By Jacques Dixmier
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Extra resources for C-star-algebras
For C we have m = 3, n = 2, and r = 0. Thus dim((C)(A)) dim((C)(AT )) dim((N)(A)) dim((N)(AT )) = = = = 0 0 n−r =2 m− r = 3. Problem 9 Part (a): First lets consider the equivalence of the ranks. The rank of A alone is equivalent A because we can simply subtract each row of A from the correto the rank of B ≡ A sponding newly introduced row in the concatenated matrix B. Effectively, this is applying the elementary transformation matrix E= I 0 −I I A to produced A we can again multiply by E above obtaining to the concatenated matrix EC = I 0 −I 0 , A .
The dimension of the nullspace is zero and contains on the zero vector. The dimension of the left nullspace is zero and contains only the zero vector. For b we have m = 3, n = 6, and r = 3 then the dimension of the column space is three and has the same basis as above. The dimension of the row space is still three and has a basis given by 0 0 1 0 1 0 0 , 0 , 1 . 0 0 1 0 1 0 1 0 0 The dimension of the nullspace is 6 − 3 = 3 and a basis can be obtained from 0 0 1 0 1 0 1 0 0 −1 , 0 , 0 .
5 0 xˆ = The projection matrix is given by P2 = a1 aT 1 aT 1 a1 and equals P2 = 1 0 0 0 . 1 5 1 2 2 4 = So that P2 P1 is given by 1 0 0 0 P2 P1 = 1 5 1 2 0 0 . Which is not a projection matrix since it would have to be written proportional to a row which it can’t be. Problem 11 Remembering our projection theorems AT Aˆ x = AT b and p = Aˆ x we can evaluate the various parts of this problem. Part (a): We find that AT A is given by 1 0 0 1 1 0 AT A = and AT b is given by 1 1 0 1 = 0 0 2 3 = 4 1 0 0 1 1 0 AT b = 1 1 1 2 2 5 , .
C-star-algebras by Jacques Dixmier