Abstract Algebra: True or False?

Assume that the underlying inner product spaces are finite-dimensional.


(a) Operators and their adjoints have the same eigenvectors.


(b) If T is an operator on an inner product space V, then T is normal iff [T]_β is normal, where β is any ordered basis for V.


(c) A real complex matrix A is normal iff L_A is normal.


(d) Every normal operator is diagonalizable.





This is what I have:


(a) False. Becuase T and T* are pretty much completely different operators.


(b) False. This is becuase if T is normal it doesn't necessarily mean that [T]_β will be becuase β was only an ordered basis. I'm guessing that if β had consisted of eigenvectors of T then it would have been true?


c) True. L_A = Ax so taking adjoint it becomes A* x, and then obviously (A*x)(Ax) = (Ax)(A*x). And other direction, AA* = A*A, and multiplying both sides by xx it follows that L_A is normal.


d) False. That's what I think but not sure how to show.





I would really appreciate it if someone could help me =)

Abstract Algebra: True or False?
(a) is false, but it's better just to give a counterexample, say


[1 1]


[0 2].





(b) I'm guessing that [T]_β means the operator T expressed with respect to the basis β. So here's what to do to check it out: a change of basis takes T to S = PTP^(-1). Is it true that T normal iff S normal?





(c) Real or complex doesn't matter; just work over the arbitrary field F. Note that (L_A)* = L_(A*). L_A normal means that (L_A) (L_A)* =(L_A)* (L_A); i.e. that


(L_A) (L_A*) =(L_A*) (L_A). Let x ∈ F^n be arbitrary.





If A is normal, then (L_A) (L_A*) x = (L_A)(A*x) = AA*x = A*Ax = A*(L_A)x = (L_A)* (L_A)x. Since x was arbitrary, we have (L_A) (L_A)* =(L_A)* (L_A); i.e.L_A is normal.





Now you do the opposite direction.





(d) True; that's what the spectral theorem says.

statice