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Unformatted text preview: Chapter 3 Inner Products In Linear Algebra 1, we briefly looked at the dot product function for vectors in R n . We saw that the dot product function has an important relationship to the length of a vector in R n and to the angle between two vectors. Moreover, the dot product has some important applications; in particular, it gave us an easy way of finding the projection of a vector onto a plane. Our goal in this chapter is to extend these ideas to general vector spaces. 3.1 Inner Product Spaces Recall that we defined a vector space so that the operations of addition and scalar multiplication had the essential properties of addition and scalar multiplication of vectors in R n . Thus, to generalize the concept of the dot product it makes sense to include the essential properties of the dot product in our definition. DEFINITION Inner Product Let V be a vector space. An inner product on V is a function ( , ) : V V R that has the following properties: For every vectorv,vectoru, vectorw V and s, t R we have I1 ( vectorv,vectorv ) 0 and ( vectorv,vectorv ) = 0 if and only if vectorv = vector (Positive Definite) I2 ( vectorv, vectorw ) = ( vectorw,vectorv ) (Symmetric) I3 ( svectorv + tvectoru, vectorw ) = s ( vectorv, vectorw ) + t ( vectoru, vectorw ) (Bilinear) DEFINITION Inner Product Space A vector space V with an inner product ( , ) on V is called an inner product space . In the same way that a vector space is dependent on the defined operations of addition and scalar multiplication, an inner product space is dependent on the definitions of addition, scalar multiplication, and the inner product. So, when defining an inner product space, one must specify which inner product is being used. 1 2 Chapter 3 Inner Products EXAMPLE 1 The dot product (often called the standard inner product) defines an inner product on R n . EXAMPLE 2 Which of the following defines an inner product on R 3 ? (a) ( vectorx, vector y ) = x 1 y 1 + 2 x 2 y 2 + 4 x 3 y 3 . Solution: We have ( vectorx,vectorx ) = x 2 1 + 2 x 2 2 + 4 x 2 3 so ( vectorx,vectorx ) 0 and ( vectorx,vectorx ) = 0 if and only if vectorx = vector 0. ( vectorx, vector y ) = x 1 y 1 + 2 x 2 y 2 + 4 x 3 y 3 = y 1 x 1 + 2 y 2 x 2 + 4 y 3 x 3 = ( vector y,vectorx ) and ( svectorx + tvector y,vector z ) = ( sx 1 + ty 1 )( z 1 ) + 2( sx 2 + ty 2 )( z 2 ) + (4 sx 3 + ty 3 ) z 3 = s ( x 1 z 1 + 2 x 2 z 2 + 4 x 3 z 3 ) + t ( y 1 z 1 + 2 y 2 z 2 + 4 y 3 z 3 ) = s ( vectorx,vector z ) + t ( vector y,vector z ) Thus ( , ) is an inner product. (b) ( vectorx, vector y ) = x 1 y 1- x 2 y 2 + x 3 y 3 . Solution: Observe that if vectorx = 1 , then ( vectorx,vectorx ) = 0(0)- 1(1) + 0(0) =- 1 < Hence, this is not an inner product....
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