March 7, 2017

C*-algebras Volume 2: Banach Algebras and Compact Operators by Corneliu Constantinescu

By Corneliu Constantinescu

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Example text

Hence, there is a P E IR[t]\{0} with P(x) =0. There are two finite families (c~,),et , (/~),et in IR and an a E IRk{0 } such that P ( t ) = ~ 1-[(t ~ + . , t + ~,). tEI Then 17I(~ ~ + , , ~ + ~,1) = P ( ~ ) = 0. 7). Since E is a division algebra x 2 + a,x +/~L 1 = O. If a, _> 4/~ then there are 7, ~ E IR with (x + 71)(x + 51) = x 2 + a , x + ~, = O. 6), which contradicts the hypothesis. Hence 2 4fl~, > c~,. 19 43 Let E be a f i n i t e - d i m e n s i o n a l real unital algebra. Take x, y E E linearly i n d e p e n d e n t with X 2 ___ y 2 = _ 1.

By a) and the above considerations, o~(~) =o~(~). 15 ( 0 ) such that for (x,y) E E x F, m Let E be a n algebra a n d F a subalgebra o f E , xy E F whenever xy - yx. Then oE(~) = OF(~) for every x E F. Take a E lK\{0}. 11, 1 a ~ aE(x) r 3 y E E , ay - -x a q~ a y ( x ) 3y e F, r ay- 1 -x- - xy = O , xy = xy , x y = O, x y = x y , ce On the other hand, xy = yx, c~y- for every y G E . Thus I -x I - x y = O ==~ y = - ( - x I + xy) E F fg. Banach Algebras 26 r OE(~) *=* ~ r o~(~). Assume that 0 ~ aE(X).

Hence y x E R which proves that R is an ideal of E . 23 left ideals F ( 4 ) Let E be a unital algebra and ~ be the set of of E f o r whzch 1 - x ts invertible w h e n e v e r x E F . radical of E is the greatest element of ~ . T h e n the 30 ~. Banach Algebras Let R be the radical of E and take x E R . 21 a =~ b, there is a y E E with y(1 - x) = 1. Thus 1 - y = -yx E R. 21 a =~ b, again, there is a z E E with 1 = z ( 1 - (1 - y ) ) = z y . 3, y is invertible and y-l=l-x. Hence 1 - x is invertible and R E ~.

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