectors If is the angle beween these two vectors, then by elementry trigonometry is the length of the projection of the vector u onto the vector . (See the figure on the next page). Consequently, the projection of u onto [denoted proj (u)] is the vector

That shows

The projection of a vector u onto a vector is pictured belowThis motivates the following definition.

Definition 4.1.6 If < ,> is an inner product on we define the orthogonal projection of u onto v by

Observe in the picture above that the vector is perpendicular to the vector v. This is a general fact about inner product spaces, whose (purely algebraic) proof is given next.Lemma 4.1.7 Whenner s real inner product on V, the vectors are orthogonal.Proof : We compute, using the properties of a real inner product :

In euclidean space, Rn , we have that . We use this same idea and define the norm associated with an arbitary inner product.Definition 4.1.8 If V is a vector, is an inner product on V, we define This function is called the norm associated with < , >.The norm associated with an inner product is fundamentally related to the inner product. One fact we shall use often is that whenever v and w are orthogonal, . This fact is easily chekced: . We show in Theorem 4.1.10 that satisfies all the conditions of Definition 4.1.1 for being a norm on V. In order to prove Theorem 4.1.10, we need the following theorem, which is a useful result in its own right.Theorem 4.1.9 (Schwarz Inequality) Suppose that V is a real inner product space with inner product < , >. For , ,

Equality holds if and only if for some Proof : If the result is easy. Thus we may assume that As is a scalar multiple of v, Lemma 4.1.7 shows that Using this , the positive definite property of < , > together with the linearity of inner products, we have:

Hence ,

Multiplying this inequality by the positive real number shows that . The inequality of the theorem follows taking by square roots.In case where then and . In case , then the chain of equalities above shows that By the positive definite property of the inner product we have so But then u is a scalar multiple of that is, If then Hence and the theorem is proved.We now show that the norm associated with a real inner product astisfies all the consitions given in Definition 4.1.1. This result is well know for the standard inner product on Rn. Many of the important applications of this result are to the infinite-dimensional case. We give a glimpse of such uses in Secs. 7.3 and 7.4.Theorem 4.1.10 If < , > is a real inner product on V, then defined by is a norm on VProof : (i) For we find that

(ii) Let . Then (where the inequality follows by Definition 4.1.8 and Theorem 4.1.9).Taking square roots given the triangle inequality, and this proves the theorem.Example 4.1.11 The inner product defined on V= C([0,1],R) by

Must satisfy Schwarz inequality. Written out explicitly, we obtain that for all

In order to see how useful the techniques of linear algebra are, we invite the reader to try to prove this result using only the tools of calculus. The norm associated with this inner product is the norm described in Example 4.1.2 (iv).We now turn to the case of inner products on complex vector spaces. The situasition is remarkably similar to the case of real inner products. In fact, in the remaining sections of this chapter, whenever possible these two cases will be treated simultaneously. After giving the definition of complex inner product, we will repeat the proofs od Lemma 4.1.7 and Theorems 4.1.9 and 4.1.10. we do this so that the reader can spot precisely where the difference are.If where recall that the complex conjugate of .Two important properties of complex conjugation we need are that for all Definition 4.1.3 Let V be a complex vector space. A complex (or hermitian) inner product on V is a complex-valued function on VxV, usually denoted by < , > which satiesfies:(i) (ii) (iii) (iv) It is customary to refer to the vector space V, together with some complex inner product < , > as a hermitian inner product space. A finite-dimensional complex inner product space is called a unitary space.