
#1
Dec2712, 08:18 AM

P: 91

I was looking for a hint on a problem in my professor's notes (class is over and I was just auditing).
I want to show that if [itex]T:V→V[/itex] is a linear operator on finite dimensional inner product space, then if [itex]T[/itex] is diagonalizable (not necessarily orthdiagonalizable), so is the adjoint operator of [itex]T[/itex] (with respect to the inner product). I think I should show that the eigenspaces of λ and [itex]\overline{λ}[/itex] have the same dimension (I know they are not the same since this is only true for normal operators), but I'm not sure if this is the right way to go. Any small push in the right direction would help. Thanks very much. EDIT: The definition here of diagonalizable is that there exists a basis, [itex]\chi[/itex], such that [itex][T][/itex]_{[itex]\chi[/itex]} is a diagonal matrix (i.e. the matrix representation of [itex]T[/itex] with respect to the basis is a diagonal matrix). 



#2
Dec2712, 08:44 AM

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PF Gold
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You should post your definition of "diagonalizable".




#3
Dec2712, 01:41 PM

P: 91

I included the definition in the edit I made above.




#4
Dec2712, 02:05 PM

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PF Gold
P: 8,992

Adjoint Operator Help
OK, that is one of several definitions that we can work with. (Another one is that there exists an invertible matrix P such that ##P^{1}TP## is diagonal). When we're using your definition, I think the easiest way is to just write down an explicit formula for the ij component of ##[T]_\chi## (with i and j arbitrary), that involves the basis vectors and the inner product. Can you do that?
Once you've done that, you just use the definition of the adjoint operator, and you're almost done. In case you're wondering why I don't just tell you the complete answer, it's because the forum rules say that we should treat every textbookstyle problem as homework. So I can only give you hints. 



#5
Dec2712, 05:12 PM

P: 91

Not to worry, I didn't want an answer, just a hint. 



#6
Dec2712, 05:24 PM

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#7
Dec2712, 05:28 PM

P: 91

How are you defining "diagonal" when you say [itex]PTP[/itex]^{1} is diagonal?




#8
Dec2712, 07:12 PM

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PF Gold
P: 8,992

Unfortunately, this seems to make the answer to the question "Is T diagonal?" depend on the choice of the isomorphism F. So maybe this was a bad idea. 



#9
Dec2712, 07:22 PM

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P: 16,565

How about showing that there exists a basis of eigenvectors for A*?
If [itex]\lambda[/itex] is an eigenvalue is an eigenvalue of A, then [itex]\overline{\lambda}[/itex] is an eigenvalue of A*. So like in the OP, it suffices to show that the eigenspaces have the same dimension. 



#11
Dec2712, 08:56 PM

P: 91





#12
Dec2712, 09:49 PM

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#13
Dec2812, 06:42 AM

P: 91




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