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By T. Kuhn

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For instance, a typical operator on is represented by the matrix C2 (l + i 2 1-i 3 )=A C2 The algorithm for determining Av, given A and v,is the same as before, asis the procedure for finding the matrix AB, the product of A and B, given those matrices. For example, let A be as above, and let Vector Spaces Then 33 ( 1 +2 i 1 -3 i) ( 1 +1 i) - (((21 )(+1 )z)(l)+ (1+-(3)(z)(l1 ++ zz))) = ( 4(1 4+ z)) = 4v _ Av - We see that v is an eigenvector of A, with corresponding eigenvalue 4. It turns out that the vector 1 = is also an eigenvector of A.

Then (ul v) = (2)(3) + (1)(4) = 10. How does this number acquire a geometrical significance? Consider the case when u=v= (�) In this case (ul v) = (ulu) = x 2 + y 2 • Here the geometrical significanceis clear; by Pythagoras' theorem, (ulu)is equal to the square of the length of the vector u. We denote the length of the vector v by l vl and observe that (1. 9) l vl = "'(vl v) If l vl = 1, then we say that vis normalized. Given any vector v, we can always produce a normalized vector collinear with v by dividing v by its own length; in other words v/ l vis l a normalized vector along the same line as v.

The elements with special properties are 0 and 1. The structure in question is that of afield, ;J = (F, +, ,- 1 , 0, 1 ), where the operations and designated elements have familiar properties, such as a + 0 = a and (a)(a - 1 ) = 1, for all a in F (see MacLane and Birkhoff, 1979, chaps. 3 and 8). Both the set of real numbers and the set of complex numbers have the structure of a field. Omitting the formal definition of a field, I will move straight on to the definition of a vector space over a field.

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