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I succeeded with F=C, but I can't get it for the general case.

- Thread starter Treadstone 71
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- #1

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I succeeded with F=C, but I can't get it for the general case.

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matt grime

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I guess you've tried considering

<Nx,Nx>=<x,N*Nx>=<x,NN*x>=<N*x,N*x.

What did it get you?

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AKG

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|Tx| = |T*x|

T - cI is normal for all c in F

If Tx = cx, then T*x = c*x, where c* denotes the conjugate of c.

If c and d are distinct eigenvalues of T with corresponding eigenvectors x and y, then x and y are orthogonal

Exercise 12 says to prove, for any linear operator T on any inner product space V

a) the orthogonal complement of R(T*) is N(T)

b) If V is finite dimensional, then R(T*) is the orthogonal complement of N(T)

It gives a hint to use another exercise which says:

V is an inner product space, S and S' are subset of V, W is a finite dimensioanl subspace of V. Prove:

a) S' contained in S implies the orthogonal complement of S is contained in the orthogonal complement of S'

b) S is contained in the orth. compl. of the orth. compl. of S, so span(S) is containd in the orth. compl. of the orth. compl. of S

c) W = the orth. compl. of the orth. compl. of W

d) V = W direct sum with the orth. compl. of W

For c, use the fact that if W is a f.d. subspace of an i.p.s V, and y is in V, then there exist unique u in W and z in orth. compl. of W such that y = u + z. For d, prove the fact that if W and W' are subspaces of a vector space V, then V is their direct sum iff for all v in V, there exist unique w and w' in W and W' respectively such that v = w + w'.

- #4

Tom Mattson

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My guess is that he's usingmatt grime said:What did you do for the F=C case, not that you've said what F is (underlying field, perhaps? or the space itself?)

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