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My linear algebra textbook gives the definition of the Adjoint Operator and then says,

You should verify the following properties:

  • Additivity: $(S + T)^* = S^* + T^*$
  • Conjugate homogeneity: $(aT)^* = \overline{a}\,T^*$
  • Adjoint of adjoint: $(T^*)^* = T$
  • Identity: $I^* = I$, where $I$ is the identity operator on $V$.

I've stared at the pages for a couple hours now. How do you verify this?

Here's my attempt at a proof for Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T

Is that correct reasoning?

BTW, this is NOT homework. Just reading for pleasure.

Thanks!

  • 0
    You need to write out each side of the equations using the given definition and then show that equality really does hold. What have you tried so far?2012-12-20
  • 0
    What should $T^*$ satisfy to be an adjoint? Try to check this for $S^*+T^*$.2012-12-20
  • 1
    The adjoint is uniquely determined by a certain condition (if you haven't proven uniqueness you should prove that too). All of these properties assert that the adjoint of some operator can be described as some other operator, so what you need to verify is that that other operator satisfies the condition that uniquely determines the adjoint.2012-12-20
  • 0
    Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T2012-12-20
  • 0
    Is what I posted accurate?2012-12-20
  • 0
    @Megan: no. "$(T^*)^* = (T^*v,w)^*$" makes no sense. On the LHS we have an operator, on the RHS we have something that doesn't make sense.2012-12-20
  • 0
    @kahen I don't understand...2012-12-20
  • 0
    (T*v, w) is not a matrix, it's a number (assuming you mean the inner product of the two vectors.)2012-12-20
  • 0
    So then how would you prove Adjoint of Adjoint?2012-12-20

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