- #1
sammycaps
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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 orth-diagonalizable), 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).
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 orth-diagonalizable), 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).
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